Kepler's 3rd Law

[Reverie29]

1. The problem statement, all variables and given/known data
A satellite is in a circular orbit very close to the surface of a spherical planet. The period of the orbit is 2.49 hours.
What is density of the planet? Assume that the planet has a uniform density.


2. Relevant equations

T^2 = (4(pi)^2*r^3) / GM

3. The attempt at a solution
Okay, so I coverted the period into seconds and got 8964 seconds.
Then I rearranged the equation to get
M/r^3 = 4(pi)^2 / GT^2, assuming that M/r^3 would get me density.
So then according to that Density = 4(pi)^2 / (6.67e-11 N*m^2/kg^2)(8964 s)^2 which gives 7366 kg/m^3 which is not correct.

Or am I missing something about density? Density is mass divided by area. Should I be finding a radius to find the area and then find the mass somehow... I don't know. I'm confused on what to do.

[rock.freak667]

\rho = \frac{M}{V}

Assuming the planet is a perfect sphere, V=\frac{4}{3} \pi r^3

So

\rho = \frac{M}{\frac{4}{3} \pi r^3} = \frac{3M}{4\pi r^3}

[Janus]

Your problem is in assuming that M/r^2 gives you density.

Density is mass divided by volume. So what is the formula for the volume of a sphere?

[Reverie29]

Okay.
The density of a sphere is = 3M / 4(pi)r^3. And I have already solved for M/r^3. I tried multiplying by 3 and dividing by 4pi, but still got an incorrect answer. I got 17,356 kg/m^3.

Should I be looking at another equation?

[Dick]

3/(4pi)<1. Why did your answer increase?

[rock.freak667]

T^2=\frac{4\pi r^3}{GM}

\frac{1}{T^2}=\frac{GM}{4\pi r^3}

\frac{1}{T^2}=\frac{G}{3} \frac{3M}{4\pi r^3}

\frac{1}{T^2}=\frac{G}{3} \rho

and then you got \rho to be that value? If so and you calculated correctly...that should be the answer.

[Reverie29]

I have no idea why it increased! I guess I must be calculator retarded. I've got it now, thanks!!!