Chapter 3.

Force and Gravity

> *Being in orbit is like being infatuated -- you are constantly
> falling, but you aren't getting closer.*

Gravity surprises

When orbiting the Earth, the head of an astronaut is "weightless." When
he sneezes, it will snap back ...

> *exactly as much as it does when he is sitting on the Earth\'s
> surface*

At an altitude of 100 km, you are virtually at the edge of space, since
more than 99.999% of the Earth's atmosphere is below you. At this
altitude, the force of gravity is less than at the Earth\'s surface. It
is smaller by ...

> *about 1.6%.*

There is a contest underway to see whether a private individual or
company can send a rocket to an altitude of 100 km. The first to do this
will win \$10 million. The energy the rocket uses to get something to an
altitude of 100 km, compared to the energy needed to put that same thing
into Earth orbit, is ...

> *thirty times less*.

If the Sun suddenly turned into a black hole, but its mass didn\'t
change (i.e. it stayed the identical mass (about 300,000 times the mass
of the Earth) the orbit of the Earth ...

> *wouldn\'t change*.

If there were no surprises for you in those sentences, maybe you already
know what is in this chapter. For most people, gravity, like many
subjects in physics, is full of surprises. If you want to think clearly
about space, satellites, and their applications, then it helps to have a
deeper understanding.

The Force of Gravity

Any two objects that have mass attract each other with a force we call
gravity. You probably never noticed this for small objects, because the
force is so weak. But the Earth has lots of mass, and so it exerts a big
gravitational force on you. We call that force your *weight*. The fact
that gravity is actually a force of attraction is not obvious. Prior to
the work of Isaac Newton, it was assumed that gravity was simply the
natural tendency of objects to move downward.

If you weigh 150 lb, and are sitting about 1 meter (3.3 feet) from
another person of similar weight, then the gravitational force of
attraction between the two of you is 10^-7^ lb. This seems small, but
such forces can be measured; it is about the same as the weight of a
flea.

You weigh less when you stand on the Moon, because the force of
attraction is less. If you weigh 150 lb on the Earth,you would weigh
only 25 lb on the Moon. You haven't changed (you are made up of the same
atoms), but the force exerted on you is different. Physicists like to
say that your *mass* hasn't changed, only your weight. Think of mass as
the amount of material, and weight as the force of attraction of the
Earth (or whatever other planet or satellite you are standing on).

Mass is commonly measured in kilograms. If you put a kilogram of
material on the surface of the Earth, the pull of gravity will be a
force of 2.2 lbs. So a good definition of a kilogram is an amount of
material that weighs 2.2 lbs when placed on the surface of the Earth.
That number worth remembering.[^1] Go to the surface of Jupiter, and you
will weigh nearly 400 lbs. On the surface of the Sun you will weigh
about 2 tons, at least for the brief moment before you are fried to a
crisp. But in all cases your mass will be 68 kg.

The equation that describes the pull of gravity between two objects was
discovered by Isaac Newton. It says that the force of attraction is
proportional to the mass -- double the mass and the force doubles. The
force also depends on the distance. It is an *inverse* *square* law. It
is inverse because when the distance gets larger, the force gets
smaller. It is a square law because if you triple the distance, the
force decreases by nine; if you make the distance increase by 4, then
the force goes down by 16, etc.

> The Equation of Newton's Law of Gravity (optional)
>
> "Newton's Law of Gravity" gives the gravity force between two objects
> with masses m and M separated by distance D:
>
> $$F = kG\frac{mM}{D^{2}}$$
>
> *G* is called the Gravitational Constant, and has the value 6.67
> x10^-11^. *k* is a constant that makes the force come out in pounds
> (rather than the physicist unit of Newtons[^2]); it has the value *k*
> = 0.224. (We'll use this constant a lot.)
>
> Let's go back to the example given in the text: two 150 lb people
> separated by 1 meter. The mass of each person is 150/2.2 = 68 kg.
> Putting these into the formula gives F ≈ 10^-7^ lbs.
>
> Newton's Law of Gravity actually gives the force only between two
> *small* objects. If one of the objects is a sphere (such as the Earth)
> then it turns out that you can still use the formula, but you must use
> the distance to the center of the sphere as the value for *D*. As an
> example, let's put in numbers for a 1 kg object sitting on the surface
> of the Earth. Then the force of attraction is given by the gravity
> equation with *m* = 1 kg, *M* = the mass of the Earth = 6 x10^24^ kg,
> and *D* = radius of the Earth (that's the distance to the center of
> the sphere). This distance is D = 6371 km ≈ 6.4 x10^6^ meters. Without
> plugging in the numbers, can you guess what the answer will turn out
> to be? Guess, and then check this footnote[^3] to see if you guessed
> correctly.
>
> Suppose you weigh 150 lbs on the Earth. Then you mass is 150/2.2 = 68
> kg. What will you weigh on the Moon? We can calculate that by using
> Newton's Law of Gravity, and putting in the *M* = the mass of the Moon
> = 7.3 x10^22^ kg, *D* = the radius of the moon = 1.74 x10^6^ meters.
> The answer is *F* = 25 lb. That means you will weigh 25 lb. on the
> surface of the Moon.

Newton's first law

Here is something that might surprise you: if you weigh 150 lb, not only
is the Earth attracting you with a force of 150 lb, but you are
attracting the Earth with a force of 150 lb too. This is an example of
something called "Newton's first law" -- if an object exerts a force on
you, then you exert the same force back on it.

But if you are so small, how can you exert such a large force on the
Earth? The answer is that, even though you are small, your mass exerts a
force on every piece of the Earth, simultaneously. When you add all
those forces together, the sum is 150 lb. Soo you are pulling up on the
Earth exactly as much as the Earth is pulling down on you.

Think of it in the following way. If you sit push on some else's hand,
they feel your force. But you feel the force too. You push on them; they
push back on you. The same thing works with gravity. The Earth pulls on
you; you pull on the Earth.

The "weightless astronaut" paradox

Imagine an astronaut in orbit in a capsule 200 km (125 miles) above the
surface of the Earth. What does he weigh? Because he is further away
from the mass of the Earth, the force is slightly lower. From Newton's
Law of Gravity, we can calculate that he is weighs 142 lb, i.e. he is 8
lb lighter.

> **Optional calculation:** We can calculate the astronaut's weight by
> using Newton's Law of Gravity. Assume his mass is 68 kg. (That is his
> mass if he weighs 150 lb when standing on the surface.) The distance
> between him and the surface of the Earth is 6371 + 200 km (his
> altitude + the Earth's radius) = 6.57 x10^6^ meters. Plug that and you
> get *F* = 142 lbs. On Earth he weighed 150 lbs. The astronaut weighs 8
> lb less because he is further from the Earth. 200 km seems like a lot,
> but added to the radius of the Earth, it made only a 3% change in *D*,
> and that's why the weight didn't change by much.

But wait a minute -- aren't orbiting astronauts weightless? Movies show
them inside space ships floating around. How can they do that if they
weigh almost as much as they do when they are on the surface of the
Earth?

To understand the answer to this paradox, we have to think about what it
means to be weightless. Suppose you are in an elevator, and the cable
suddenly breaks. The elevator and you fall together. During those few
seconds before you crash into the ground, you will feel weightless. You
will float around inside the elevator. You will feel no force on your
feet, and your shoulders will not feel the weight of your head. (Your
head falls with your chest, at the same rate, so the muscles in your
neck need exert no force to keep the head above the chest.) In those
brief seconds you have the same "weightless" experience as the
astronauts. All the time, of course, the Earth is pulling you rapidly
towards it.[^4] You have weight, but you feel weightless. A movie made
of you in the elevator would show you floating around, apparently
without weight, while you and the elevator fell together. You would look
just like the astronauts floating around in the Space Station.

Now imagine that instead of falling, you and the elevator are shot
together out of a gun, and fly 100 miles before hitting the ground.
During that trip, you will again feel weightless. That's because you are
in motion along with the elevator. You and it fly in the same arc.[^5]
Your head and chest are both moving in that arc together; there is no
force between them and your neck muscles can be completely relaxed. Your
head will seem to have no weight. Prior to the space program, potential
astronauts were flown in airplanes following such arcs in order to see
how they responded to the sensation of weightlessness -- and to get them
used to it.

When you and the elevator are moving together under the force of gravity
(either falling or shot in an arc) there seems to be no gravity. From
that alone, you might think you were far out in space, far away from the
gravity of any planet, star, or moon. From inside the elevator, you
can't tell the difference.

Now imagine that at the top of a very tall tower (200 km high) is a
large gun, pointing horizontally. We shoot the elevator and you
horizontally. If we picked a low velocity (e.g. 2 km/sec) you and the
elevator would curve towards the Earth, and you would crash into it, as
in path *A* in the figure below. But instead we pick a high velocity: 8
km/sec. You and the elevator are shot from the gun, and you curve
towards the Earth, but because of your high velocity, you miss the edge
of the Earth, as in path *B*. You keep on curving downward, but you
never hit. You are in orbit. The force of gravity makes the path of the
elevator -- let's call it a space capsule now -- curve downwards. But if
that curvature matches the curvature of the Earth, then it misses the
surface, and stays at a constant height.[^6]

![](media/image1.png){width="2.160416666666667in"
height="2.2180555555555554in"}

capsule shot into space from a tower

This may seem preposterous, but it is reasonable to think of an
astronaut in orbit around the Earth as being in a state of perpetual
falling. That's why he feels weightless.

You can think of the Moon as doing the same thing. It is attracted to
the Earth by gravity, but it has high sideways motion. Even though it is
falling towards the Earth, is always misses.

The velocity for LEO (Low Earth Orbit)

To stay in a circular orbit just a few hundred miles above the surface,
the velocity of the satellite must be about 8 km per second, which is
about 18,000 miles per hour. (The actual value depends slightly on the
altitude; we'll derive this number from a calculation later in this
chapter.) At this velocity, the satellite orbits the 24,000 mile
circumference of the Earth in about 1.5 hours.

If the astronaut want to land, he does *not* point his rockets downward
and fire his rockets away from the Earth; he fires his rockets towards
the direction he is headed -- in the forward direction! That slows down
the satellite, so it is no longer going fast enough to miss the edge of
the Earth. Gravity brings the satellite back to Earth. If the satellite
moves faster than 8 km per second, it leaves the circular orbit and
heads out into space. At about 11 km per second it will have sufficient
velocity to reach the Moon and beyond. This velocity is called the
escape velocity. We'll discuss this concept further later in this
chapter.

Analogy with a rock and sling

There is another way to think about Earth satellites. Forget gravity for
a moment. Imagine that you have a rock tied at the end of a string, and
you are spinning it in a circle above your head. The string provides the
force that keeps the rock from flying away, that keeps the rock in
circular motion. If the string breaks, the rock flies off in a straight
line. Gravity does the same thing for an Earth satellite: it provides
the force that keeps the satellite in a circular orbit.

An old weapon called the "sling" is based on this principle. A rock is
held by a leather strap, and spun in circles over the head. Arm motion
helps it pick up circular speed. It is the strap that keeps the rock in
circular motion. When the strap is released, the rock flies in a
straight line towards its target. Such a sling was the weapon that,
according to the Bible, David used to kill the giant Goliath.

In a similar manner, if we could suddenly "turn off" the force of
gravity, the Moon would leave its circular orbit, and head off in a
straight line. Likewise for all the satellites now in orbit around the
Earth. And with the Sun's gravity turned off, the Earth would head out
into space too, at its previous orbital speed of 30 km/sec (67,500 mph).

Geosynchronous satellites

Weather satellites and TV satellites have a very special orbit: they are
"geosynchronous." This means that they stay above the same location of
the Earth at all times. That means that the same weather satellite will
be able to watch the development of a storm, or of a heat wave,
continuously. It means that if you are receiving a signal for your TV,
you never have to re-point the antenna. The satellite remains in the
same direction above your house at all times.

