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Homework Help: Orbiting of Planet

  1. Dec 19, 2008 #1
    1. The problem statement, all variables and given/known data
    When a moon orbits a planet, it can be shown that the period of the moon's orbit depends only on the mass of the planet and the radius of the moon's orbit.
    a) Draw and label a diagram to illustrate the variables.
    b) Derive a formula for the period of a moon's orbit. The only variables in the final answer should be the mass of the planet and the radius of the orbit. All other values should be constants.


    2. Relevant equations
    Fg= Gm1m2/r^2
    g=Gm1/r^2


    3. The attempt at a solution
    Okay, I don't really understand how to make formula's to get an answer like this. I figure that the variables in the formula would be the period of orbit, mass and radius. The constants would probably be pi and G.

    Could someone give me a hint at how to do this?
     
  2. jcsd
  3. Dec 19, 2008 #2

    LowlyPion

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    Homework Helper

    If it's orbiting then you know that centrifugal is balanced by gravitational attraction.

    You also know that v = ωr = 2πf*r = 2πr/T
     
  4. Dec 19, 2008 #3
    I don't get it still. How do I derive a formula out of that?
     
  5. Dec 19, 2008 #4
    F=ma right? What else does F equal? What is the formula for Force of gravity?

    Set them equal to each other and lets see what happens.
     
  6. Dec 19, 2008 #5
    Fg= Gm1m2/r^2...
     
  7. Dec 19, 2008 #6
    Yep... and? Like I said set it equal to ma and let's see what happens.

    [tex]m_{moon}a=\frac{GM_{planet}m_{moon}}{r^2}[/tex]

    now what?
     
  8. Dec 19, 2008 #7
    So I cancel out the two moons and get a=GM/r^2.
     
  9. Dec 19, 2008 #8
    Yes. And when a "particle" is experiencing radial acceleration, how else can a be written?
     
  10. Dec 19, 2008 #9
    v^2/r=GM/r^2 And then if I cancel out the 2 r's I would get v^2=Gm/r...

    Would I use v=d/t here? If so, then for (d/t)^2=GM/r^2 what would cancel out the d?
     
  11. Dec 19, 2008 #10
    You are very close, but not exactly. Instead of distance we should use something else.... what is the distance called when we are talking about a circle?
     
  12. Dec 19, 2008 #11
    Circumference, which is 2pir?

    Then (2pir/t)^2=GM/r^2
    Which would end up as

    2pir^3/Gm=T^2

    Would that be right?
     
  13. Dec 19, 2008 #12
    [tex](\frac{2\pi r}{T})^2=\frac{2^2*\pi^2*r^2}{T^2}=\frac{GM}{r}[/tex]

    Don't forget to square everything in the parentheses. And then you've got it.
     
  14. Dec 19, 2008 #13
    Thank you sooo much for going through this with me C: I really appreciate it
     
  15. Dec 20, 2008 #14
    No probs :wink:
     
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