Gravitation, Orbits, and Space

1)  The schwarzshield radius is the radius at which nothing, including
    light, can escape from the gravitational force of the mass inside.

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a)  What is the formula for the schwarzshield radius?

The formula for the schwarzshield radius is an escape velocity problem.
At what radius is light unable to overcome the gravitational potential?

½ m~1~v^2^ = [Gm~1~m~2~]{.underline} Set the kinetic energy of one
object equal to gravitational potential

r energy between the two objects.

½ v^2^ = [Gm~2~]{.underline} You can divide m~1~ into both sides, giving
this new equation.

r

r = [2Gm]{.underline} Solve the equation for r, and set the velocity to
be the speed of

c^2^ light to get the formula for the Schwarzshield radius.

b)  If the sun has a mass of 1.99 \* 10^30^ kg, how small would the sun
    have to be compressed to become a black hole (what is the sun's
    schwarzshield radius).

Use the formula for the Schwarzshield radius from part a)

> r = [2Gm]{.underline}
>
> c^2^

r = [2 \* 6.67\*10^-11^ \* 1.99\*10^30^]{.underline} Plug in the
numbers: 6.67\*10-11 for G, 1.99\*1030

> (3\*10^8^)^2^ for m, and 3\*10^8^ for c.

r = 2949.62 m Calculate the answer to the problem above.

c)  What is the schwarzshield radius of the black hole at the center of
    our galaxy, Sagittarius A\*, with a mass 4 Million times that of the
    sun?

Use the formula for the Schwarzshield radius from part a)

> r = [2Gm]{.underline}
>
> c^2^

r = [2 \* 6.67\*10^-11^ \* 1.99\*10^30^ \* 4\*10^6^]{.underline} Plug in
the numbers: 6.67\*10-11 for G,

> (3\*10^8^)^2^ 1.99\*1030 for m, and 3\*10^8^ for c.

r = 1.80 \*10^10^ m Calculate the answer to the problem above.

d)  What is the gravitational force between the earth and the sun? What
    is the gravitational force between the earth and the sun after the
    sun becomes a black hole?

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^ \* 9.57\*10^24^]{.underline} Plug all
the numbers in: 6.67\*10^-11^ for G,

> (6.38\*10^8^)^2^ 1.99\*10^30^ for m~1~ because the mass of the sun is
> 1.99\*10^30^ kg, 9.57\*10^24^ for m~2~, which is the mass of Earth,
> and for r, 6.38\*10^8^ meters is the distance between the Earth and
> the sun.

F = 3.12 \* 10^27^ N Solving the math problem results in this answer

Calculating the force between earth and the sun as a black hole will get
the same answer, because the mass and distance would be the same.

e)  How would the sun turning into a black hole effect our orbit?

There would be no real change to our orbit, since the gravitational
force would remain the same between the sun and the earth before and
after it became a black hole.

2)  Find the Schwarzschild radius for a star with mass equal to that (a)
    of Earth and (b) our Sun

Answer: The Schwarzschild radius is given by .

> (*a*)
> $R_{\text{Earth}} = \frac{2\text{GM}_{\text{Earth}}}{c^{2}} = \frac{2\left( 6\text{.}\text{67}´\text{10}^{- \text{11}}N \cdot m^{2}/\text{kg}^{2} \right)\left( 5\text{.}\text{98}´\text{10}^{\text{24}}\text{kg} \right)}{\left( 3\text{.}\text{00}´\text{10}^{8}m/s \right)^{2}} = 8\text{.}\text{86}´\text{10}^{- 3}\text{ m}»\begin{matrix}
> 8\text{.}\text{9 mm}
> \end{matrix}$
>
> (*b*)
> $R_{\text{Sun}} = \frac{2\text{GM}_{\text{Sun}}}{c^{2}} = \frac{2\left( 6\text{.}\text{67}´\text{10}^{- \text{11}}N \cdot m^{2}/\text{kg}^{2} \right)\left( 1\text{.}\text{99}´\text{10}^{\text{30}}\text{ kg} \right)}{\left( 3\text{.}\text{00}´\text{10}^{8}m/s \right)^{2}} = \text{2950}\text{ m} = \begin{matrix}
> 2\text{.}\text{95 km}
> \end{matrix}$

3)  What is the Schwarzschild radius for a typical galaxy, which has a
    mass of about 3 x 10^41^ kg?

Answer: The Schwarzschild radius is given by .