How can this be, since satellites must orbit the Earth to avoid crashing
back down? The answer is elegant: the satellites orbit the Earth at such
a high altitude (where the gravity is weak) that they go at a low
velocity, and take 24 hours for each orbit. Since the Earth rotates once
in that period, they stay above the same location. Both are moving --
your home with the TV dish, and the satellite -- but their angle with
respect to each other doesn't change.

Geosynchronous satellites orbit the Earth at the very high altitude of
22,000 miles. At this altitude, over 5 times the radius of the Earth.
Remember that the force between two objects had a 1/D2 in it? If the
distance is 5 times larger, then that factor makes the force 25 times
smaller. Moreover, at the high altitude the distance to make a circular
orbit is longer. These factors combine to make the time to circle the
Earth equal to 24 hours, rather than the 1.5 hours of a LEO.

There is a catch. If the satellite is to stay in the same location in
our sky relative to the ground, it must orbit above the equator. Can you
see why that is true? A geosynchronous satellite moves in a circle
around the center of the Earth. If the satellite is not in an equatorial
orbit, then it will spend half of its orbit in the Northern Hemisphere,
and half in the Southern. Only if it orbits above the equator can it
stay precisely above the same Earth location at all times.

As a result, all geosynchronous satellites are right above the equator.
If you look at them up in the sky, they all line up in a narrow arc. In
fact, there is so little room left, that international treaties are
required to divide up the space. (If satellites are too close to each
other, their radio signals can interfere.)

If you are kidnapped, and don't know where you have been taken, try to
spot a satellite dish. If the dish is pointing straight up, then you
know you are on the Equator. If it is pointing horizontally, then you
know you are at the North pole.[^7]

Spy satellites

Spy satellites are satellites that carry telescopes to look down on the
surface of the Earth and see what is going on. They were once used
exclusively by the military, to try to see the secrets of adversaries,
but now they are widely used by government and industry to look at
everything from flooding and fires to the health of food crops.

The ideal spy satellite would stay above the same location all the time.
But to do that, it must be geosynchronous, and that means that its
altitude is 22000 miles. At those distances, even the best telescopes
can't see things smaller than about 200 meters. (We'll derive that
number when we discuss light.) That means that such a spy satellite
could see a football stadium, but couldn't tell if a game was being
played. Such satellites are good enough to watch hurricanes and other
weather phenomena, but are not useful for fine details, such as finding
a particular terrorist.

Thus, to be useful, spy satellites must be much closer to the Earth.
That means they must be in LEO (low earth orbit), no more than a few
hundred miles above the surface. But if they are in LEO, then they are
not geosynchronous. In LEO they zip around the 24,000 mile circumference
of the Earth in 1.5 hours; that gives them a velocity relative to the
surface of 16,000 miles per hour. At this velocity, they will be above a
particular location (within ± 100 miles of it) for only about 7.5
minutes.[^8] This is a very short time to spy. In fact, many countries
who want to hide secret operations from the United States keep track of
the positions of our spy satellites, and make sure their operations are
covered over or hidden during the brief times that the spy satellite is
close enough to take a photo.

LEO satellites cannot hover. If they lose their velocity, they fall to
Earth. If you want to have continuous coverage of a particular location,
you must use a circling airplane, balloon, or something else that can
stay close to one location.

GPS -- Medium Earth Orbit satellites (MEO)

One of the wonders of the last decade is the GPS satellite system. GPS
stands for "Global Positioning System", and if you buy a small GPS
receiver (cost under \$100), it will tell you your exact position on the
Earth within a few meters. I've used such a receiver in the wilderness
of Yosemite, in the souks of Fez, and in the deserts of Nevada. You can
buy a car with a built-in GPS receiver that will automatically display a
map on your dashboard showing precisely where you are. The military uses
GPS to make its smart bombs land at just the location they want.

The GPS receiver picks up signals from several of the 24 orbiting GPS
satellites. It is able to determine the distance to each satellite by
measuring the time it took the signal to go from the satellite to the
receiver. Once it knows the distance to three satellites, it can then
calculate precisely where on Earth it is.

The GPS satellites were not put in geosynchronous orbit, because the
great distance would require that their radio transmitters have much
more power to reach the Earth. They were not put in low orbit (LEO)
because they would then often be hidden from your receiver by the
horizon. So they were placed in a medius Earth orbit (MEO) about 12,000
miles high. They orbit the Earth every 12 hours.

To understand how GPS works, consider the following puzzle. A person is
in a U.S. city. He is 800 miles from New York City, 900 miles from New
Orleans, and 2200 miles from San Francisco. What city is he in?

Look on a map. There is only one city that has those distances, and that
is Chicago. Knowing three distances uniquely locates the position. GPS
works in a similar manner, but instead of measuring distances to cities,
it measures distances to satellites. And even though the satellites are
moving, they their location when they broadcast their signals are known,
so the computer in your GPS receiver can calculate its position.

Using gravity to search for oil

I said earlier that every object exerts a small gravitational force on
every other object. Remarkably, measurement of such small forces has
important practical applications. If you are standing over an oil field,
the gravity you feel will be slightly less than if over solid rock, for
the simple reason that oil weighs less, and so its gravity isn't as
strong.[^9] Such small gravity changes can even be measured from
airplanes flying above the ground. An instrument can make a "gravity
map" that shows the density of the material under the ground. Maps of
the strength of gravity, taken by flying airplanes, are commonly used by
businesses to search for oil and other natural resources.

A more surprising use of such gravity measurements was to make a map of
the buried crater on the Yucatan peninsula, the crater left behind when
an asteroid killed the dinosaurs. The crater was filled in by
sedimentary rock that was lighter than the original rock, so even though
it is filled, it shows a gravity "anomaly," i.e. a difference from what
you would get if the rock were uniform. An airplane flying back and
forth over this region made sensitive measurements of the strength of
gravity, and they produced the map shown below. In this map, the tall
regions are regions in which the gravity was stronger than average, and
the low regions were locations where the gravity was slightly weaker.

![](media/image2.png){width="4.097222222222222in" height="4.0in"}

The crater shows several concentric circles, with the largest over 100
km in diameter. The inner rings probably formed when the huge crater
initially filled in as material from under the crater was forced upward,
partially filling it.

manufacturing objects in a weightless environment

When the space program began, many people thought that there would be
significant advantages to being in a weightless environment. In a
satellite, things wouldn't sag under their own weight. It might be
possible to make better (rounder) ball bearings, or to grow more perfect
crystals (used in computers and other electronics) than can be done on
the ground.

This promise has been largely unfulfilled. The additional cost of doing
the work in a satellite has not turned out to be worth it. It costs
about \$10,000 to launch a kilogram of anything into orbit. In the near
future, commercial companies hope to reduce that price will come down to
\$1,000 per kilogram. It's hard to make a profitable factory in space
when it costs that much just to get there. There is no reason in
principle why getting to space *must* be expensive; we'll show in a
later section that the energy required is only 15 Cal per gram. Some
time in the future, if travel to space becomes as cheap as an airplane
ride, then the idea of factories in orbit may be more likely.

Optional: Measuring the mass of the Earth

> Have you wondered how we know the mass of the Earth? You can't just
> put it on a scale. The first person to determine it was Henry
> Cavendish, in 1798. He did it in a very indirect way. Even though
> Newton had discovered the equation of gravity, at the time of
> Cavendish nobody knew the value of the constant *G*. Cavendish
> determined it by taking several masses in his laboratory and measuring
> the force of gravitational attraction between them. (It was not an
> easy experiment; as I said earlier, the force you get is about the
> same as the weight of a paramecium.) Once he knew *G*, he could figure
> out the mass of the Earth from the known distance to the moon, and the
> fact that it goes around once every 28 days. When asked what he was
> doing in his lab, his answer (some people claim) was "weighing the
> Earth!"
>
> Problem: How could you use similar principles to determine the mass of
> the Sun?

Gravity on the moon and asteroids

Our Moon has about 1/81 of the mass of the Earth. So you might think its
gravity would be 81 times less. But its radius is only 1/3.7 that of the
Earth. Remember that gravity is an inverse square law, so from the small
radius you would expect the force to be (3.7)^2^ = 13.7 times larger. If
you combine these two effects, you get the surface gravity is 13.7/81 =
1/6 that on the Earth. And that is the value the astronauts found when
they landed there. The gravity was so weak that they seems to bound
around in slow motion. When they jumped, they went high, and when they
came back down, the came down slowly.

What is the surface gravity of an asteroid, which has a radius of only 1
km? We can't say, since we don't know the mass. However, if you assume
that the density is the same as for the Earth, then we can derive a
simple result: the surface gravity is proportional to the radius.[^10]
Since the radius of the Earth is 6378 km, this means that the surface
gravity on an asteroid is 1/6378 of the value on the Earth. If you weigh
150 lb on Earth, you would weigh 150/6378 = 0.023 lb, or about a third
of an ounce. That's about the weight of three pennies on the Earth.

As we'll show in a later section (on escape velocity) have to be very
careful, because your escape velocity would be very low. To jump into
space from the Earth requires a velocity of 11 km/sec, but from the
asteroid you would only require a velocity of 2 m/s. (The same jump
speed would get you less than a foot high on the surface of the Earth.)
You could easily launch yourself into space by jumping. This low escape
velocity was a problem for a U.S. space probe called "NEAR" (stands for
Near Earth Asteroid Rendezvous). If the satellite landed at a velocity
of 2 m/s or more, then it might have bounced right back out into space.

Gravity in Science Fiction

One of the most common "errors" in Science Fiction movies is the
implicit assumption that all planets in all solar systems have a gravity
about equal to that on the Earth. There is no reason why that should be
so. Pick a random planet, and you are just as likely to be a factor of 6
lighter (and bouncing around like astronauts on the Moon) or six times
heavier and unable to move because of your limited strength. Imagine a
person who weighs 150 lb on the Earth, trying to move on a planet where
he weighs 900 lb.)

> **Numerical exercise**: use the simple radius formula to estimate the
> surface gravity on the moon. How close is your answer to the correct
> one that we calculated? Why aren't the answers the same? Can you guess
> how far wrong the approximate formula must have been when applied to
> the asteroid?

Falling to Earth

Let's now talk about everyday gravity, the gravity that you feel when
you are near the surface of the Earth. When you jump off a diving board
(or a bungee tower) gravity pulls you downwards. The force of gravity
acts on all parts of your body, and makes them fall faster and faster.
You accelerate, but in a very remarkable way: every second that you
fall, you pick up an additional 9.8 meters per second of velocity. Put
in the form of an equation, your velocity v after a time t is

*v = g t*

where the constant *g* = 9.8 is usually called the *acceleration of
gravity*.

By *acceleration* we mean the rate at which your velocity changes. For
gravity, the acceleration is constant. After one second, your velocity
is 9.8 meters/sec. After two seconds, it is 19.6 m/s. After three
seconds, 29.4 m/s. After 4 seconds, 39.2 m/s. Every second you pick up
an additional 9.8 m/s.

Here is a useful conversion factor:

1 m/s = 2.24 mph

After falling for 4 seconds, our velocity was 39.2 m/s. To convert this
to mph, just multiply by 2.24. So the velocity after 4 seconds is 39.2
*x* 2.24 = 88 miles per hour. That may give you a better sense of how
fast you're moving. How far will you fall in this time? The equation can
be derived using calculus.[^11] The answer is:

$$D = \frac{1}{2} >^{2}$$

You are not required to memorize this equation; I show it so you can see
how we do these calculations. If we put in *t* = 4 seconds, this
gives[^12] D = 78 meters (257 feet. That's how far you'll fall in 4
seconds.

We can also use the equations backwards to see how long it takes to fall
a certain distance. In 1939, when King Kong fell off the Empire State
Building, it was 330 meters high. [^13] So how long should it have taken
him to fall from the top to the bottom? Take our equation for *D* and
solve for *t*. This gives:

*t* $= \sqrt{\frac{2D}{g}}$

Put in *D* = 330, *g* = 9.8, to get that *t* = 8.2 seconds. Watch the
movie (the 1939 version) and see if this matches what is in the movie.

