Forces

1)  When black holes merge, the resulting black hole sometimes begins
    traveling through the universe. The fastest a black hole can be sent
    traveling by this process is 176 km/s.

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a)  The nearest black hole to earth is V 4641, a 3 million solar mass
    black whole, which is 1,600 light-years away. If this black hole
    merged with another and started traveling towards us at the maximum
    speed, how long would it take to reach earth?

We start this by using the distance formula from Kinematics

x = v~0~t + ½at^2^ v~0~ is the initial velocity, x is the distance, a is
the acceleration, and t is the time.

1600 \* 9.46\*10^15^ = 176 \* 10^3^ t plug in the numbers. 1600 \*
9.46\*10^15^ goes in for x, because the distance is 1600 light-years,
and you need to convert 1600 light-years to meters, 176\*10^3^ goes in
for velocity, because the velocity is 176 km/s, and you need to convert
the velocity from km/s to m/s.

8.6 \* 10^13^ seconds = time solve the problem for time

2.7 \* 10^6^ years converted into years (to aid in understanding)

b)  In order to stop the black hole from reaching earth, we hook up our
    most powerful rocket to it. This rocket fires 24.02 MN of thrust.
    How long will it take this rocket to stop the black hole? If the
    rocket started firing as soon as the black hole moved, would it stop
    before reaching earth?

First, we need to figure out what the acceleration is on the black hole
from this rocket.

F = ma F is the force, m is the mass, and a is the acceleration

24.02\*10^6^ = 1.99\*10^30^ \* 3\*10^6^ a Plug in the numbers.
24.02\*10^6^ for force, because you need to convert the units to
newtons, and 1.99\*10^30^ \* 3\*10^6^ for m, because that the black
hole's mass is 3 million times the sun's mass

4.02\*10^-30^ m/s^2^ Solve for acceleration

Now, we plug that acceleration into the kinematics equation for
velocity.

v = v~0~ + at v~0~ is the initial velocity, x is the distance, a is the
acceleration, and t is the time.

0 = 176000 -- 4.02\*10^-30^ t Plug in the numbers. 0 for velocity,
because we

> want to find the time when the black hole is stopped
>
> 176000 for initial velocity, because we converted it into m/s, and
> -4.02\*10^-30^ for acceleration, because we found it in the last
> problem and it pushes the opposite direction of the velocity.

4.37\*10^34^ seconds= t Solve for t

c)  What acceleration would we need to apply to the black hole to get it
    to stop before it reached earth?

Using the velocity equation, we can solve for the time in terms of the
acceleration

v = v~0~ + at v~0~ is the initial velocity, x is the distance, a is the
acceleration, and t is the time.

0 = 176000 + at plug in 0 for velocity, because we want the black hole
to stop, and 176000 for initial velocity, because we need to convert it
to m/s.

t = -176000/a Solve for the time in terms of acceleration

Now, we use the distance equation

x = v~0~t + ½at^2^ v~0~ is the initial velocity, x is the distance, a is
the acceleration, and t is the time.

1600 \* 9.46\*10^15^ = -176000 \* -176000/a -- ½a\*(-176000/a)^2^ Plug
in all the

> information we have for the problem:
>
> 1600 \* 9.46\*1015 goes in for x, because it is the distance to the
> black hole converted into meters. --176000 goes in for v, because it
> is the velocity converted into m/s, and we use the time we found above
> for t.

1.02 \* 10^-9^ m/s2 = a solve the equation for acceleration

d)  What force would it be necessary to apply the acceleration required
    to stop the black hole from reaching earth?

F = ma

F = 1.99\*10^30^ \* 3\*10^6^ \* 1.02 \* 10^-9^ plug in the numbers, a is
the acceleration we

> found in the last part, 1.02 \* 10^-9^, and m is the mass of the black
> hole, 1.99\*10^30^ \* 3\*10^6^

F = 6.09\*10^27^ N solve for the force

e)  How many rockets would it take to supply the force necessary to stop
    the black hole?

