# Planets and satellites, law of periods circular orbit.

1. Jul 19, 2015

### J-dizzal

1. The problem statement, all variables and given/known data
A 15 kg satellite has a circular orbit with a period of 4.4 h and a radius of 3.5 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 1.2 m/s2, what is the radius of the planet?

2. Relevant equations

3. The attempt at a solution

I think im on the right track, but im stuck finding the radius of the planet. in my equation at the bottom for radius what would i use for M and m?

2. Jul 19, 2015

### Nathanael

In your equation $r=\sqrt{\frac{GMm}{F_g}}$ r is the distance between the two objects, not the radius of the planet.

Try to use the centripetal acceleration to find the mass of the planet (only 1 mass will cause this orbit) and then try to find the radius (with the mass known, only 1 radius will cause that value of surface gravity).

Last edited: Jul 19, 2015
3. Jul 19, 2015

### TSny

Considering Newton's 2nd law, what does Fg /m represent?

4. Jul 19, 2015

### J-dizzal

would it work if M is the mass of the planet and m is a mass on the surface of the planet? the distance between them would be about the radius of the planet?

acceleration. But in my equation for r; Fg is the force of acceleration on the surface of the planet. would that work with the given a=1.2m/s2

5. Jul 19, 2015

### Nathanael

Oh my bad, yes then it would work. You just need to find the mass of the planet first.

6. Jul 19, 2015

### J-dizzal

Does my equation for M look ok, toward the top of my page M=946496.4kg?
If its ok, then would any value for m work in my equation for r?

7. Jul 19, 2015

### Nathanael

Sorry for ignoring your initial work, apparently I'm not so good at skimming pictures

The equation is right but you didn't cube the radius when you solved it.

8. Jul 19, 2015

### J-dizzal

In my equation for r, im letting Fg=a. where a=1.2m/s/s is a given in the problem statement. Im guessing this is wrong because its the force on the satellite at distance r from the surface.

9. Jul 19, 2015

### Nathanael

a is 1.2 m/s/s, not newtons. Consider an object on the surface of mass m. If it accelerates at 1.2m/s/s, then what must the force on that object (of mass m) be?

10. Jul 19, 2015

### TSny

Fg = GMm/r2 gives the force of gravity on m as a function of r. Using this, you can derive an expression for FG /m as a function of r. Thus, you have an expression for the acceleration due to gravity as a function of r. Apply this for a point at the surface of the planet.

11. Jul 19, 2015

### J-dizzal

F=(15kg)(1.2m/s/s)=18N

Fg/M = Gm/r2 so this would be the acceleration of M?

12. Jul 20, 2015

### TSny

Well, yes that would give the acceleration of the planet (M) due to the attraction of the satellite (m). But that would be extremely small and it is not of relevance to this problem.

However, Fg /m gives the acceleration of the satellite due to the attraction of the planet. This is what you would call the "acceleration due to gravity of the planet". You are given the acceleration due to gravity of the planet at the surface of the planet.

So you want to find an expression for Fg /m rather than Fg /M. Then apply the expression to a point near the surface of the planet.

13. Jul 20, 2015

### J-dizzal

im getting R= sqrt(GM/a) =1.337x106 same answer i got before...

14. Jul 20, 2015

### TSny

R = sqrt(GM/a) is correct. What did you get for the mass of the planet, M, after correcting for cubing r as pointed out by Nathanael in post #7?

15. Jul 20, 2015

### J-dizzal

I got 3.219x1022kg

16. Jul 20, 2015

### J-dizzal

ok i got it correct now thanks again.