How fast was Kong moving when he hit the ground? We just calculated that
*t* = 8.2 seconds. We can put that into our velocity equation to get

*v = g t* = 9.8 *x* 8.2 = 80 m/s

Remember that 1 m/s is 2.24 mph. That means that Kong was falling at
2.24 *x* 80 = 180 miles per hour when he hit. No wonder he was killed.

Dropping food, terminal velocity, and parachutes

In 2002, during the U.S. war in Afghanistan, food was dropped out of
airplanes to feed the Afghan people. The food was dropped from an
altitude of 3000 meters (about 10,000 ft). How fast was it going when it
hit the ground?

Let's use the equations to figure this out. The time it took to fall
was:

*t =* $= \sqrt{\frac{2D}{g}}$ =
$= \sqrt{\frac{2x\text{300}}{9\text{.}8}}$ = 24.7 seconds

and so its velocity would have been

*v = g t* = 9.8 x 24.7 = 2425 m/s

= 5335 miles per hour

That is incredibly fast. The speed of sound is only 330 meters per
second, so according to these equations it was going 7.3 times the speed
of sound, also called *Mach 7.3*.

Suppose the object slows down by friction with the air, and that all its
energy goes into heating the object. In Chapter 2 we show that the
temperature it reaches at Mach number *M* will be *M*^2^ x 300 K. So
from friction with the air, the food should have heated up to 7.3^2^ x
300 K = 16,000 K, more than twice as hot as the surface of the Sun! It
all should have been vaporized. Clearly that didn't happen to the food
dropped on Afghanistan. Our calculation was wrong. But what did we do
wrong?

We made the mistake of neglecting the force of air as the packages were
dropped. Every time the package hits a molecule of air, it transfers
some of its energy. The force F of air depends on the area A of the
package (more area means it hits more air molecules). That's not
surprising. But more interesting is the fact that the force depends on
the square of your velocity. Double your velocity, and the force goes up
four times; go 3 times faster, and your force increase by 9. This means
that air resistance becomes extremely important at high velocities. This
fact is the key to understanding not only dropped food, but also
parachutes, space capsule reentry, and automobile gasoline efficiency.

> **optional**: the air resistance equation is:

*F* ≈ (1/2) *k* *A ρ v^2^*

> The symbol ρ is the density of air, 1.25 kilogram per cubic meter, *A*
> is the front area of the object (as seen from the direction it is
> heading; if it is falling, then this is the area looking up at it),
> *v* is the velocity (in meters per second), and *k* is the same
> constant we used earlier in this chapter. If *k* = 0.224, then *F*
> will be in pounds; if *k* = 1, then *F* is in Newtons.

As the object falls faster and faster, the force of air resistance gets
greater and greater. Gravity is pulling down, but the air is pushing up.
The force of the air resists the gravity; it opposes it. If the object
keeps on accelerating, eventually the force of air will match the weight
of the object. When that happens, gravity and air resistance are
balanced. The object doesn't stop moving, but it stops accelerating,
that is, it no longer gains additional velocity. When this happens we
say the falling object has reached *terminal velocity*.

For the food dropped over Afghanistan, the terminal velocity was about 9
mph. For a falling person it is typically 70-100 mph. That's fast, but
people have survived falls from great heights into water. (Try to
imagine this next time you are going in a car or train at 74 mph.) If a
falling person spreads out his arms and legs (like sky divers) that
increases the effective area, and the person will fall even slower. A
person using a parachute is not much heavier, but the parachute area A
is large, so the terminal velocity is only about 15 mph. For King Kong
(about the same area as a parachute, but much heavier) the terminal
velocity was several hundred mph; he hit the ground without ever
reaching it.

It is interesting to think about why it goes at this speed. If the
falling object went any faster, the upward force (from the air) would be
greater than the weight, and that net force would slow the fall. If the
packet fell any slower than this terminal velocity, then the downward
pull of gravity would be stronger than the upward force of air
resistance, and the packet would fall faster.

> **Optional: the terminal velocity equation**. If we take the air
> resistance equation and set *F =k m g* (where *m* is the mass of the
> falling object) and solve for *v,* we get the equation for terminal
> velocity:

$$v = \sqrt{\frac{2\text{mg}}{A\rho}}$$

> A food packet in Afghanistan had a mass of about *m* = 0.1 kg, and an
> area of about *A* = 0.3 m x 0.3 m = 0.1 square meter; *g* = 9.8.
> Putting in these numbers we get for the food packets:

$v = \sqrt{\frac{2´0\text{.}1´9\text{.}8}{0\text{.}1´1\text{.}\text{25}}}$
≈ $\sqrt{\text{16}}$ = 4 m/s = 9 mph

> That's about the speed at which you jog. So the food floats down
> relatively slowly, not at *Mach 7.3* as we had obtained when we
> neglected air resistance.
>
> Let's calculate the terminal velocity for a person. We need to
> estimate the person's area and his mass. The only area that counts is
> the area hit by air as he is falling. If he is diving, that would be
> the area of the top of his head. If he is doing a swan dive, it would
> be the area of the front of his body, roughly his height times his
> width. To make the numbers simple, let's assume he is 2 meters tall
> and 1/2 meter wide, giving an area of 2 x 1/2 = 1 square meter. Take
> his weight to be M = 160 lb = 70 kg. Then, according to our formula,
> his terminal velocity will be:

$v = \sqrt{\frac{2\text{mg}}{A\rho}}$$= \sqrt{\frac{2´\text{70}´9\text{.}8}{1´1\text{.}\text{25}}}$*=*
33 m/s

> We can use the conversion factor that 1 meter per second is 2.24 mph
> to get

*v* = 2.24 *x* 33 = 74 mph

> With a parachute, the mass of the falling person is about the same
> (parachutes are light and don't contribute much to the total weight),
> but the area subject to air resistance can be 30 times larger than the
> area of a person's body. Plug that into the equation and see how much
> it slows the falling person. (Hint: it will be sqrt(30) ≈ 5.5 times
> slower. Do you see why?)
>
> What about King Kong? Did he slow down? The answer is no -- and the
> reason is interesting. If he were 10 times taller than a person, his
> weight would be 1000 times more. (He is not only 10 times taller, but
> also 10 times wider and 10 times thicker; that makes his volume 1000
> times larger.) But his area would have been only 100 times greater.
> (The area that the air it hitting, to slow him down, is width times
> height; it doesn't depend on his thickness.)
>
> His weight is much bigger than that of a human, and his area (the
> feature that responds to air resistance) is not nearly that much
> bigger. Plug in those numbers (M = 70,000 kg and A = 100 square
> meters) and you'll see that King Kong would not have slowed down very
> much. That aspect of the movie was accurate. (Maybe they had a PffP
> student consulting for them.)

Automobile fuel efficiency

A moving car feels the force of air on its front, and that tends to slow
the car down. To keep going at the same velocity, the engine must make
up the lost energy. We'll show that much of the gasoline used by the car
is to overcome the force of this air resistance. That is an important
fact to know. At a velocity of 30 m/s = 67 mph, the force of air on the
front of the car will be about 500 lb!

> **optional**: calculate air resistance on the front of a car. We apply
> the air resistance equation:
>
> *F = (1/2) k A ρ v^2^*
>
> The area *A* that we need to use in this equation is the area of the
> front of the car, equal to the height of the car times its width. Just
> to get a rough idea, assume the car is about 2 meters by 2 meters,
> giving *A* = 4 square meters. For the velocity let's take 67 mph = 30
> m/s. The density of air is *ρ* = 1.25 kg per cubic meter. Plugging in
> these values, we get that the force on the front of the automobile is:
>
> F = (1/2) *x* 0.224 *x* 4 *x* 1.25 *x* (30)^2^ ≈ 500 lb.

To keep the car from being slowed down by this force, the engine must
exert an equal and opposite force. That takes a lot of gasoline. At
these high velocities, more than 50% of the gasoline is used to overcome
this air resistance. As a result, car designers have worked hard to
"streamline" the shape of automobiles. If, instead of hitting the air
with a flat surface (as in the old autos from the 1920s) you tilt the
front surface, then the force can be much less, since the molecules can
bounce off obliquely instead of hitting the front and bouncing straight
back. In such a car, the force can be as low as 100 lb.

Many truck drivers are in business for themselves, and they have to pay
for the extra gasoline used to overcome air resistance. Maybe you've
noticed the smooth curves that some truck drivers have added to the cabs
of their trucks to do the same thing. Reducing the force can save
substantial money on gasoline.

![](media/image3.png){width="3.4722222222222223in"
height="2.7222222222222223in"}

Figure: Aerodynamic design for a truck cab

The top of the cab has had a contoured shape added to it to make the air
bounce off smoothly, at an angle, instead of hitting the flat face of
the truck head on. (The added shape is called "fairing.") This is shape
change is sometimes called "aerodynamic smoothing" and it saves
gasoline. It makes the effective value of "*A*", the area in the air
resistance equation, smaller.

Note also that at half that speed (i.e. if v = 15 m/s = 33 mph) the
force is four times less. That means that you use 4 times less gasoline
to overcome air resistance. So you save even more gasoline by driving
slower.

Force and energy

The relationship between energy and force is remarkably simple: energy
is force times distance. (To get the units right, we have to use our *k*
factor again. For *F* in pounds, *k* = 0.224.) The equation is:

*E = F D /k*

This means that if you push something with a force *F* for a distance
*D*, then the energy it takes is *E~,\ ~*joules. How much energy does it
take to go up a flight of stairs? Suppose you go up 4 meters (about 13
feet) and you weigh 160 lb. Then the work you do is

*E* = 160 x 4 /0.224

≈ 3000 joules

= 0.7 Calories[^14]

That's why it's hard to lose weight through exercise. Drink one 12 ounce
can of cola, which typically has 42 calories, and you can work it off by
going up 60 flights of stairs. Food contains a great deal of energy.
That's why you can get all the calories you need per day from about 2 lb
of food. (1 kg of whole wheat bread contains about 2500 Cal.)

Force and Acceleration

If you push on an object, and there is no friction to hold it back, then
it gains velocity. If you push twice as hard, it gains twice as much
velocity. The acceleration is proportional to the force. But it also
depends on the mass of the object. If the object has twice the mass, it
needs twice the force to get it moving. These two facts are summarized
the following equation:

Newton's Second Law:

*F = k m a*

In this equation, *m* is the mass (in kg), *a* is the acceleration (in
m/s per second), and *k* is a constant. For the force *F* to be in
pounds, *k* is 0.224. If you prefer the force in the physicist units
called *Newtons*[^15], then *k* = 1.

This equation tells you how much things speed up -- or slow down \--
when you apply a force.

This is Newton's Second Law. You may be wondering what Newton's *First*
Law is. It states that unless an object has an outside force on it, it
will tend to keep its motion unchanged. But that is just a special case
of the Second Law, since if F = 0, there is no acceleration, and that
means no change in velocity. Nobody would ever teach the First Law these
days if not for the fact that students sometimes wonder what came before
the Second Law.

The g

In this book, accelerations are typically measured in m/s per second.
For automobiles, we frequently measure acceleration in miles per hour
every second, sometimes written as mph/sec. For example, if a car
salesman tells you that a car will "go from zero to sixty in ten
seconds" what he is really telling you is the acceleration: 60 mph in 10
seconds. That is the same as 6 mph every second = 6 mph/s.

Another very useful unit for acceleration, used in the military and by
NASA, is the "*g*", pronounced "gee." One *g* is 9.8 meters/sec every
second, i.e. it is the acceleration of gravity. Suppose an automobile
accelerated at one *g*. How fast would it be going in 10 seconds? The
answer is found from our velocity equation by setting *a* = 9.8:

*v = a t*

= 9.8 x 10

= 98 meters/sec

We can convert this to miles per hour by multiplying by 2.24:

*v* = 98 x 2.24 miles per hour

= 219 mph

According to a NASA web site, when the Space Shuttle is launched, the
acceleration reaches a maximum of 3 g. This is usually pronounced as
"three gees." You've probably heard this term used in news programs and
in movies. A pilot in an F-18 fighter plane sometimes accelerates as
much as 10 g, i.e. "ten gees." Ten g would be 10 x 9.8 = 98 meters/sec
every second. Weightlessness is often described as "zero g." That's when
there is no acceleration -- at least with respect to your space capsule.