Each rocket supplies 24.02\*10^6^ N of force, so you divide the force
needed by the amount of force from each rocket.

6.09\*10^27^ / 24.02\*10^6^ = 2.53\*10^20^ rockets needed to stop the
black hole.

2)  What is the gravitation force of attraction between the Milky Way
    galaxy and its nearest neighbor galaxy, which is a dwarf galaxy
    (named Canis Major after the constellation it is found in and
    discovered in 2003)? Canis Major galaxy is only 42,000 light years
    from the centre of the Milky Way, so close that the Milky Way is
    gradually consuming it by pulling in its stars.. (Canis Major galaxy
    is only about 25,000 light years from Earth.) Take the mass of the
    Milky Way to be 1.2 x 10^42^ kg and assume Canis Major is 1/50 that
    mass (not yet determined; typically dwarf galaxies range from 10 to
    1000 times smaller than a typical galaxy).

> Answer. $\begin{array}{r}
> F = G\frac{m_{1}m_{2}}{r^{2}} = \left( 6\text{.}\text{67} \times \text{10}^{- \text{11}}Nm^{2}/\text{kg}^{2} \right)\frac{\left( 1\text{.}2 \times \text{10}^{\text{42}}\text{kg} \right) \times \frac{1}{\text{50}} \times \left( 1\text{.}2 \times \text{10}^{\text{42}}\text{kg} \right)}{\left\lbrack \left( \text{42000}\text{ly} \right)\left( \frac{9\text{.}\text{46} \times \text{10}^{\text{15}}m}{1\text{ ly}} \right) \right\rbrack^{2}} = 1\text{.}\text{22} \times \text{10}^{\text{31}}\text{ N} \\
> \begin{matrix}
> 1\text{.}\text{22} \times \text{10}^{\text{31}}N
> \end{matrix}
> \end{array}$

3)  Messier 31 (M31, NGC 224) is the famous Andromeda galaxy, our
    nearest large neighbor galaxy, it is about 2.9 x 10^6^ ly away. If
    both the Milky Way has a mass of 1.2 x 10^42^ kg and Andromeda has a
    mass about 80% of the Milky Way, with what gravitational force does
    each galaxy attract the other?

Answer. $\begin{array}{r}
F = G\frac{m_{1}m_{2}}{r^{2}} = \left( 6\text{.}\text{67} \times \text{10}^{- \text{11}}Nm^{2}/\text{kg}^{2} \right)\frac{\left( 1\text{.}2 \times \text{10}^{\text{42}}\text{kg} \right) \times 0\text{.}\text{80} \times \left( 1\text{.}2 \times \text{10}^{\text{42}}\text{kg} \right)}{\left\lbrack \left( 2\text{.}9 \times \text{10}^{6}\text{ly} \right)\left( \frac{9\text{.}\text{46} \times \text{10}^{\text{15}}m}{1\text{ ly}} \right) \right\rbrack^{2}} = 1\text{.}\text{02} \times \text{10}^{\text{29}}\text{ N} \\
\begin{matrix}
1 \times \text{10}^{\text{29}}N
\end{matrix}
\end{array}$

4)  How much force do you exert on the black hole at the center of the
    Milky Way? (Use the mass calculated above.) The Sun is approximately
    26,000 light-years away from the black hole in the center of the
    Milky Way galaxy.

Answer: Use Universal Gravitation:

$$F = G\frac{m_{1}m_{2}}{r^{2}} = \left( 6\text{.}\text{67} \times \text{10}^{- \text{11}}Nm^{2}/\text{kg}^{2} \right)\frac{\left( 7\text{.}2 \times \text{10}^{\text{37}}\text{kg} \right) \times \left( \text{70}\text{kg} \right)}{\left\lbrack \left( \text{26000}\text{ly} \right)\left( \frac{9\text{.}\text{46} \times \text{10}^{\text{15}}m}{1\text{ ly}} \right) \right\rbrack^{2}} = 5\text{.}6 \times \text{10}^{- \text{12}}\text{ N}$$

5)  Tidal forces and spaghettification .

    a.  Calculate the acceleration of gravity (the gravitational field
        strength) near the Schwarzchild radius of a star with mass equal
        to that of our Sun (use 3 km). Express your answer in m/s^2^ and
        in g's.

