# Kepler's Laws and orbiting satellite

1. Nov 11, 2009

### G-reg

1. The problem statement, all variables and given/known data
A orbiting satellite stays over a certain spot on the equator of (rotating) Mercury. What is the altitude of the orbit (called a "synchronous orbit")?

Mercury mass = 3.3022 e23 kg

2. Relevant equations

$$T^2$$ = (4$$\Pi^2$$/GM) * r^3

and that's all I can think of which is why I'm stuck

3. The attempt at a solution

I'm sorry that I have no attempt, I'm really lost on this one and would greatly appreciate any help!

2. Nov 11, 2009

### mgb_phys

Do you know what that equation means?

What T are you looking for?

3. Nov 11, 2009

### G-reg

Yes, the equation is telling you the period of the motion squared.

But we're not looking for the period or the mass or the radius..we're looking for the altitude but i can't find an equation in my text book that would help me out with that.

4. Nov 11, 2009

### mgb_phys

What's 'r' in that equation?

5. Nov 11, 2009

### G-reg

Are you hinting that the period is the same thing as the altitude?

6. Nov 11, 2009

### ideasrule

T is the time it takes for the satellite to make one orbit. r isn't the radius; it's the distance from the center of Mercury to the satellite. So to use that equation, you'll need to find Mercury's rotation period.

7. Nov 12, 2009

### G-reg

Oh ok, I think I understand now. So we are looking for 'r'.
And we're using the "law of periods" to solve for 'r'.
Correct?

8. Nov 12, 2009

### G-reg

Well I guess this guess is wrong because I got a wrong answer..
So where should I go from here?

9. Nov 12, 2009

### buffordboy23

What is the definition of altitude? In other words, where is the reference point for which we determine the altitude?

10. Nov 12, 2009

### G-reg

So it should be..

GM * (T/2pi)^2 = x