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Density of a planet using orbit of a satellite

  1. Apr 16, 2008 #1
    [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 [tex]\rho=\frac{3\pi}{GT^2}(1+\frac{h}{R})^3[/tex]

    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 [tex]T^2=(\frac{4\pi^2}{GM})(R+h)^3[/tex]. I used R=H to account for the radius plus the altitude of the satellite. I do some rearranging and get [tex]M=\frac{4\pi^2(R+h)^3}{T^2G}[/tex]

    I then use [tex]V=\frac{4\piR^2}{3}[/tex]. When I place M over V, I get that the density is equal to [tex]\frac{3\pi}{GT^2}(R+h)^3[/tex]
    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. jcsd
  3. Apr 16, 2008 #2


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    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 [itex]\frac{4}{3}\pi r^3[/itex], but divided the right side by only [itex]\frac{4}{3}\pi[/itex].
  4. Apr 16, 2008 #3
    Yes, I see what you mean. But now I get that the density is [tex]\frac{3\pi}{GT^2R^2}(R+h)^3[/tex]

    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 ?
  5. Apr 16, 2008 #4


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    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.
  6. Apr 16, 2008 #5
    [tex](\frac{R+h}{R})^3[/tex] can be simplified to [tex](1+\frac{h}{R})^3[/tex] correct? If so, I've got the right answer correctly. Thanks so much!!
  7. Apr 16, 2008 #6


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    Yes, that looks perfect.
  8. Apr 16, 2008 #7
    Thanks again!
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