Answer: Use Universal Gravitation:

$$g_{1} = G\frac{m}{r^{2}} = \left( 6\text{.}\text{67} \times \text{10}^{- \text{11}}Nm^{2}/\text{kg}^{2} \right)\frac{1\text{.}\text{99} \times \text{10}^{\text{30}}\text{kg}}{\left\lbrack \left( 3 \times \text{10}^{3}m \right) \right\rbrack^{2}} = 1\text{.}5 \times \text{10}^{\text{13}}\text{ m/s}^{2} = 1\text{.}5 \times \text{10}^{\text{12}}\text{ g's}$$

b.  Calculate the acceleration of gravity (the gravitational field
    strength) experienced by an astronaut located 300 km from this black
    hole. Express your answer in m/s^2^ and in g's.

$$g_{2} = G\frac{m}{\left( \text{100}r \right)^{2}} = \frac{g_{1}}{\text{10}^{4}} = 1\text{.}\text{4748111} \times \text{10}^{9}\text{ m/s}^{2} \approx 1\text{.}5 \times \text{10}^{8}\text{ g's}$$

c.  If an astronaut is located at this position 300 km from the black
    hole and her feet 1.6 meters closer to the black hole compared to
    her head, determine the difference in gravitational acceleration
    between her feet and her head. Express your answer in m/s^2^ and in
    g's.

> This differential force on the astronaut is called a tidal force and
> it also occurs for the Earth and Moon and the Earth and the Sun
> because one side of the Earth is closer to the Moon (or Sun) than the
> other side and hence, the closer side experiences a stronger
> gravitational field strength.

Let
$g_{\text{feet}} = g_{2} = 1\text{.}\text{4748111} \times \text{10}^{9}\text{ m/s}^{2}$be
the gravitational field strength at her feet.

$$g_{\text{head}} = \left( 6\text{.}\text{67} \times \text{10}^{- \text{11}}Nm^{2}/\text{kg}^{2} \right)\frac{1\text{.}\text{99} \times \text{10}^{\text{30}}\text{kg}}{\left\lbrack \left( 3 \times \text{10}^{3}\text{m } + 1\text{.}\text{5 m} \right) \right\rbrack^{2}} = 1\text{.}\text{4747954} \times \text{10}^{9}\text{ m/s}^{2}$$

d.  Using the above results describe the fate of a 70-kg astronaut who
    is located near the Schwarzchild radius of this star. Note: a human
    femur has a cross-sectional radius of about 1 cm and a tensile
    strength of about 130 x 10^6^ N/m^2^.

Answer: The gravitational field strength between her head and her feet
differs by about 1.6 x 10^4^ m/s^2^ or about 1600 g's. The astronaut
would be pulled apart unless her bones are strong enough to inhibit this
differential acceleration. We can approximate the amount of force by
using the frame of reference relative to a stationary center of mass of
the astronaut. Thus there is a force pulling at either end of the
astronaut of

F = ma = 70 kg x (1.6 x 10^4^ m/s^2^) = 1 x 10^6^ N

The maximum force an adult femur, one of the strongest bones in the
body, can withstand can be found from the cross-sectional area and the
tensile strength:
$\text{strength} \times \text{Area} = \text{130 x 10}^{6}\text{ N/m}^{2} \times \pi\ \left( 0\text{.}\text{01}\ m \right)^{2} = 4 \times \text{10}^{4}\ N$.

Hence, the astronaut will be pulled apart.

Note: as the astronaut's molecules get's closer to the black hole, even
the molecules get ripped apart.