When something accelerates at a = 2 g, we'll sometimes say it in text as
"the acceleration is two gs." That can be said aloud as "the
acceleration is two gees."

> Optional: Why is the acceleration of gravity a constant?
>
> Why does gravity give a constant acceleration g? It turns out that
> this law is a consequence of combining Newton's Law of Gravity and
> Newton's Second Law. The force of gravity near the surface of the
> Earth is given by the equation
>
> $$F = kG\frac{mM}{D^{2}}$$
>
> In this equation, D is the distance to the center of the Earth, and M
> is the mass of the Earth. According to Newton's Second Law, this force
> will causes an acceleration given by *F =k m a*. Let's set these two
> equations for F equal to each other:
>
> $$F = kma = kG\frac{mM}{D^{2}}$$
>
> If we solve for *a* we get:
>
> $$a = G\frac{M}{D^{2}}$$
>
> But the right hand side of the equation has nothing but constants: *G*
> is the gravitational constant, *M* is the mass of the Earth, and *D*
> is the radius of the Earth. If we put in these numbers we get:
>
> *a* = 9.8
>
> That is just the value that we previously called *g*. It is the
> "acceleration of gravity."
>
> So here is an "explanation" of why all objects fall with the same
> acceleration. More massive objects need a bigger force to get
> accelerated; that is Newton's second Law. But they have a bigger force
> -- their weight! So big objects will accelerate at the same rate as
> less massive objects.
>
> Now the equation *F = k m a* becomes *W = k m g*. Put in *k* = 0.224,
> *m* = 1 kg, and *g* = 9.8, to get *W* = 2.2 lb., confirming the value
> I gave at the beginning of the chapter.

The g-rule

There is a very good reason to think of acceleration in terms of gs: it
enables you to solve important physics problems in your head. Suppose I
told you that I am going to accelerate you in the a horizontal direction
by 10 gs. How much force will that take? The answer is simple: 10 times
your weight! When astronauts are accelerated by 3 gs, the force on them
to do this must be 3 times their weight. I call this the "g-rule": **the
force to accelerate an object is equal to the number of gs times the
weight**

To get the number of gs, just calculate the acceleration in m/s per
second, and divide by 9.8. So, for example, we write that an
acceleration of *a* = 19.6 m/s per second = 2 *g*. This is an
acceleration of two gs. To accelerate something to 2 gs takes a force
equal to twice the weight of that something.

Optional math: why the g-rule is true

> The force required to accelerate an object of mass m is given by
> Newton's Second Law:
>
> *F = m a*
>
> The number of gs is
>
> *N = a/g*
>
> so
>
> *N g = a*
>
> The weight of the object is
>
> *w = m g*
>
> We now use all of these equations:
>
> *F = m a*
>
> *= m g N*
>
> *= w N*
>
> and the last equation is the g-rule.

Optional: the distance equation

> In the beginning of this chapter, we said that the distance you fall
> in a time t, under the influence of gravity, is
>
> *D = (1/2) g t^2^*
>
> The same formula will apply when you undergo any constant acceleration
> *a*, if you simply replace the symbol *g* with the symbol *a*. So the
> distance you travel will be:
>
> *D = (1/2) a t^2^*
>
> We can write this in terms of the final velocity v by substituting *t
> = v/a*:
>
> *D = (1/2) a t^2^*
>
> *= (1/2) a (v/a)^2^*
>
> *= (1/2) v^2^/a*
>
> If we solve this equation for a, we get
>
> *a = v^2^/(2 D)*
>
> This equation tells you the acceleration you need to reach a velocity
> *a* in a distance *D*.

Shooting an astronaut into space: the rail gun

To go into orbit around the Earth, a satellite must have a velocity of 8
km/sec. Why not give it this velocity in a gun? Could we literally
"shoot" an astronaut into space?

The answer is: you might be able to do this, but the astronaut would be
killed by the force required to accelerate him. Let's assume we have a
very long gun, an entire kilometer long. If we use the (optional)
equation in the last section, we derive that the required acceleration
is *a* = 408 *g*, i.e. the acceleration is 408 times the acceleration of
gravity.

> **optional calculation: 1 km gun shooting to 8 km/sec.**
>
> We can calculate this by using the last formula in the last section.
> The hardest part is getting the units right. The distance D = 8 km =
> 8000 meters, and the velocity v = 8 km/sec = 8000 m/sec. Plugging
> these in, we get:
>
> *a = v^2^/(2 D)*
>
> = (8000^2^)/(2 x 8000)
>
> = 4000 m/s per sec
>
> = 4000/9.8 *g* = 408 *g*
>
> **math exercise**: Calculate the number of gs the astronaut will
> experience if the length of the gun is 100 meters (the length of a
> soccer field) rather than 1000 meters.

That's a lot of gs. Remember the g-rule: the force that it will take to
accelerate you this fast will be 408 times your weight. So if you weigh
150 lb, the force that must be applied to you is 408 x 150 = 60,000 lb =
30 tons. That is enough to crush your bones.

Suppose we want to experience no more than 3 gs -- that what the Space
Shuttle does. How long would the gun have to be to get you going at v =
8 km/s? The gun would have to be 1000 km long! That is, of course,
ridiculously impractical.

> **optional calculation**: We first must convert 3 gs to metric units:
> 3 gs = 3 x 9.8 = 29.4 m/s per sec. Now put this into the distance
> equation:
>
> *D = (1/2) v^2^/a*
>
> = (1/2) x (8000)^2^ /29.4 = 1,1 x10^6^ meters = 1100 km

Of course, acceleration over such long distances is not impractical --
it's what the Space Shuttle does. It takes off, accelerates at 3 gs, and
it must go over 1000 km to reach orbital speed. So it is like a very
long gun, but without a barrel. It actually takes the Space Shuttle a
further than 1000 km since 3 gs is only the peak acceleration; for most
of the flight, the acceleration is less.

Acceleration during airplane take off

The take-off speed for a commercial airplane is about 160 mph. What
acceleration is needed to achieve this speed in a 1 km runway? Let me
make that question a little more personal. You are sitting in such an
airplane. In a few seconds, you will be at the other end of the runway,
and you will be moving (along with the airplane) at 160 mph. What force
does the seat have to apply to your back to accelerate you?

We show below that this requires the airplane to accelerate at 2.5 m/s
per second. We can convert this number to *g* units by dividing by 9.8,
to get that the acceleration is 2.5/9.8 = 0.26 *g*. So the force that
the airplane must push on you to get you going that fast is 0.26 times
your weight, i.e. about a quarter of your weight. If you weigh 150 lb,
then the push you feel on your back will be about 39 lbs. Think of this
next time you are on an airplane taking off. Does this number feel about
right?

> **optional calculation:** To calculate the acceleration of the
> airplane, we first convert to metric units. We divide the lift-off
> speed of 160 mph by 2.24 to get v = 71 m/s. We take *D* = 1 km = 1000
> meters and put it into the acceleration equation:
>
> *a = v^2^/(2 D)*
>
> = (71)\^2/(2000)
>
> = 2.5 m/s per sec

Circular acceleration

Physicists like to define velocity as having magnitude and direction. If
your velocity changes, we call it acceleration. But suppose you only
change your direction and not the actual number of m/s? We still call
that acceleration. [^16] The reason we do that is that many of the
equations we've been using still work.

The most important example of this kind of acceleration is when you go
in a circle. If the magnitude of your velocity is *v*, and this doesn't
change (you keep going the same number of m/s or mph), and the circle
has radius *R*, then we say that the circular acceleration is

*a = v^2^/R*

This kind of acceleration is very important for the fighter pilot who is
trying to change his direction rapidly. For example, if he is moving at
velocity of 1000 mph, and is turning in a circle of radius R = 2 km,
then his acceleration turns out to be 10 *g*. That's about as much as a
fighter pilot can tolerate.

> **optional calculation**: acceleration of fighter pilot. First we
> convert these to metric units. We convert mph to m/s by dividing by
> 2.24, giving v = 1000/2.24 = 446 m/s. We also convert R to meters: R =
> 2000 meters. Plugging these in gives:
>
> *a* = (446)\^2/2000
>
> = 100 m/s per sec
>
> = 100/9.8 ≈ 10 g

The problem with accelerating more than 10 gs is that a human's blood
pressure is not great enough to keep the blood in the brain, and that
causes the pilot to faint. Even specially chosen and physically fit
fighter pilots cannot take more than 10 *g*.

Fighter pilots and astronauts are usually tested and trained in a
spinning cylinder, rotated fast enough for them to experience 10 g
circular acceleration.

High g, from circular acceleration, is also a method used to separate
the components of uranium for making a nuclear weapon. Such a device is
called a centrifuge. The heavier parts of the uranium feel a greater
force, and they are pulled more strongly towards the outer parts of the
spinning cylinder. Such centrifuges are much in the news. In 2004, Libya
disclosed the fact that it had purchased large centrifuge plants for the
production of nuclear weapons material. Parts of such centrifuges have
been found hidden in Iraq. We'll talk more about these centrifuges when
we get to the chapter on nuclear weapons.

gravity in space -- according to science fiction

Science fiction movies sometimes show a large rotating satellite. There
is then "artificial gravity" from the rotation, and astronauts can walk
around on the outside. This actually makes sense. The astronauts will
feel a force on their feet (which point towards the outside) that will
appear to them indistinguishable from gravity. Such a satellite is shown
in the class movie *2001 -- A Space Odyssey*. We'll show below (in the
optional calculation) that the edge of a 200 meter-radius satellite must
be moving at about 44 meters/sec. That means it will rotate once every
40 seconds.

> **optional calculation:**
>
> If the satellite has radius *R* and is rotating with a rim velocity
> *v*, then the force on the astronauts fee will be
>
> *F = m a = m v^2^/r*
>
> To make this equal to the astronaut's weight, we set F = mg. Solving
> for v, we get
>
> *v =* sqrt*(g R)*
>
> For a satellite with a radius of 200 meters, the rim velocity must be
>
> *v* = sqrt(9.8 x 200) = 31 meters per second.

artificial gravity in space -- without circular motion

Many science fiction movies show space voyagers in the space ships
walking around as if on the Earth. Is that nonsense? Where does that
gravity come from?

I can make sense out of it. I just assume that the space ships are not
moving at constant speed. If the ship engine accelerates the ship by *a*
= 9.8 meters per second, then the ship will put a force on any
astronauts inside to accelerate them too. A person of mass m will feel a
force *F = ma*. But since the engines have set a ≈ g, the force on the
astronaut will be *F* = *m g*. That is exactly his Earth weight. If he
places his feet on the backside of the ship (opposite in direction to
the path the ship is taking) then he will feel like he is standing on
Earth. In fact, he can't even tell the difference; the acceleration
serves as a "virtual" gravity.[^17]

Here is an interesting number: the distance the ship would travel in a
year if it had constant acceleration of g. I do the calculation below.
The answer is 5 x10^15^ meters, which is about a half light year (the
distance that light travels in a year). The distance to the nearest star
(not counting the sun) is about 4 light years.

> **optional calculation:** The distance that you travel at acceleration
> *g* is given by our distance equation: *D =* (1/2) *g t^2^*. We put in
> *g* = 9.8. For mks units, we need to have t in seconds. We can
> calculate the number of seconds in a year as follows: 1 year = 365
> days = 365 x 24 hours = 365 x 24 x 60 minutes = 365 x 24 x 60 x 60
> seconds = 3.16 x10^7^ seconds. Plugging in this value we finally get D
> = (1/2) x 9.8 x (3.16 x10^7^)^2^ = 5 x10^15^ meters. One light year is
> the speed of light (3 x10^8^ m/s) times the number of seconds in a
> year (which we just showed was 3.16 x10^7^ seconds). That gives 9.5
> x10^15^ meters.