> $$R = \frac{2\text{GM}}{c^{2}} = \frac{2\left( 6\text{.}\text{67}´\text{10}^{- \text{11}}N \cdot m^{2}/\text{kg}^{2} \right)\left( 3´\text{10}^{\text{41}}\text{kg} \right)}{\left( 3\text{.}\text{00}´\text{10}^{8}m/s \right)^{2}} = \begin{matrix}
> 4´\text{10}^{\text{14}}m
> \end{matrix}$$

4)  The Schwarzchild radius for a star with mass equal to that of our
    Sun is about 2.95 km. What is its corresponding density? Compare
    this density to the density of the nucleus of hydrogen. Assume
    hydrogen's nuclear radius is about 10^--15^ m.

Answer: Mass of Sun is 1.99 x 10^30^ kg.\
Black hole density =
$\rho = \frac{M}{(43\pi R^{3})}\text{ = }\frac{1\text{.}\text{99 x 10}^{\text{30}}\text{ kg}}{43\pi \times \left( 2\text{.}\text{95} \times \text{10}^{3}m \right)^{3}}\text{ =1}\text{.}\text{8 } \times \text{10}^{19}\text{kg/m}^{3}$

Hydrogen atom mass \~ 1.67 x 10^--27^ kg.

Hydrogen nuclear density =
$\rho = \frac{M}{(43\pi R^{3})}\text{ = }\frac{1\text{.}\text{67 x 10}^{\text{-27}}\text{ kg}}{43\pi \times \left( 1 \times \text{10}^{- \text{15}}m \right)^{3}}\text{ =}4 \times \text{10}^{17}\text{kg/m}^{3}$

5)  Find the mass of a black hole whose Schwarzchild radius is the same
    as the Earth\'s radius.

Answer.
$M_{\text{Earth}} = \frac{R_{\text{Earth}}c^{2}}{2G} = \frac{\left( 6\text{.}\text{371} \times \text{10}^{6}\ m \right)\left( 3\text{.}\text{00} \times \text{10}^{8}\ m/s \right)^{2}}{2\left( 6\text{.}\text{67} \times \text{10}^{- \text{11}}Nm^{2}/\text{kg}^{2} \right)} = 4\text{.}3 \times \text{10}^{\text{33}}\text{ kg}$

6)  The Sun orbits the black hole in the center of the Milky Way galaxy.
    It takes approximately 225 million years for the Sun to make one
    revolution, and the Sun is approximately 26,000 light-years away.
    Estimate the mass of the galaxy contained inside the Sun's orbit. If
    there is roughly 5 times more mass in the Milky Way galaxy outside
    of the Sun's orbit than inside, estimate the total mass galaxy.

Answer: To find the mass of the black hole mass plus all the matter
within 26 000 ly of the center of the galaxy use Kepler's third law M =
4π^2^r^3^ / GT^2^

T = period = 225 years = 7.10030834 × 10^15^ s r = 26,000 light-years =
2.45973739 × 10^20^ m

M = 4π^2^r^3^ / GT^2^ = 4π^2^(2.5 x 10^20^ m)^3^ / (6.67x10^--11^ N
m^2^/kg^2^ )x (7.1 x 10^15^ s)^2^) = 1.83 x 10^41^ kg

Mass of Sun is 1.99 x 10^30^ kg. So this corresponds to about 10 ^11^
Suns. Multiply by 6 to get the total galaxy mass/

7)  Determine the mass of the black hole lurking at the center of our
    Galaxy. Conveniently, Nature has arranged for a number of stars to
    be orbiting the black hole. Star S0­19 has an orbital period of 37
    years, and a semimajor axis of 1720 AU. Assume circular orbit and
    compute the mass of the black hole. Express your answer in solar
    masses, M~sun~.

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.

8)  Can you draw a triangle on the surface of a sphere for which the sum
    of the angles is (a) 359°, and (b) 180°.

> ![](media/image1.png){width="1.5138888888888888in"
> height="1.5138888888888888in"}Answer. (*a*) If we take the vertices of
> the triangle as the North pole and two points on a latitude line on
> opposite sides of the Earth, the sum of the angles would be 360°.
> Simply take the two latitude points as not quite directly opposed;
> that is take them 1° off.
>
> (*b*) You need a two-dimensional triangle. The Earth appears flat over
> small lengths. A small triangle will do.

9)  What is the average number of nucleons per cubic meter in the
    universe assuming the universe is at the critical density for
    closure of the universe ( ρ~c~ ≈10^--26^ kg/m^3^)?

> Answer.
> $\text{10}^{- \text{26}}\frac{\text{kg}}{m^{3}}´\frac{1\text{ nucleon}}{1\text{.}\text{67}´\text{10}^{- \text{27}}\text{ kg}} = \begin{matrix}
> 6\text{nucleons}/m^{3}
> \end{matrix}$