Answer: To find the mass of the black hole mass use Kepler's third law M
= 4π^2^r^3^ / GT^2^

T = period = 37 years = 1.16760626 × 10^9^ s r = 1720 A.U. = 2.5730856 ×
10^14^ m

**M = 4π^2^r^3^ / GT^2^ = 4π^2^(2.6 x 10^14^ m)^3^ / (6.67x10^--11^ N
m^2^/kg^2^ )x (1.2 x 10^9^ s)^2^) = 7.2 x 10^37^ kg**

Mass of Sun is 1.99 x 10^30^ kg. So this corresponds to about 3.6
million solar masses.

6)  The black hole Sagitarrius A\* is a super-massive black located in
    the center of the

> Milky Way Galaxy, 26,000 light-years away. This black hole has a mass
> approximately 4 million times the mass of our sun.

a)  Given that the mass of the sun is 1.99 \* 10^30^ kg, what is the
    force of gravity between this black hole and a 75 kg person?

The equation which measures the gravitational force between two objects
is:

F = [Gm~1~m~2~]{.underline} Where G is the gravitational constant, m~1~
is the mass of one object, m~2~ is

r^2^ the mass of the other object, and r is how far apart the two
objects are.

F = [6.67\*10^-11^ \* 1.99\*10^30^ \* 4\*10^6^ \* 75]{.underline} Plug
all the numbers in: 6.67\*10^-11^ for G,

> (26000 \* 9.46 \* 10^15^)^2^ 1.99\*10^30^ \* 4\*10^6^ for m~1~ because
> the mass of the sun is 1.99\*10^30^ kg, and the black hole is as
> massive as 4 million suns, 75 for m~2~, and, for r, you need to
> convert 26000 light years to meters, since G is in meters, so you
> multiply 26000 \* 9.45\*10^15^, and that goes in for r

F = 1.65 \* 10^-19^ N Solving the math problem results in this answer

b)  What is the force of gravity between that same 75 kg person and our
    sun, which is 1.5 \* 10^8^ km away from earth?

F = [Gm~1~m~2~]{.underline} Where G is the gravitational constant, m~1~
is the mass of one object,

> r^2^ m~2~ is the mass of the other object, and r is how far apart the
> two objects are.

F = [6.67\*10^-11^ \* 1.99\*10^30^ \* 4\*10^6^ \* 75]{.underline} Plug
all the numbers in: 6.67\*10^-11^ for G,

> (1.5\*10^8^ \* 10^3^)^2^ 1.99\*10^30^ for m~1~ because the mass of the
> sun is 1.99\*10^30^ kg, 75 for m~2~, and, for r, you need to convert
> 1.5\*10^8^ km to meters, since G is in meters, so you multiply
> 1.5\*10^8^ by 10^3^, and that goes in for r.

F = 4.60 \* 10^-1^ N Solving the math problem results in this answer

c)  What is the force of gravity between two 75 kg people on opposite
    sides of the earth (6478 km apart)

F = [Gm~1~m~2~]{.underline} Where G is the gravitational constant, m~1~
is the mass of one object,

> r^2^ m~2~ is the mass of the other object, and r is how far apart the
> two objects are.

F = [6.67\*10^-11^ \* 75 \* 75]{.underline} Plug all the numbers in:
6.67\*10^-11^ for G,

> (6478 \* 10^3^)^2^ 75 for m~1~, 75 for m~2~, and, for r, you need to
> convert 6478 km to meters, since G is in meters, so you multiply 6478
> by 10^3^, and that goes in for r.

F = 8.94 \* 10^-21^ N Solving the math problem results in this answer

7)  An object starts at rest outside of the black hole Sagitarrius A\*

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a)  What is the acceleration of the object 10^8^ meters from the center
    of Sagitarrius A\*, a 4 million solar mass black hole?

We're using the gravitational force equation.

F = [Gm~1~m~2~]{.underline} Where G is the gravitational constant, m~1~
is the mass of one object, m~2~ is

r^2^ the mass of the other object, and r is how far apart the two
objects are.