Escape to space

Suppose you want to completely leave the Earth. Just going into orbit
isn't enough; you want to get away completely, maybe to take a trip to
the Moon, Mars, or a distant star. That takes more energy than just
going into orbit. How much more? The answer is surprisingly simple:
exactly twice as much. You can get that much kinetic energy by going
1.414 times faster than orbital speed. (That's because energy depends on
the square of the velocity, and 1.414^2^ = 2.) Since orbital velocity is
8 km/sec, that means that escape velocity is 1.414 *x* 8 ≈ 11 km/sec =
11000 m/sec. The formula is given in an optional footnote. [^18]

> **optional: equations for escape energy and escape velocity**.
>
> If you know some calculus, then you can understand that the energy is
> E = ∫ F dr . If you put in *F = GMm/r^2^* and integrate from *r =
> R~E~* (the Earth's radius) to *r* = infinity, then you the result *E =
> GMm/R~E~*. This is the equation for the escape energy. For the escape
> velocity, set this energy equal to *E = (1/2) m v^2^* and solve for
> *v*. This gives the escape velocity:
>
> *v* = sqrt(2*GM/R~E~*)
>
> If we put in the values for *G, M* (the mass of the Earth) and *R~E~*,
> then we get the value that *v* = 11000 m/s = 11 km/s.

At the escape velocity, the kinetic energy of a 1 kg object is equal to
15 Calories per gram.[^19] That is only a little bigger than the energy
of gasoline -- but as soon as you are a few kilometers above the surface
of the Earth, you begin to run out of oxygen, and gasoline needs oxygen.
If you carry the oxygen with you (and rockets do that), it greatly
increases the weight. That's why it is so hard to get objects into
space. If you use chemical fuel, then the fuel weighs substantially more
than the object you are sending up. Typically, you have to use 20 grams
of fuel to launch 1 gram of satellite; equivalently, you have to use 20
tons of fuel to launch 1 ton of satellite. That's why rockets are so
much larger than the satellites they launch.

It is very interesting that the escape velocity does not depend on the
mass. More massive objects do require more energy to escape, but at the
same velocity, more massive objects already have more energy! Remember
*E = (1/2) m v^2^*. At escape velocity v = 11 km/sec, and object with 2
kg mass has twice the energy as an object of 1 kg.

Black holes[^20]

Big planets have high escape velocities. For Jupiter, it is 61 km/sec.
For the sun it is 617 km/sec. Are there any objects for which the escape
velocity is higher than the speed of light, 3 x 10^5^ km/s = 3 x10^8^
m/s? The surprising answer is *yes*. We call such objects black holes.
They get their name from the fact that even light cannot escape, so we
can never see their surface; they are black. We'll show in Chapter 11
that no ordinary object (made out of mass as we know it) can go faster
than the speed of light. We'll see in Chapter 11 that nothing can go
faster than light. That means that nothing could escape a black hole.

If it's invisible, how do we know it's there? The answer is that even
though it can't be seen, we can see the effects of its very strong
gravity. When there is nothing visible, but the gravity is so strong
that we know there must be something of great mass present, then we
deduce it must be is a black hole.

To be a black hole, you have to do one of two things: either have a lot
of mass, or pack a moderate amount of mass into a very small radius.
Several black holes are known to exist; even though we can't see them
(no light leaves the surface) we know that they are there because of
their strong gravitational force they exert. The known black holes are
all as massive as a star -- or greater. But if you were to take the mass
of the Earth and pack it inside a golf ball (in principle this is
possible), then it would be a black hole. The mass would be the same,
but the radius would be so small that the gravitational force on the
surface of the golf ball would be enormous.

The sun is more massive, so you don't have to pack it so tightly. It
would be a black hole if you packet it inside a sphere with radius 2
miles.

I calculated these sizes by using the "black hole equation." This
equation comes from taking the equation for the escape velocity and
setting it equal to the speed of light, *c* = 3 x 10^8^ m/sec.

> **optional: the black hole equation**
>
> If R is the radius of the black hole (usually called the
> "Schwartzchild radius") and M is its mass, then we get the black hole
> equation by taking the escape velocity equation and setting *v = c*,
> the speed of light. (Strictly speaking, this derivation is not correct
> since we are assuming that the mass is independent of velocity. We are
> also assuming ordinary "Euclidean" geometry). When we do this we get:
>
> *c = sqrt(2GM/R)*
>
> Solving for R, this gives the Black Hole Equation:
>
> *R = 2GM/c^2^*
>
> Any object of mass M that has all that mass in a sphere of radius R or
> smaller, will be a black hole. If we take M = M~E~ (the mass of the
> Earth), then we conclude that the Earth would be a black hole if its
> radius were less than
>
> *R = 2GM~E~/c^2^*
>
> = 2 x 6.68x10^-11^ x 6x10^24^/(3x10^8^)^2^
>
> = 9x10^-3^ m = 0.9 cm
>
> Thus, the Earth would be a black hole if all its mass were stuffed
> into a golf ball. For the Sun to turn into a black hole, its mass
> (*M~S~* = 2x10^30^ kg) would have to be stuffed in a sphere of radius
>
> *R = 2 G M~S~/c^2^*
>
> = 2 x 6.68 x10^-11^ x 2E30/(3 x10^8^)^2^
>
> = 3000 m = 3 km ≈ 2 miles
>
> So, if the Sun were squeezed into a radius of 2 miles it would be a
> black hole.

There are several black holes that were created, we believe, when the
inner part of a star collapsed from its own weight into a very small
radius. The object in the sky known as Cygnus X-1 is thought to be a
black hole from such a collapse. If you have spare time, look up Cygnus
X-1 on the web and see what you find.

Many people now believe that large black holes exists at the center of
the large collections of billions of stars known as galaxies, such as
the Milky Way galaxy. We presume these were created when the galaxy
formed, but we know almost none of the details about how this happened.

Even more remarkably, the Universe itself may be a black hole. That's
because the black hole radius for the Universe is about 15 billion light
years, and that is approximately the size of the observable Universe. In
other words, the Universe appears to satisfy the black hole equation.
We'll discuss this further in Chapter 11. But you probably won't be
surprised at one inescapable conclusion that follows[^21]: we can never
escape from the Universe.

Momentum -- and Newton's third law

If you shoot a powerful rifle, then the rifle puts a large force on the
bullet sending it forward. But the bullet puts a backward force on the
rifle, and that is what causes the "kick." The rifle can suddenly go
backwards so rapidly that it can hurt your shoulder. If you don't have
your feet firmly planted on the ground, you will be thrown backwards.

This effect is sometimes given the fancy name "Newton's Third Law", and
it is stated as "for every action there is an equal and opposite
reaction." But we no longer use this old terminology of "action" and
"reaction." Instead we say that if you push on an object (such as a
bullet), then the bullet also pushes back on you for exactly the same
amount of time. Of course, the bullet is lighter, so it is accelerated
much more than you are.

Based on the fact that the bullet pushes on the rifle for the same time
that the rifle pushes on the bullet, we can derive[^22] an extremely
important equation, sometimes called the conservation of momentum. For
the rifle (subscripts R) and the bullet (subscripts B), the equations
are:

*m~B~ v~B~ = m~R~ v~R~*

The equation for the rifle recoil is simple: the mass of the bullet,
times its velocity, is the same as the mass of the gun times its recoil
velocity. Of course, the velocities are opposite; this is sometimes
indicated by putting a minus sign in front of one of the velocities. (I
didn't do that here.) When the gun is stopped by your shoulder, then you
recoil too -- but less, because you have more mass.

The product *m v* is called the "momentum." One of the most useful laws
of physics is called "the conservation of momentum." Before the rifle
was fired, the bullet and gun were at rest; they had no momentum. After
the rifle fired, the bullet and the rifle were moving in opposite
directions, with exactly opposite momenta (the plural of momentum), so
the total momentum was still zero.

Here's another way to say that: when you fire a gun, the bullet gets
momentum. You and the rifle you are holding get an equal and opposite
momentum. If you are braced on the ground, then it is the Earth that
recoils with that momentum. Because the Earth has large mass, its recoil
velocity is tiny, and difficult to measure.

If the objects are in motion prior to the force acting, then what
happens is that the *changes* in momentum must be equal and opposite.
Let's apply that to the comet that crashed into the Earth and killed the
dinosaurs. To make the calculations easy, assume that before the
collision the comet with mass m~C~ was moving at v~C~ = 30 km per second
(a typical velocity for objects moving around the Sun). Assume the Earth
was at rest. After the collision, the total momentum would be the same.
The Earth would have mass m~E~ (that now included the mass of the comet)
and have velocity v~E~. So we can write

*m~C~ v~C~ = m~E~ v~E~*

Solving for *v~E~* we get

*v~E~ = m~C~ v~C~/m~E~*

The mass of the comet is about 10^19^ kg.[^23] Everything else is known,
so we can plug into this equation and get the v~E~, the velocity of
recoil of the Earth:

*v~E~* = (10^19^)(30000)/(6x10^24^)

= 0.5 meters per second

That's not much of a recoil, at least when compared to the usual
velocity of the Earth, which is 30 km/sec = 30,000 meters/sec. So the
Earth was hardly deflected. It's orbit changed, but only by a tiny
amount.

Let me estimate how much a truck recoils when it hits by a mosquito.
Assume (from the point of view of the truck, the mosquito (weighing 2.6
milligrams) is moving at 60 mph = 27 meters per second. Assume the truck
weighs 5 metric tons = 5000 kg. (Hint: use the same equation, except let
the 2.6 milligram mosquito represent the comet. Don't forget to convert
everything to the same units: kilograms and meters.)

Although the conservation of momentum is one of the most important laws
of physics, it is violated in many action movies. For example, if the
hero in the Matrix punches the villain, and the villain goes flying
across the room, then the hero should go flying backwards (unless he is
braced on something big and massive). Likewise, small bullets, when they
hit a person, seem able to impart very large velocities to the person
that they hit, so the person goes flying backwards.[^24]

Rockets

Imagine trying to get into space by pointing a gun downward and firing
bullets so rapidly that the recoil pushes you upward. Sound ridiculous?
Yet that is exactly how rockets work.

Rockets fly by pushing burned fuel downward. If the fuel has mass m~F~
and is pushed down with a velocity v~F~, then the rocket (which has mass
m~R~) will gain an extra upward velocity v~R~ given by the same kind of
equation we used for the rifle:

v~R~ = v~F~ m~F~/m~R~

Compare this to the rifle equation, and to the comet/Earth collision
equation.

Because the rocket weighs so much more than the fuel which is expelled
every second (i.e. m~F~/m~R~ is tiny), the amount of velocity gained by
the rocket is much less than the fuel velocity. As a consequence,
rockets are a very inefficient way to gain velocity. We use them to go
into space only because in space there is nothing to push against except
expelled fuel. Another way to think of this is as follows: rockets are
inefficient because so much of the energy goes into the kinetic energy
and heat of the expelled fuel, rather than in the kinetic energy of the
rocket.

The equation above gives the velocity *change* when a small amount of
fuel is burned and expelled. To get the total velocity given the rocket,
you have to add up a large number of such expulsions. Meanwhile, the
mass of the rocket (which is carrying the unused fuel) is changing as
fuel is used up. A typical result is that the rocket must carry huge
amounts of fuel. The mass of fuel used is typically 25 to 50 times
larger than the payload put into orbit, as I show in the optional
calculation below.

> **optional calculation: rocket fuel**
>
> If we assume that the rocket started at rest, then the final velocity
> of the rocket is given by a the rocket equation, which is derived in
> the footnote[^25]
>
> v = 2.3 v~F~ log~10~(m~R~/m~F~)
>
> This can also be written in terms of the amount of fuel needed:
>
> m~F~ = m~R~ x 10^(v/2.3\ v^F)
>
> In this equation m~F~ is the initial mass of the rocket and fuel
> (mostly fuel) and m~R~ is the final mass (mostly rocket).
>
> Assume[^26] now that the burned fuel expulsion velocity is 3 km/sec,
> and that orbital velocity that we want to achieve is 8 km/sec. Then,
> according to this equation, the mass of fuel needed to reach orbital
> velocity is
>
> m~F~ = m~R~ x 10^(8/(2.3\ *x*\ 2.5)^
>
> = m~R~ x10^1.4^ = 24 m~R~
>
> Thus, for this example, the fuel carried by the rocket must weigh 24
> times as much as the payload!

