1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Kepler's 3rd Law

  1. May 2, 2008 #1
    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.
  2. jcsd
  3. May 2, 2008 #2


    User Avatar
    Homework Helper

    [tex]\rho = \frac{M}{V}[/tex]

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


    [tex] \rho = \frac{M}{\frac{4}{3} \pi r^3} = \frac{3M}{4\pi r^3}[/tex]
  4. May 2, 2008 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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?
  5. May 2, 2008 #4
    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?
  6. May 2, 2008 #5


    User Avatar
    Science Advisor
    Homework Helper

    3/(4pi)<1. Why did your answer increase?
  7. May 2, 2008 #6


    User Avatar
    Homework Helper

    [tex]T^2=\frac{4\pi r^3}{GM}[/tex]

    [tex]\frac{1}{T^2}=\frac{GM}{4\pi r^3}[/tex]

    [tex]\frac{1}{T^2}=\frac{G}{3} \frac{3M}{4\pi r^3}[/tex]

    [tex]\frac{1}{T^2}=\frac{G}{3} \rho[/tex]

    and then you got [itex]\rho [/itex] to be that value? If so and you calculated correctly...that should be the answer.
  8. May 2, 2008 #7
    I have no idea why it increased! I guess I must be calculator retarded. I've got it now, thanks!!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook