Calculating the Period of an Orbit around Earth

[gemini2904]

Homework Statement

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)?

Homework 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

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?

 

[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

 

[gemini2904]

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

 

[LowlyPion]

The 30,000 km is the radius.

If they had meant in addition to the radius it would have said above the surface.

 

[Lyuokdea]

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

~Lyuokdea

 

[LowlyPion]

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

Kepplers 3rd Law is:T = 2π√(r³/GM )

 

[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??

 

[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?

 

[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

 

[LowlyPion]

There is a Standard Gravitational Parameter for Earth that is the product of G*M because it comes up so often.

Link:
http://en.wikipedia.org/wiki/Standard_gravitational_parameter

μ = G*M

The units of the standard gravitational parameter are km³s^-2

Earth ≈ 400,000 km³s^-2

 

[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 into spread their Christmas cheer - Joy! Bah humbug!

Best wishes,
Claire