For a long time, this huge fuel to payload ratio led people to believe
that rockets into space were impossible; after all, how could you even
hold the fuel if it weighed 24 times as much as the rocket? The problem
was solved by using rockets with multiple stages, so that the heavy
containers that held the initial fuel never had to be accelerated to the
final orbital velocity. For example, for the space shuttle, the final
payload (including orbiter weight) is 68,000 kg = 68 tons, and the
amount boosters plus fuel weighs[^27] 1931 tons, a factor of 28 times
larger. Of course, the booster never gets into orbit, only the much
smaller shuttle.

balloon rockets and astronaut sneezes

When you inflate a balloon, and then release the end, the balloon goes
whizzing around the room, driven by the air being pushed out the end.
What is happening is nearly identical to the way a rocket flies. Before
you release the balloon opening, the balloon and the air have zero total
momentum. When you let the air come out, it rushes out with high speed.
It is pushed by the compressed air in the balloon, and it pushes back.
It pushes back on that air, and that air pushes on the balloon. The
released air goes one way, and the balloon containing its remaining air
goes the other way, and the two momenta cancel.

When you sneeze, the sudden rush of air outward likewise can push your
head backwards. In the opening of this chapter, there was a puzzle: how
much does the astronauts head snap back. I read in a newspaper once that
the head would snap back at a dangerous speed because the head was
weightless in space. That's not true, of course. The writer of the
article had confused weight with mass. The astronaut's head has no
weight but it has every bit as much mass as it did on Earth. The force
of the sneeze will accelerate the head backwards by an amount given by
*F = k m a*, but since m is the same as on Earth, the acceleration a
will be no greater.

Skyhook

Ponder the Space Shuttle. To put 1 kg payload into space requires 28 kg
(fuel + container + rocket). Suppose, instead, we built a tower[^28]
that went all the way up to space. How much energy would it take to haul
the ton up to the top, using an elevator? According to the section
"Escape to Space", the energy required take a gram of material to
infinity is 15 Calories. That's roughly the energy in 1.5 grams of
gasoline (not including the oxygen). So if we had a tower with the
elevator, ran the motor below (where it could use air) getting to space
would take 28/1.5 = 19 times less fuel.

Many people have pondered the fuel waste from rockets. Although a tower
to space seems impossible, it may make sense to hang a cable down from a
geosynchronous satellite and use it to haul payload up, an idea once
referred to as "project skyhook." The recent discovery of very strong
carbon nanotubes has revived the idea. Arthur C. Clarke used this idea
in his 1977 science fiction novel "Fountains of Paradise."

A more likely idea is to "fly" to space on an airplane. Airplanes have
two attractive features: they use oxygen from the atmosphere as part of
their fuel (so they don't have to carry it all, as do rockets) and they
can push against the air, instead of having to push against their own
exhaust. Although it is possible in principle, the technology to achieve
8 to 12 km/sec with airplanes does not yet exist. We'll talk more about
airplanes in a mement.

Earlier in this chapter we mentioned the possibility of using "rail
guns." These are long devices that achieve their high projectile
velocities by using electric and magnetic forces to push on the
projectile. But recall their limitation: they must be very long in order
to avoid huge accelerations that could kill humans.

Ion rockets

The inefficiency of rockets comes from the fact that typical chemical
fuels only have enough energy to give their atoms a velocity of 2-3 per
second. Rockets have been proposed that overcome this limitation by
shooting out *ions*, a name for atoms that have an electric charge. You
can find out a lot about these on the web. Like the rail gun, the ions
can be given their high velocity through electric forces, so they are
not limited to the 2-3 km/sec typical of rocket fuel. For example, a
proton expelled at an energy of 100,000 eV has a velocity of 4400
km/sec. This makes ion rockets potentially much more efficient than
chemical rockets, but so far nobody has figured out how to make the mass
of the expelled ions sufficiently great to be able to launch a rocket
from the Earth. They might be used for long duration trips between
planets, or (if we ever do that) between stars.

Flying: airplanes, helicopters, and fans

Airplanes fly by pushing air downwards.[^29] Every second, the airplane
tends to pick up downward velocity from the Earth's gravity. It stays at
the same altitude by pushing enough air downward that it overcomes this
velocity.

The fact that wings push air down is most readily observed in a "rotary
wing" aircraft, otherwise known as a helicopter. (Helicopter pilots
called the ordinary airplane a "fixed wing" aircraft.) In fact, the
helicopter blades are designed to have a shape identical to wings, and
air is pushed past them when they spin. They push the air down, and that
is what pushes the helicopter upward. If you stand under a helicopter
rotor when it is spinning, you can feel the air being forced downward.

It is probably more convenient, however, to observe how wing-shaped
blades push air by standing in front of a fan. Fans work the same way as
a helicopter blades, and as airplane wings. Movement through the air
forces air perpendicular to the direction of motion of the blade.

For the airplane and the rocket, the v~R~ needed is the velocity to
overcome the pull of gravity. In once second, gravity will give any
object a velocity of

*v = g t* = 10 m/s

This falling velocity must be cancelled by accelerating upward, and this
is done in an airplane by pushing air downwards. Air is typically a
thousand times less dense than the airplane (1.25 kg/m^3^), so to get
enough air (i.e. to make mair large) the wings must deflect a large
amount of air downward.

The wake of a large airplane consists of this downward flowing air,
often in turbulent motion. It can be very dangerous for a second plane
if it encounters this wake, since the amount of air flowing downward is
large.

Flying: balloons

The first way that humans "flew" was in hot air balloons, in 1783 above
Paris. These make use of the fact that hot air expands. It takes more
volume compared to an equal mass of cool air. Another way to say this is
that the density of hot air (the mass per volume) is less than that of
cool air.

In a liquid or gas, things that are less dense tend to float. That's why
wood floats on water (if it has a density less than one gram per cubic
centimeter; some woods sink). The heavier fluid "falls" and flows under
the less dense object, pushing it upward. Boats float only if their
average density (metal hull plus empty space inside) is less than that
of water. That's why a boat will sink if it fills with water.

If you fill a balloon with hot air, it will rise until it reaches an
altitude at which the density of the surrounding air matches that of the
balloon. (Of course, you have to include the mass of the balloon, and
any weight that it carries, along with the mass of the air inside.)

Better than hot air are light gases such as hydrogen and helium. Air
weighs about 1.25 kg per cubic meter. The same volume of hydrogen weighs
about 14 times less, i.e. about 0.089 kg = 89 grams.[^30] If we fill a
one-cubic-meter balloon with hydrogen, it will tend to float. In fact,
the "lift" (upward force) of the balloon is just equal to the difference
in weights of air -- hydrogen. Putting this into numbers, it means that
a cubic meter balloon will have a lift of 1.25 -- 0.089 = 1.16 kg. That
means that if you hang an object from it that weighs less than one
kilogram (including the mass of the balloon skin), it will still go
upwards.

Helium gas isn't quite as light as hydrogen gas, so if the balloon is
filled with helium, the lift isn't quite as much. Here's the
calculation: helium gas weighs about 0.178 kg per cubic meter. So the
lift on a helium balloon would be 1.25 -- 0.178 = 1.07 kg. Note that
even though the helium is twice as dense as hydrogen, the lift is almost
as good.

But 1.16 kg of lift for a cubic meter of balloon is not really much.
That's why, despite the cartoons you may have watched on TV, even a
large packet of balloons are not enough to lift a 25 kg child.

Hot air balloons have even less lift. If you heat the air to 300 C, then
it's temperature in absolute scale is 600 K. That's twice its normal
temperature, so its density is half of its usual density. The lift of a
cubic meter would be 1.25 -- 1.25/2 = 0.62 kg per cubic meter. Notice
that that is significantly worse than hydrogen or helium. To lift a
person who weighs 100 kg (including basket, balloon skin, cables) would
take 100/0.62 = 161 cubic meters of hot air. If the balloon were shaped
like a cube, the side of such a balloon would be ^3^√161 = 5.4 meters =
18 feet. That's why hot air balloons have to be so big, and why they
don't lift much.

Floating on water

The same principles apply to floating on water. Salt water is denser
than fresh water, so the density difference between it and you is
greater; that's why you float higher in salt water. Even strong swimmers
take advantage of their buoyancy (the fact that they are less dense, on
average, than water). If water becomes filled with bubbles, its average
density can become less than yours, and you will sink. The New York
Times on August 14, 2003, had an article describing how a group of boys
drowned because they were in bubbly water.[^31] Undersea volcanic
eruptions have led to bubbly water in the oceans, and in such water even
ships will sink.

Submarines can adjust their depth below the ocean surface by taking in
and expelling water into their ballast tanks. When they take in water,
the air in the tanks is replaced by the heavier water, and the average
density of the submarine increases. This makes the submarine sink. The
only thing that will stop the sinking is expulsion of water from the
tanks. If the submarine sinks too far, then it is crushed by the weight
of the water above it; that makes it even more dense, and so it sinks
faster. That is called the hull crush depth. In the movie "Crimson
Tide," that depth was 1800 feet, about 1/3 mile down.\
The submarine in that movie (the fictional USS Alabama) was able to save
itself by getting its engine running, and using its short "wings" to get
lift in way similar to the way that an airplane does, by moving forward
and pushing water downward. A submarine can also push compressed air
into its ballast tanks, driving out the water, and decreasing its
average density.

Sperm whales are said to be able to dive as deep at 2 miles. Diving deep
is easy, since as the whale goes deeper, any air in its lungs (remember,
a whale is a mammal) is compressed, and that makes the whale less
buoyant. So once the whale is denser than water, it will sink. Coming up
is the hard part. Whales save enough energy in their effortless dives,
to be able to swim back up to the surface.

Air pressure on mountains, outside airplanes and satellites

Air pressure is simply the weight of the air above you. In any fluid or
gas, the pressure is evenly distributed, so that the air at sea level
will push an equal amount up, down, and sideways. As you go higher,
there is less air above you, so the pressure decreases. At an altitude
of 18,000 feet (3.4 miles, 5.5 km) the pressure is half of what it is at
sea level. Go up another 18,000 feet -- to 36,000 feet, and the pressure
is reduced by another factor of 2 -- to one quarter the pressure at sea
level. That's a typical altitude for jet airplanes. And that rule
continues; for every additional 18000 feet, the pressure (and the
density of the air[^32]) drops by another factor of two. You can not
live at such a low density of air, and that is why airplanes are
pressurized. You can live at that pressure if the "air" you are given is
pure oxygen (rather than 20%), and that is what you would get from the
emergency face masks that drop down in airline seats if the cabin ever
loses pressure.

The equation for this can be written in a simple way. Let P be the
relative pressure at an altitude H, compared to the pressure at sea
level. So for H = 0, P = 1. Then the equation is:

P = (1/2)^H/18000^ for H in feet

P = (1/2)^H/5.5^ for H in km

To work this out, take the altitude (say in km), divide it by 5.5, and
write down your answer. Then multiply 1/2 by itself that many times.
That can easily be done on a calculator. For example, consider a "low
earth orbit" satellite, at 200 km above sea level. H/5.5 = 200/5.5 =
36.36 ≈ 36. Now multiply (1/2) by itself 36 times to give:

P = (1/2)^36^ = 1.45 x10^-11^ = 0.00000000001

That's a pressure of 10 trillionths times as small as at sea level.[^33]
Satellites need this low pressure to avoid being slowed down by
collisions with air.

Because the density of air decreases with altitude, a helium balloon
will not rise forever. Eventually it reaches an altitude at which the
outside air has the same density as the helium (with the weight of the
gondola averaged in), and then it stops rising. I noticed this as a
child, when I was disappointed to see that the helium-filled balloon I
had released did not go all the way to space, but went high up and then
stopped rising.

That's why balloons are not a possible way to get to space.

