# Orbiting of Planet

1. Dec 19, 2008

### ahrog

1. The problem statement, all variables and given/known data
When a moon orbits a planet, it can be shown that the period of the moon's orbit depends only on the mass of the planet and the radius of the moon's orbit.
a) Draw and label a diagram to illustrate the variables.
b) Derive a formula for the period of a moon's orbit. The only variables in the final answer should be the mass of the planet and the radius of the orbit. All other values should be constants.

2. Relevant equations
Fg= Gm1m2/r^2
g=Gm1/r^2

3. The attempt at a solution
Okay, I don't really understand how to make formula's to get an answer like this. I figure that the variables in the formula would be the period of orbit, mass and radius. The constants would probably be pi and G.

Could someone give me a hint at how to do this?

2. Dec 19, 2008

### LowlyPion

If it's orbiting then you know that centrifugal is balanced by gravitational attraction.

You also know that v = ωr = 2πf*r = 2πr/T

3. Dec 19, 2008

### ahrog

I don't get it still. How do I derive a formula out of that?

4. Dec 19, 2008

F=ma right? What else does F equal? What is the formula for Force of gravity?

Set them equal to each other and lets see what happens.

5. Dec 19, 2008

### ahrog

Fg= Gm1m2/r^2...

6. Dec 19, 2008

Yep... and? Like I said set it equal to ma and let's see what happens.

$$m_{moon}a=\frac{GM_{planet}m_{moon}}{r^2}$$

now what?

7. Dec 19, 2008

### ahrog

So I cancel out the two moons and get a=GM/r^2.

8. Dec 19, 2008

Yes. And when a "particle" is experiencing radial acceleration, how else can a be written?

9. Dec 19, 2008

### ahrog

v^2/r=GM/r^2 And then if I cancel out the 2 r's I would get v^2=Gm/r...

Would I use v=d/t here? If so, then for (d/t)^2=GM/r^2 what would cancel out the d?

10. Dec 19, 2008

You are very close, but not exactly. Instead of distance we should use something else.... what is the distance called when we are talking about a circle?

11. Dec 19, 2008

### ahrog

Circumference, which is 2pir?

Then (2pir/t)^2=GM/r^2
Which would end up as

2pir^3/Gm=T^2

Would that be right?

12. Dec 19, 2008

$$(\frac{2\pi r}{T})^2=\frac{2^2*\pi^2*r^2}{T^2}=\frac{GM}{r}$$

Don't forget to square everything in the parentheses. And then you've got it.

13. Dec 19, 2008

### ahrog

Thank you sooo much for going through this with me C: I really appreciate it

14. Dec 20, 2008