A satellite is located 214.2 km above the surface of the earth. Using the given values for the earth, what is the period of the satellite? (Assume the period of the Moon is 27.423 days.)
- Numbers for Earth:
Earth

6.38E6m = Radius

5.98E24kg = Mass

1.50E11m = Orbital Radius.

If the period of the moon is 28 days, what is the period (in hours ("hr")) of a 47.5 kg satellite with an orbital radius of 73.3 x103 km around the Earth?

Numbers for Moon:
Moon

1.74E6m = Radius

7.35E22kg = Mass

3.84E8m = Orbital Radius

Got 8 out of the 10... Just the last two are too complicated for me. Please help

Anyone please ?

From :
http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion



Problem 1 (satellite 214.2 km above surface of Earth) :

4pi² / T² = G x 5.98.10^24 / (6.38.10^6 + 214.2.10^3)^3

either you already know a value of G = 6.6738.10^(-11) m^3.kg^(-1).s^(-2),

or you can express it given the numbers for the Moon :
4pi² / (27.423x24x3600)² = G x 5.98.10^24 / (3.84.10^8)^3
-> solve for G

Answer for problem 1 :

T² = 4pi² x (6.38.10^6 + 214.2.10^3)^3 / (6.6738.10^(-11) x 5.98.10^24)
T² ~ 28364275.5
T ~ 5326
5326 s ~ 1h 28min 46 s

Problem 2 leaves me confused. Are you talking about a satellite around the Moon, or around the Earth ?
What's the use of mass of Moon and radius of Moon if it's an Earth satellite ?

Also, it seems to me that mass of satellite (47.5 kg) is useless for your problem.

At 73.3.10^3 km above surface of Earth, using same formulas :

T ~ 223700 s
T ~ 2 days 14h 8min 20 s