Convection -- thunderstorms and heaters

When air is heated near the ground, its density is reduced, and it tends
to rise, just like a hot air balloon. unconstrained by any balloon, the
air expands as it rises, and an interesting result is that it will
remain less dense than the surrounding air until it reaches the
"tropopause", the level in which ozone absorbs sunlight causing the
surrounding air to be warmer. When it reaches the tropopause, its
density is no longer less than that of the surrounding air, so the air
stops rising.

On a summer day, when thunderstorms are growing, it is easy to spot the
tropopause. It is the altitude at which the thunderstorms stop rising,
and begin to spread out laterally. The tropopause is a very important
layer in the atmosphere. It is the location of the "ozone layer" which
protects us from cancer-producing ultraviolet light. We'll talk more
about his layer in Chapter 8 Waves (because of the important effect it
has on sound), and in Chapter 10 Invisible light, when we discuss
ultraviolet radiation and its effects.

Convection is the name we give to the process of hot air expanding and
rising. When you have a heater in a room near the floor, the rising hot
air forces other air out of the way; that results in a circulation of
the air. This is a very effective way to warm a room -- much faster than
heat conduction through the air. But it invariably results in the
warmest air being near the top of the room. If you put the heater near
the top, the warm air just stays up there. On a cold day, in a room with
a heater near the floor, stand up on a stepladder and feel how much
hotter it is near the ceiling.

Angular momentum and torque (optional)

In addition to ordinary momentum (mass times velocity), there is another
kind of momentum that physicists and engineers find enormously useful in
their calculations called angular momentum. Angular momentum is similar
to ordinary momentum but it most useful for motion that is circular,
i.e. rotation. If an object of mass *M* is spinning in a circle of
radius *R,* moving at velocity *v,* then its angular momentum *L* is:

*L = M v R*

What makes angular momentum so useful is that, like ordinary momentum,
when there are no external forces on an object, it is "conserved", i.e.
its value doesn't change. Have you ever spun on ice-skates? Actually,
ice skates are not necessary -- just stand in a spot and start spinning
with your arms stretched out. If you've never done this, then I strongly
recommend you try it right now. Then rapidly pull your arms in. (You'll
never forget the experience, and you can have great fun entertaining
children with it. ) To the surprise of most people, they will suddenly
spin much faster. You can predict that from the angular momentum
equation. If the angular momentum L is the same before and after the
arms are pulled in, and the mass of the arms *M* is the same, then *v R*
must be the same. If *R* gets smaller, then *v* must get bigger.

Angular momentum conservations also explains why water leaving a tub
through a narrow drain begins to spin. In fact, it is very unlikely that
the water in the tub wasn't already spinning, at least a little bit. But
when the distance to the drain (R, in the equation) gets small, the v in
the equation gets very big. A similar effect occurs in hurricanes and
tornadoes. Air being sucked into the center (where there is a low
pressure due to air moving upward) spins faster and faster, and that's
what gives the high velocities of the air in these storms. The air in
hurricanes gets its initial spin from the spin of the Earth, and then
they amplify it by the angular momentum effect. So hurricanes really do
spin in different directions in the Southern and Northern Hemispheres.

It is not true that sinks or tubs drain differently in the northern and
southern hemispheres. Their direction depends on the small residual
rotation in the water left over from the filling of the tub, or from a
person getting out.

Conservation of angular momentum can be used to understand how a cat,
dropped from an upside down position, can still land on his feet. (Don't
try this one at home! I never have, but I've seen a movie\...) If he
spins his legs in a circle, his body will move in the opposite
direction, keeping his total angular momentum equal to zero. That way he
can bring its legs underneath him. Astronauts do this trick when they
want to reorient themselves in space. Spin an arm in a circle, and your
body will move in the opposite direction. You can try that trick on ice
skates too.

The conservation of angular momentum has other useful applications. It
helps keep a bicycle wheel from falling over, when the wheel is
spinning. It can also cause a difficulty: if kinetic energy is stored in
a spinning wheel (usually called a "flywheel"), then the angular
momentum makes it difficult to change the direction that the wheel is
spinning. This makes its use for energy storage in moving vehicles, such
as buses, problematical. It is often addressed by having two flywheels
spinning in opposite directions, so although energy is stored, the total
angular momentum is zero.

Angular momentum can be changed by a suitable application of an outside
force. The required geometry is that the force must act obliquely, at a
distance. We define torque as the tangential component of the force (the
oblique part) times the distance to the center. So, for example, to
start a bicycle wheel spinning, you can't just push on the rim in a
radial manner. You have to push tangentially. That's called torque. The
law relating torque and angular momentum is this: the rate of change of
angular momentum is numerically equal to the torque.

You can probably see why mastery of the equations of angular momentum is
very useful to engineers and physicists in simplifying their
calculations.

END OF CHAPTER

Quick review

Weight is the force of gravity acting on your mass. The force of gravity
is an inverse square law, so when the distance increases by (for
example) 3, the force becomes 9 times weaker. It is this force that
keeps the Moon in orbit around the Earth, and the Earth in orbit around
the Sun. If gravity were turned off, satellies would move in straight
lines rather than in circles. Even at great distances, the force of
gravity never goes completely to zero. The sensation of weightlessness
felt by astronauts is really the sensation of continuous falling.

All satellites must keep moving, or they fall to Earth; they cannot
hover. In LEO, the satellite moves at 8 km/sec, and orbits the Earth in
1.5 hours. LEO satellites are useful for Earth observations, including
spying. Geosynchronous satellites are useful for applications where the
positon of the satellite must stay fixed with respect to the ground. A
medium Earth orbit (MEO) is in between. GPS satellites are MEO. A GPS
receiver determines its location by measuring the distance to three or
more of these satellites.

Close to the Earth (where we live) objects fall with constant
acceleration, i.e. their velocity *v = g t*, until they are slowed down
by air resistance. The distance they fall is *h* = (1/2) *g t^2^*. Air
resistance increases with the square of the velocity. If you go 3 times
faster, the force of air resistance gets 9 times larger. Air resistance
limits the velocity of fall to no more than the "terminal velocity."
This depends on the area and mass of the object, and is different for
people (60-100 mph), parachutes (15 mph), and large objects such as King
Kong. Air resistance also limits the fuel efficiency of automobiles.
Satellites must fly high (\> 200 km) to avoid air resistance.

A force on an object makes it accelerate by an amount (given in m/s per
sec) of

*F = k m a*

This is known as "Newton's Second Law." (*k* = 0.224 is a constant that
makes the units come out in pounds.) Acceleration can also be measured
in units of *g*, the acceleration of gravity. The g-rule says that to
accelerate an object to 10 g requires a force 10 times as great as that
object's weight. That's about the limit for a human to endure, so higher
accelerations (e.g. the rail gun) can't be used. The Space Shuttle never
accelerates more than 3 *g*. Circular motion can also be considered to
be acceleration, with Newton's Second Law applying, even if the
magnitude of the speed doesn't change. Based on this, we can calculate
the velocity a satellite must have to stay in a circular orbit at
different altitudes above the Earth.

The surface gravity on other planets and on asteroids is very different
than it is on the Earth, science fiction movies notwithstanding.

The energy it takes to push an object with a force F (measured in kg)
over a distance D is

*E = F D/k*

To escape to space completely requires an energy of about 15 Calories
per gram. This is enough energy to lift you up a big enough elevator, if
one could be built ("skyhook"). If you have a velocity of 11.2 km/sec,
then your kinetic energy is sufficient to escape. Black holes are
objects whose escape velocity exceeds the speed of light.

Gravity measurements have practical applications. Since oil is lighter
than rock, it has a weaker gravity, and that fact has been used to
locate it. Gravity measurements give us the best image of the Chicxulub
crater.

When a gun is fired, the bullet goes forward and the gun goes backward.
This is an example of the conservation of momentum. Other examples:
Rockets go forward (very inefficiently) by shooting burnt fuel
backwards. Airplanes and helicopters fly by pushing air downwards.

Objects float when their density is less than that of the fluid or gas
they are in. That includes boats and balloons. Hot air rises because it
is less dense than the surrounding air, and that happens for hot air
balloons and thunderstorms. The density and pressure of air decreases
with altitude according to a halving rule:

P = (1/2)^H/6.9^

where the height H is expressed in km.

Angular momentum (a momentum that applies to circular motion) is also
conserved, and that causes contracting objects to speed up. Examples
include sink drains, hurricanes, and tornadoes.

Qualitative questions

At high velocities, most of the fuel used in an automobile is used to
overcome

> ( ) gravity
>
> ( ) momentum
>
> ( ) air resistance
>
> ( ) buoyancy

Airplanes fly by

> ( ) pushing fuel downward
>
> ( ) pushing fuel backwards
>
> ( ) pushing fuel upwards
>
> ( ) pushing air downwards

At an altitude of 200 km, the downward force of gravity on an Earth
satellite is

> ( ) the same as on the surface of the Earth
>
> ( ) a little bit weaker
>
> ( ) zero
>
> ( ) a little bit stronger

In the southern hemisphere, sinks drain

> ( ) usually clockwise
>
> ( ) the same way they do in the Northern Hemisphere
>
> ( ) the answer is different in the Eastern and Western Hemispheres.

A rail gun

> ( ) accelerates rails
>
> ( ) can shoot bullets faster than ordinary guns
>
> ( ) doesn't really accelerate things
>
> ( ) uses ions for propulsion

The Sun would be a black hole if it were squeezed into a radius of

> ( ) 2 miles
>
> ( ) 2,000 miles
>
> ( ) 2 million miles
>
> ( ) It already is a black hole

The method that would take the least energy to get something to space
is:

> ( ) lift it by elevator (if the elevator existed)
>
> ( ) launch it in a one-stage rocket
>
> ( ) fly it in a balloon
>
> ( ) launch it in a three-stage rocket

If an artificial satellite orbited the Earth at the distance of 240,000
miles (the present distance to the moon), it would orbit the Earth in a
period of

> ( ) 90 minutes
>
> ( ) 1 day
>
> ( ) 1 week
>
> ( ) 1 month

The time it takes a geosynchronous satellite to orbit the Earth is:

> ( ) 90 minutes
>
> ( ) 1 day
>
> ( ) 1 week
>
> ( ) 1 month

A typical terminal velocity for a high diver is closest to:

> ( ) 1 mph
>
> ( ) 10 mph
>
> ( ) 100 mph
>
> ( ) 1000 mph

The force of gravity between two people standing next to each other is

> ( ) zero
>
> ( ) too small to measure
>
> ( ) small but measurable
>
> ( ) greater than 1 kg

For a low Earth satellite (altitude 200 km), the force of gravity is

> ( ) zero
>
> ( ) a few percent less than on the Earth's surface
>
> ( ) exactly what it is on the Earth's surface
>
> ( ) about half of what it is on the Earth's surface

The altitude of a geosynchronous satellite is closest to:

> ( ) 200 miles
>
> ( ) 26,000 miles
>
> ( ) 2,000,000 miles
>
> ( ) 93,000,000 miles

Rockets fly by

> ( ) using antigravity
>
> ( ) pushing air downwards
>
> ( ) pushing fuel downwards
>
> ( ) being lighter than air

Balloons rise until

> ( ) the surrounding air becomes too dense
>
> ( ) the surrounding air becomes too thin ( not dense)
>
> ( ) the surrounding air becomes too cold
>
> ( ) they reach space

You can see where the tropopause is by

> ( ) looking at Earth satellites
>
> ( ) seeing how high birds fly
>
> ( ) looking at the tops of thunderstorms
>
> ( ) seeing where lightning comes from

Ice skaters spin faster by pulling in their arms. This illustrates that

> ( ) momentum is conserved
>
> ( ) angular momentum is conserved
>
> ( ) energy is conserved
>
> ( ) angular energy is conserved

Someone firing a rifle is pushed back by the rifle. This illustrates
that

> ( ) momentum is conserved
>
> ( ) angular momentum is conserved
>
> ( ) energy is conserved
>
> ( ) angular energy is conserved

Internet research topics

See what you can find on the internet about the following topics:

> Manufacturing objects in space
>
> Project skyhook
>
> Spy satellites
>
> Rail guns
>
> Black holes
>
> Ion propulsion
>
> GPS

Essay questions

Automobiles and trucks are often designed with a tapered ("aerodynamic")
rather than a blunt front surface. Why is this? What does the tapered
front accomplish? How important is it?

