# Orbital period questions continued

1. Dec 8, 2008

### Inertialforce

1. The problem statement, all variables and given/known data
A 9.0x10^3kg satellite with an orbital radius of 3.20x10^7m orbits the Earth at an altitude of 2.56x10^7m. What is the orbital period?

2. Relevant equations
ΣFc = mac
Kepler's third law equation

3. The attempt at a solution
For this question I know that it is asking me to find the orbital period, but it has also given me an altitude and mass for the satellite. Are they there just to through me off because you are supposed to use Kepler's third law equation to find the orbital period?

Or are they important and I am supposed to use the equation:

Fg = 4(pie)^2r/T^2

Basically what I am wondering is which equation do we use?

Last edited: Dec 9, 2008
2. Dec 9, 2008

### Lyuokdea

What does Kepler's third law give you? In relation to what you are looking for? What are the inputs required to use Kepler's Law?

~Lyuokdea

3. Feb 19, 2009

### skiing4free

Keplers third law: "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." i don't think it is necessary to use keplars law to solve this question, this is how i would figure it out;

ok so you can solve how fast the satellite is moving in relation the Earth: v=(GMe/r)^.5
(G=6.67x10^-11 gravitational constant) so plugging the numbers in gets you a velocity of 3536.4176......ms^-1, next we use the formula Ac=4$$\pi$$^2r/T^2 like you had before but you can solve the acceleration by using Ac=v^2/r now that you have the velocity. This gives you 0.39082m/s^2 now put this back into your original formula which gives you a period of 15.79 hours

hope this helps