# Period orbit around earth

1. Dec 9, 2008

### gemini2904

1. The problem statement, all variables and given/known data

A space station is placed in an orbit of radius 30 000km. What is the period of the station's orbit around the Earth?

(I previously worked out the period of rotation of the space station as 22 secs but I didn't think it was relevant to this part of the question besides it's probably wrong anyway!!!! For that question the wheel shaped station, with radius 20m, rotating around a central hub. astronaunt standing on rim head facing forward , feels force on feet, to stimulate 1/6 of the magnitude of gravity on earth)???

2. Relevant equations

I was looking to use T = 2pi$$\sqrt{}Rearth/g$$

but Instead of Rearth do I need to use the radius 30 000? Or have I got the wrong equation again!?!?

Many thanks,

Claire

3. The attempt at a solution

T = 84mins opps not 5!!! I don't understand how the radius of 30000 fits in, should I take it as distance?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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

### Lyuokdea

What do you know about how force from perfect spheres can be modeled? What is the distance between the space shuttle and the "origin" of the gravitational force?

~Lyuokdea

3. Dec 9, 2008

### gemini2904

I guess the distance is the 30 000km but I still don't know what to do with it? Help!!!

4. Dec 9, 2008

### LowlyPion

The 30,000 km is the radius.

5. Dec 9, 2008

### Lyuokdea

Good point, I'm reading too fast...

~Lyuokdea

6. Dec 9, 2008

### LowlyPion

But the formula you are trying to use is not exactly correct.

Kepplers 3rd Law is:

T = 2π√(r³/GM )

7. Dec 11, 2008

### gemini2904

Hi,

Thanks for replying. Unfortunately were not given that equation at this stage and so the answer needs to be worked out some other way, which is the confusing bit!

all were given is the above equation and for kepplers laws K = T^2/a^2 which I don't think is any help??

8. Dec 11, 2008

### LowlyPion

You can derive it easily enough.

You know centripetal force. You know gravitational force.

They balance for orbit.

GM*m/r² = m*v²/r = m*ω²r

ω² = (2π/T)² = GM/r³

T² = (2π)² *r³ /GM

Anything look familiar?

9. Dec 11, 2008

### gemini2904

Hi,

Thanks very much for replying. I kind of get it but to ask a really stupid question what do I use for GM. I only have gravity and radius of the earth? I'm not given the mass or told to look it up???

Cheers,
Claire

10. Dec 11, 2008

### LowlyPion

11. Dec 18, 2008

### gemini2904

Hi,

Thanks very much for all your help and for including the link, I get it now, finally!!!

Sorry for the delay in replying, I've been really busy lately with Christmas preparations and unexpected visitors popping in to spread their Christmas cheer - Joy! Bah humbug!!!

Best wishes,
Claire