Since F = ma, we change this to:

m~1~a = [Gm~1~m~2~]{.underline} By substituting m~1~a in for F

r^2^

a = [Gm~2~]{.underline} Dividing both sides by m~1~, we get this new
equation

r^2^

a = [6.67\*10^-11^ \* 1.99\*10^30^ \* 4\*10^6^]{.underline} Substitute
in the numbers, 6.67\*10-11 for G,

> (10^8^)^2^ 1.99\*10^-30^ \* 4\*10^6^ for m~2~, which is the mass of
> the black hole, and 10^8^ for r, which is how far away the object is
> from the center of the black hole.

a = 5.31\*10^10^

b)  Assuming the acceleration were constant, how long would it take for
    the object to reach the center of the black hole?

Start with the distance equation from kinematics

x = v~0~t + ½at^2^ v~0~ is the initial velocity, x is the distance, a is
the acceleration, and t is the time.

10^8^ = ½ \* 5.31\*10^10^ t^2^ Plug in the numbers. 10^8^ is x, the
distance the objects travels, the initial velocity is 0, and the
acceleration is 5.31\*10^10^, from the last problem.

6.1\*10^-2^ seconds = t Solve for time

c)  What is the object's velocity when it reaches the center. Compare
    this to the speed of light.

Start with the velocity equation from kinematics

v = v~0~ + at v~0~ is the initial velocity, v is the velocity, a is the
acceleration, and t is the time.

v = 5.31\*10^10^ \* 6.1\*10^-2^ Plug in the numbers, 0 for initial
velocity, 6.1\*10^-2^ for the time, and 5.31\*10^10^ for the
acceleration.

v = 3.24 \* 10^9^ m/s Calculate the velocity.

This velocity exceeds the speed of light.

8)  An interesting thing happens to objects as they are pulled into a
    black hole. Say you have a piece of iron (56g/mol) going into a
    black hole. The atoms in the piece of iron are 2.5 \* 10^-10^ m
    apart. Let's looks at the force on the atoms in this piece of iron
    as you get close to the center of Sagittarius A\*, the black hole at
    the center of our galaxy.

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a)  What is the gravitational force on one atom when it is 0.01 meters
    away from the center of the black hole?

We're using the gravitational force equation.

F = [Gm~1~m~2~]{.underline} Where G is the gravitational constant, m~1~
is the mass of one object, m~2~ is

r^2^ the mass of the other object, and r is how far apart the two
objects are.

We need to find out the mass of one iron atom so we can use this
equation. If iron is 56g/mole, we need to divide 56 by 6.02\*10^23^,
because there are that many atoms in a mole. This gets 9.30 \*10^-23^ g.
We will convert this to kg, so we get 9.30\*10^-26^.

F = [6.67\*10^-11^ \* 9.30\*10^-26^ \* 1.99\*10^23^ \*
4\*10^6^]{.underline} Plug all the numbers in: 6.67\*10^-11^

> (0.01)^2^ for G, 1.99\*10^30^ \* 4\*10^6^ for m~1~ because the mass of
> the sun is 1.99\*10^30^ kg and the black hole is 4 million times as
> massive as the sun, 9.30\*10^-26^ for m~2~, which is the mass of one
> iron atom, and for r, 0.01 meters is the distance from the black hole
> to the iron.

F = 0.0493766760 Calculate the answer

b)  What is the gravitational force on the next furthest atom in the
    piece of iron at that same time?

We're going to be using the gravitational force equation again, and the
mass we found of an iron atom.

F = [6.67\*10^-11^ \* 9.30\*10^-26^ \* 1.99\*10^23^ \*
4\*10^6^]{.underline} Plug all the numbers in: 6.67\*10^-11^

> (0.01 + 2.5\*10^-10^)^2^ for G, 1.99\*10^30^ \* 4\*10^6^ for m~1~
> because the mass of the sun is 1.99\*10^30^ kg and the black hole is 4
> million times as massive as the sun, 9.30\*10^-26^ for m~2~, which is
> the mass of one iron atom, and for r, 0.01 + 2.5\*10^-10^, because
> this atom is one space behind the last one we calculated.

