Question

The following information will aid you in your calculations: Mass of Earth: 5.97 x 104 kg Mass of Sun: 1.99 x 1030 kg Earth-Sun Distance: 1.496 x 108 km Mass of Moon: 7.35 x 1022 kg Earth-Moon Distance: 3.84 x 105 km 1) Calculate the altitude necessary for a satellite to be in a geostationary (geosynchronous) orbit about the Moon. You may need to look up the term geostationary or geosynchronous and some additional information before you can answer the question. 2) Calculate the average speed of the Moon in orbit around the Earth. 3) Calculate the escape velocity for an object being shot from the surface of the Moon. Part 3: Orbital Motion of the Moon Prelab Video Link #3 (Part 3) In this part of the laboratory, we will investigate whether Kepler's First Law applies to the Moon's orbit. Using the pictures of the Moon taken at different locations in orbit, you will plot the orbit of the Moon and determine tho orhitl egoontrioity Ag tho Moon mvs loser to the Farth it will annear larger and as it moves farther

Need help with one and two.

Problem 2 I did just need it double checked.

Answer 1

(1)
We know that the orbital velocity must be same as the average velocity of the satellite to be in geostationary or geosynchronous satellite. Therefore ,
$V_{o} = V$
$\sqrt{\frac{GM}{R_{m}+h}} = \frac{2\pi (R_{m}+h)}{T}$
Where R_m is radius of moon = 1737 km
M is mass of moon = 7.35*10²² kg
h is the necessary altitude for geostationary satellite
T is the time period of moon = 27.3 days = 2358720 seconds
Let us consider (R_m+h) = R , therefore
$\sqrt{\frac{GM}{R}} = \frac{2\pi R}{T}$
taking square at both ends
$\frac{GM}{R} = \frac{4\pi^{2} R^{2}}{T^{2}}$
$R^{3}= \frac{GMT^{2}}{4\pi^{2}}$
On solving we get
R = 88403.4 km  
Now we know that
R = R_m + h = 1737 + h
1737 + h = 88403.35
h = 86,666.35 km
Hence the altitude would be 86,666.35 km for geostationary satellite.
(2)
Solution is correct.