Can this gravity problem be solved without using the Earth's mass

[hndalama]

Homework Statement

A satellite orbits the Earth in a geosynchronous orbit around the equator, meaning that its period is 24 hours and it stays above the same location on Earth at all times. (G = 6.67 x 10⁻¹¹ Nm²/kg².) What is the radius of its orbit?

Homework Equations

GM/r² = v²/r = w²r

The Attempt at a Solution

I can solve this if the mass of Earth is given but since it isn't I'd like to know if there is a way to solve this without using Earth's mass

 

[kuruman]

Sometimes you may have to look things up.

 

[lychette]

Homework Statement

A satellite orbits the Earth in a geosynchronous orbit around the equator, meaning that its period is 24 hours and it stays above the same location on Earth at all times. (G = 6.67 x 10⁻¹¹ Nm²/kg².) What is the radius of its orbit?

Homework Equations

GM/r² = v²/r = w²r

The Attempt at a Solution

I can solve this if the mass of Earth is given but since it isn't I'd like to know if there is a way to solve this without using Earth's mass

You know that g = GM/r_e², rearrange this to give you the expression for M then go from there

 

[kuruman]

You know that g = GM/r_e², rearrange this to give you the expression for M then go from there

Sure, but then r_e needs to be looked up.

 

[lychette]

This is true enough and in general I think that r_e is given in data tables rather than M.(certainly in A level reference tables)
Also the combination GM usually crops up on both side of gravitational equations and therefore cancel out
In this example for the Earth g_E = GM_E/r_E² and for the satellite g_s = GM_E/r_s²
This gives g_Er_E² = g_sr_s²
knowing that g_s = v²/r_s and further substitution enables r_s to be calculated.
If r_e is not given then an answer can still be obtained in terms of r_e

 

[kuruman]

If re is not given then an answer can still be obtained in terms of r_e

This is, of course, correct. Perhaps the problem should have asked

What is the radius of its orbit as a fraction of the Earth's radius[/color]?

to make things clear.