F = 0.0493766735 Calculate the answer

c)  If the strength of covalent bonds between atoms is 2 \* 10-9 N, what
    will happen to this piece of iron as it gets this close to the black
    hole?

Before the piece of iron hits the center of the black hole, it would be
pulled apart atom by atom, because force of gravity on two atoms next to
each other is so different that the bonds between atoms won't be able to
hold.

**Newton\'s Laws**

**Black Holes**

**OTHER**

1\. The purpose of this set of problems is to engage students in
classical physics using black holes to stimulate interest. The concepts
covered in this problem set include Newton's Laws of moton. Most ideas
and vocabulary are readily understandable for instructors, but some
things may need explanation for students.

The *Accretion Disk* is a disk of gas that surrounds a black hole. As
the material spirals closer to the black hole, it accelerates and bumps
into itself creating of heat. All this energy needs to go somewhere, so
the disk actually starts to glow. The outgoing radiation can get strong
enough to exert a noticeable force. This force is called *radiation
pressure* and is mentioned in a few problems.

The numbers used in the problems are arbitrary and can be changed to fit
studen's needs.

ANS:

None.

PTS: 1

**PROBLEM**

1\. A 3.14 x 10^6^-kg spaceship is flying along the x-axis. At a point
in a spaceships flight a black hole is 30^o^ from the path of the ship.
If the black hole pulls on the ship with a force of 1.57 x 10^3^ N what
would the x- and y-components of the force be?

ANS:

x-component = 1.36 x 10^3^ N

y-component = 785 N

PTS: 1 TOP: Components of Forces

A probe is under the gravitational attraction of three black holes.
One's gravitational pull is along the negative x-axis and has a
magnitude of 280 N. Another is along the positive x-axis with a
magnitude of 325 N. The last is in the positive y direction with a force
of 879 N.

2\. Find the net external force in the x direction.

ANS:

45 N

PTS: 1 TOP: Components of Forces

3\. Find the net external force in the y direction.

ANS:

879 N

PTS: 1 TOP: Components of Forces

4\. Find the magnitude and direction of the net external force on the
probe.

ANS:

880 N at 87^o^

PTS: 1 TOP: Components of Forces

5\. If the left black hole was move to be 30 degrees north of the
negative x-axis, what would be the x-components of the net force on the
probe?

ANS:

82.5 N

PTS: 1 TOP: Components of Forces

6\. If the left black hole was moved to be 30 degrees north of the
negative x-axis, what would be the y-component of the net force?

ANS:

1019 N

PTS: 1 TOP: Components of Forces

7\. What would be the acceleration of the probe if it had a mass of 340
kg?

ANS:

2.59 m/s/s at 87 degrees

PTS: 1 TOP: Newton\'s 2nd Law

8\. What would be the acceleration of the probe if it had a mass of
3,400 kg?

ANS:

0.259 m/s/s at 87 degrees

PTS: 1 TOP: Newton\'s 2nd Law

9\. A scientific probe is heading toward a black hole. Its trajectory is
such that the motion of the accretion disk (the hot rotating gas around
a black hole) is perpendicular to the probes path. If the gravitational
attraction on the probe has a magnitude of 9.05 x 10^4^ N downward and
the accretion disk adds a 1.05 x 10^4^ N force to the right, what will
be the resultant magnitude and direction of the force on the probe?

ANS:

9.31 x 10^4^ N at 6.48^o^ counterclockwise from straight down.

PTS: 1 TOP: Components of Forces

10\. A probe studying the accretion disk of a black hole needs to fire
its thrusters. The accretion disk applies a 325 N radiation pressure
force to the probe in the y direction. The gravitational force of the
black hole is 257 N at an angle of 57.3^o^ below the x-axis. In what
direction and with what force must the thrusters fire to hold the
probe's position.

ANS:

285 N at 40.7^o^

PTS: 1 TOP: Equilibrium

11\. The radiation pressure from the accretion disk at a certain point
is 500 N. If the force of gravity from the black hole is also 500 N at
that point what is the net force on an object at that point? Is it in
equilibrium?

