Density of a planet using orbit of a satellite

1. Apr 16, 2008

XxBollWeevilx

[SOLVED] Density of a planet using orbit of a satellite

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

A satellite is in a circular orbit about a planet of radius R. If the altitude of the satellite is h and its period is T, show that the density of the planet is $$\rho=\frac{3\pi}{GT^2}(1+\frac{h}{R})^3$$

3. The attempt at a solution

I feel that I am mostly doing this problem correctly, but I think I'm leaving out something I need. I am able to derive the entire formula except for the part in the parentheses. I use the fact that the density equals M\V. To find the mass, I solved for M in the equation $$T^2=(\frac{4\pi^2}{GM})(R+h)^3$$. I used R=H to account for the radius plus the altitude of the satellite. I do some rearranging and get $$M=\frac{4\pi^2(R+h)^3}{T^2G}$$

I then use $$V=\frac{4\piR^2}{3}$$. When I place M over V, I get that the density is equal to $$\frac{3\pi}{GT^2}(R+h)^3$$
I have absolutely no clue where the 1 + h/R comes from that I must fine. I must have done something wrong in my deriving or rearranging...could someone point out a mistake? Thanks so much!

2. Apr 16, 2008

alphysicist

XxBollWeevilx,

It looks to me like you've made an algebraic error when dividing M by V. It looks like you divided the left side by $\frac{4}{3}\pi r^3$, but divided the right side by only $\frac{4}{3}\pi$.

3. Apr 16, 2008

XxBollWeevilx

Yes, I see what you mean. But now I get that the density is $$\frac{3\pi}{GT^2R^2}(R+h)^3$$

So I have another R62 in the bottom there...is there a way I can bring that up and unclude it in the R+h ?

4. Apr 16, 2008

alphysicist

I think it needs to be R^3 in the denominator. (because your dividing by (4/3)pi r^3).

Then you can combine the (r+h)^3 in the numerator and the r^3 in the denominator into one fraction that is raised to the third power.

5. Apr 16, 2008

XxBollWeevilx

$$(\frac{R+h}{R})^3$$ can be simplified to $$(1+\frac{h}{R})^3$$ correct? If so, I've got the right answer correctly. Thanks so much!!

6. Apr 16, 2008

alphysicist

Yes, that looks perfect.

7. Apr 16, 2008

XxBollWeevilx

Thanks again!