# Kepler's Laws and orbiting satellite

## Homework Statement

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

## Homework Equations

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

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

## 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!

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mgb_phys
Homework Helper
Do you know what that equation means?

What T are you looking for?

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.

mgb_phys
Homework Helper
What's 'r' in that equation?

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

ideasrule
Homework Helper
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.

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?

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

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

So it should be..

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