ANS:

0 N

Yes

PTS: 1 TOP: Equilibrium

12\. The sensor in a probe records the magnitude and direction of the
net external force acting on the probe. The probe records a
gravitational force of 4500.0 N forward while simultaneously being
forced up by the radiation pressure of the accretion disk with a force
of 130.0 N. What force will the sensor report?

ANS:

4502 N at 1.655 degrees above forward.

PTS: 1 TOP: Components of Forces

13\. A shuttle is studying a black hole and its accretion disk. The net
external force on the 7.5 x 10^4^-kg shuttle is 7.35 x 10^5^N. What is
the shuttle's acceleration?

ANS:

9.8 m/s^2^

PTS: 1 TOP: Newton\'s 2nd Law

14\. A black hole exerts a 234 N gravitational force on a 6.78 x
10^3^-kg chunk of rock in space. What will be the rock's acceleration?

ANS:

3.45 x 10^-2^ m/s^2^

PTS: 1 TOP: Newton\'s 2nd Law

15\. A 255-kg piece of unknown substance passes close to a black hole.
If the black hole exerts a net force of 2,501 N on the substance, what
will be the objects acceleration?

ANS:

9.81 m/s^2^

PTS: 1 TOP: Newton\'s 2nd Law

16\. A 2,000-kg probe sent to study the magnetic fields associated with
rotating black holes approaches a prospective subject. It travels about
1.47 x 10^11^ m, about one AU, in 3.00 x 10^10^ seconds before its
velocity is zero. What is the average force acting on the probe during
this trek?

ANS:

-0.653 nN

-6.53 x 10^-7^ N

PTS: 1 DIF: Hard TOP: Newton\'s 2nd Law

17\. A lost soccer ball comes near a black hole. When the gravitational
force on the ball is 1.00 N it accelerates at a rate of 0.481 m/s^2^
toward the black hole. What is the mass of the ball?

ANS:

2.08 kg

PTS: 1 TOP: Newton\'s 2nd Law

A 7.0-kg object undergoes an acceleration of 1.7 m/s^2^ due to the
radiation pressure of the accretion disk surrounding a seven solar mass
black hole.

18\. What is the magnitude of the net external force acting on it?

ANS:

12 N

PTS: 1 TOP: Newton\'s 2nd Law

19\. If this same force is applied to a 4.0 kg object, what acceleration
is produced?

ANS:

3 m/s/s

PTS: 1 TOP: Newton\'s 2nd Law

20\. A black hole pulls on a much smaller object with a linear force.
Newton's Third law states that the object exerts an equal and opposite
force on the black hole. Why does the object move toward the black hole?

ANS:

Equal forces mean equal accelerations only if the masses are equal as
well. The same force on the much more massive black hole results in a
much smaller acceleration as for the smaller object. We can see the
motion of the object easier than the motion of the black hole.

PTS: 1 TOP: Newton\'s 3rd Law

21\. Identify the action-reaction pairs in the scenario below:

A rocket ship applies thrust away from a black hole, but remains
stationary.

ANS:

Gravitational:

Ship on Black Hole\--Black Hole on Ship

Thrust:

Hot gasses on ship\--ship on hot gasses

PTS: 1 TOP: Newton\'s 3rd Law

22\. A space ship's engines apply a horizontal 18,000 N force to the
240,000-kg ship while passing over a black hole. The gravitational pull
of the black hole is 36,000 N and directed perpendicular to the ship's
thrust. What is the magnitude and direction of the acceleration of the
ship?

ANS:

40,249 N at 63.4 degrees below the horizontal

PTS: 1 TOP: Newton\'s 2nd Law

23\. If a ten solar mass black hole and a twenty solar mass black hole
were to collide, which one would experience a greater impact force?

ANS:

They would be the same due to Newton's Third law.

PTS: 1 TOP: Newton\'s 3rd Law