Artificial Earth satellites fly at different altitudes, depending on
what they need to do. Describe the differences between low Earth
satellites, medium Earth satellites, and geosynchronous satellites. For
which applications would you use each?

Which would fall faster, a high diver doing a "swan dive" (in which he
spreads out his arms and falls chest-first) or a diver with his arms
pointed above his head going straight down? Why? Explain the relevant
physics.

When the diver enters the water, having your head down is much less
painful than a "belly-flop." Discuss this in terms of the "area" that
hits the water.

Science fiction movies and action movies often show things that don't
fit in easily with the laws of physics. Give examples. Can you think of
additonal examples not described in this text book?

Black holes: Why can't we see them? How do we know they are there? Are
any known to exist?

Numerical problems (for those students doing the optional sections)

Calculate the force between two people walking side by side. (Estimate
the relevant distances and masses.) Express the answer in pounds. What
size object would have approximately that weight? A mosquito? (Living
material has a mass of approximately one gram per cubic centimeter.)

We can get the mass of the Sun by knowing two facts: the sun is 150
million km = 1.5 x10^11^ m from the Earth, and the Earth orbits the Sun
in 1 year. From this, calculate the circular acceleration of the Earth.
Next, set the force of gravity equal to the centrifugal force. Even
though you don't know the mass of the Earth (leave it as a symbol M) it
will cancel. Do this, and show that the mass of the Sun is approximately
2 x10^30^ kg.

When you go to an amusement park, there is a ride in which you stand in
a cylindrical structure, and the room spins. You feel the force of the
wall on your back; it is this force that keeps you moving in a circle.
Then the floor suddenly drops away, but you don't fall; you are held in
place by the friction of the wall. Suppose the radius of the cylinder is
5 meters. How fast must you move so that the force you feel is 1 g?
(Hint: look back at the problem of the fighter pilot experiencing 10
gs.)

Estimate the velocity of recoil of a truck, when a mosquito hits it.
Assume (from the point of view of the truck), the mosquito (weighing 2.6
milligrams) is moving at 60 mph = 27 meters per second. Assume the truck
weighs 5 metric tons = 5000 kg. (Hint: use the same equation we used for
the comet hitting the Earth, except let the 2.6 milligram mosquito
represent the comet.)

Which is greater, the gravitational attraction between two people who
are talking to each other (separated by 1 meter), or the gravitational
attraction between two ants that are standing next to each other
separated by 1 mm?

[^1]: A more accurate value is that there are 2.205 lb in a kilogram,
    and 0.4536 kg in a pound, but don't bother memorizing these more
    precise numbers.

[^2]: If you want the force to come out in Newtons, then set *k* = 1.

[^3]: The answer is 2.2 lb. Of course, that is what I previously said
    was the weight of a 1 kg object.

[^4]: There are rides at some amusement parks that allow you to fall for
    long distances and experience weightlessness, at least for a small
    number of seconds. We'll calculate how many seconds in a later
    section.

[^5]: Your path is a geometric curve known as a parabola.

[^6]: If your velocity is not exactly horizontal, or if your velocity is
    a little low or high, then the orbit will not be a circle but an
    ellipse.

[^7]: But be careful. Even at the equator, the satellite doesn't have to
    be overhead. The satellite could be above the Congo and you could be
    in Brazil. So you really have to determine the direction of north to
    make good use of the satellite dish information.

[^8]: At 1600 miles per hour, it will go 200 miles in 1/8 of an hour =
    7.5 minutes.

[^9]: Optional footnote: I said earlier that the gravity of a spherical
    objects acts as if it is all coming from the center of the Earth.
    That is true only if the object's mass is uniformly distributed. If
    there is an oil field, then you can mathematically think of that as
    a sum of a uniform Earth, and a little bit of "negative" mass that
    cancels out some of the gravity. If you are close to the oil field,
    you will sense the reduced gravity because this little bit of
    negative mass will not attract you as much as if it were denser
    rock.

[^10]: Optional footnote: That's because the mass of the asteroid M =
    density *x* volume. Call the density *d*. The volume of a sphere is
    (4/3) π *R*^3^. So the gravity at the surface is *F* = *GMm*/*R*^2^
    = *G m* (4/3 π *R*^3^ *d*)/*R*^2^ = *G m* (4/3 π *d*) *R* =
    constants *x R*. So the weight of a mass of *m* = 1 kg on different
    planets will depend only on their radius *R* of that planet.

[^11]: Optional footnote: According to calculus, the distance D = ∫ v dt
    = ∫ g t dt = g ∫ t dt = (1/2) g t^2^.

[^12]: D = (1/2) 9.8 (4)^2^ = 78 meters

[^13]: In 1939 the Empire State Building was only 6 years old, and did
    not yet have the 250 foot TV antenna on its top.

[^14]: We used the fact, from Chapter 1, that to convert joules to
    Calories you divide by 4200.

[^15]: Optional footnote: Most physics texts measure force in *Newtons*,
    so the equation you'll see there is *F = m a.* Some physics texts
    will take this equation as the definition of mass. In that case,
    Newton's Second Law can be stated as follows: "the mass of an
    accelerating object is approximately constant." As we will see in
    the chapter on Relativity, at extremely high velocities, this law
    breaks down; mass increases with velocity.

[^16]: For people who have studied vectors: velocity is defined in
    physics as a vector. If the velocity at time t1 is **v1**, and at t2
    is **v2**, then the acceleration vector is defined as
    (**v2-v1**)/(t2-t1). Even if the velocity is only changing in
    direction, the difference vector (**v2-v1**) is not zero. For
    circular motion, its magnitude is given by the equation in the text.

[^17]: Movies sometimes show the virtual gravity as being off to the
    side. This could be accomplished by having special sideways
    thrusters to accomplish this. Every hour or so the thrusters could
    rotate the ship around so it doesn't wind up going too far sideways.

[^18]:

[^19]: *E = (1/2) m v^2^* = (1/2) x 1 x (11000)^2^ = 6 x10^7^ joules.
    This is 15000 Calories per kg, equal to 15 Cal per gram.

[^20]: Why is the physics of black holes included in a text titled
    "Physics for future Presidents"? Is there any practical use for this
    knowledge? The answer is: no, not really. It is included, as are a
    few other things in this book, just because most people have heard
    about them and are curious. But you never know. Knowing the size of
    a black hole once won the author of this book a free guide to Paris.
    Outside of Shakespeare Books on the west bank of the Seine was a
    sign offering this prize if anyone could answer the question, "What
    size would the Earth have to be for it to be a black hole."

[^21]: pun intended

[^22]: The derivation is based on calculus. If an object is at rest, and
    it experiences a force F for a time t, then its velocity v = ∫a dt =
    ∫(F/m) dt. Write this as m v = ∫F dt. If there are two object, and
    the forces are equal and opposite, and the time is exactly the same,
    then ∫F dt is equal and opposite for the two objects, and that means
    that the quantity m v must also be equal and opposite for the two
    objects.

    If the F1 and F2 are equal and opposite, so are the quantities m1 a1
    and m2 a2. If an object expereic

[^23]: A comet with a radius R~C~ = 100 km = 10^5^ meters, has a volume
    of *V~C~ = 4/3 π R~C~^3^* (for a sphere) = 4.2x10^15^ cubic meters.
    Assuming that the comet is made mostly of rock and ice, the density
    is probably about 2500 kg per cubic meter, so the mass is 2500 x
    4.2x10^15^ = 10^19^ kg.

[^24]: As a fan of this movie, I explain to myself that according to the
    script, "reality" is just a computer program called *the matrix*.
    Therefore I can assume that whoever programmed *the matrix* simply
    forgot to put in conservation of momentum.

[^25]: If you know some calculus, then you can calculate the final
    velocity as follows. In the notation of calculus, the change in
    velocity (which we called v~R~ in the text) is written as dv, and
    the mass of fuel consumed (which we called m~F~ in the text) is dm,
    and the mass of the rocket is m. The final velocity v = ∫ dv = ∫v~F~
    dm/m = v~F~ ln(m~R~/m~F~) = 2.3 v~F~ log~10~(m~R~/m~F~).

[^26]: **Optional calculation** for those interested: If the fuel
    carries 1 Cal per gram, and all that energy is converted into
    kinetic energy, then we can calculate the velocity by assuming it
    all goes to kinetic energy. 1 Cal = 4200 joules. So 1 kg of fuel
    would have 4,200,000 joules. If it is all converted to kinetic
    energy, we would have (1/2) m v^2^ = 4,200,000. Using m = 1 kg, this
    gives v^2^ = 8,400,000. Take the square root to get v = 3000 m/sec =
    3 km/sec. Because chemical energy is not efficiently converted to
    kinetic energy (much remains as heat) the exhaust velocity of 2
    km/sec is more typical of what is achieved.

[^27]: The external tank hold 751 tons of fuel, and there are two solid
    rocket boosters that weigh 590 tons each, for a total of 1931 tons.

[^28]: In the Bible, such a tower was attempted in ancient Babylon, and
    it is also referred to as the "Tower of Babylon." It's goal was to
    reach heaven. To prevent the Babylonians from succeeding, God made
    all the workers speak different languages. Thus, according to the
    Bible, is the origin of the multitude of languages spoken by humans.
    It is also the origin of the verb "to babble."

[^29]: In most physics books, the lift on the airplane wing is explained
    by use of a principle called Bernoulli's law. The "derivation" is
    done using a diagram that typically shows the air trailing the wing
    as if it is completely undisturbed. The astute student will be
    bothered by this. How can the air put a force on the wing but the
    wing not put a force on the air? A careful analysis (done in
    advanced aerodynamics books) shows that to establish a flow with
    higher velocity above the wing than below, the far field air
    velocity distribution is not undisturbed; it fact, it has air
    deflected downward with a momentum rate equal to the upward force on
    the wings, as it must to satisfy momentum conservation.

[^30]: Nitrogen has an atomic weight (number of neutrons plus protons)
    of 14. Hydrogen has an atomic weight of 1. The number of atoms in a
    cubic meter is the same for both gases, so the factor of 14 simply
    reflects the larger nitrogen nucleus.

[^31]: Here is a quote from that article:\
              To four teenagers from the suburbs, Split Rock Falls was a
    magical place --- cool water rushing between the granite walls of a
    mountain ravine, forming pools for hours of lazy summertime
    swimming.\
    On Tuesday afternoon the four men --- Adam Cohen, 19; Jonah Richman,
    18; Jordan Satin, 19; and David Altschuler, 18 --- returned to their
    favorite childhood summer haunt to find it engorged by a summer of
    heavy rain. By the end of the day, all four men, each an experienced
    swimmer, was dead, drowned in the waters they knew well.\
              In what officials here described as one of the worst
    drowning accidents ever in the Adirondack State Park, all four died
    after Mr. Altschuler slipped off a narrow granite ledge into a
    foaming pool of water whipped into a frenzy by a tumbling waterfall.
    In a final act of friendship, Mr. Richman, Mr. Cohen and Mr. Satin,
    who had grown up together on Long Island, jumped after him to try to
    save his life, police and officials said. The laws of physics were
    against them, though.\
              \"They call it a drowning machine,\" said Lt. Fred J.
    Larow, a forest ranger with the State Department of Environmental
    Conservation, who helped recover the bodies here, about 20 miles
    east of Lake Placid. \"The water was so turbulent and aerated that
    there was no way they could stay above water. Even the strongest
    swimmer in the world couldn\'t have survived it.\"

[^32]: The air density doesn't drop quite that much because the air up
    there is cooler.

[^33]: The equation implicitly assumed that the temperature of the air
    continues to cool at higher and higher altitude. But above the
    tropopause the temperature rises, and that makes this calculation
    interesting, but not really accurate.