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Homework Help: Gravity Question

  1. Mar 14, 2010 #1
    The problem statement, all variables and known data:

    You launch an object into a circular orbit around a spherical, smooth rock, 30 cm in diameter. The rock also has the same mass density as Earth, ρ. How long does it take the object to orbit the rock one time?

    Attempt at a solution:

    I know you should start by finding the mass of the rock, which turns out to be:
    (4/3)pi*r3

    I'm not sure what to do after that. Any help?
     
    Last edited: Mar 14, 2010
  2. jcsd
  3. Mar 14, 2010 #2

    Doc Al

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    Apply Newton's 2nd law to solve for the period. What force acts on the object? What is its acceleration?

    What radius of orbit are you mean to assume? Is the object supposed to be sliding along the surface of the rock?
     
  4. Mar 14, 2010 #3
    Yeah, the problem says the object is in orbit just barely above the surface, so I think you can assume that it's the same as if it were sliding without friction.

    The force that acts on the object is gravity. The acceleration due to gravity is given by the formula -GM/r2, where I can plug in the M from above.

    I'm not sure how to relate that to F=ma.
     
  5. Mar 14, 2010 #4

    Doc Al

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    Good. So we have the radius of the orbit.

    Good.
    That's true, but unhelpful.

    Hint: What kind of motion does the object undergo? Write the generic formula for that sort of acceleration. Plug that into F=ma.
     
  6. Mar 14, 2010 #5
    The object is moving in uniform circular motion. So, F=m*v2/r

    You know r and the force is gravity, but do you know m? Or does that cancel out with the force of gravity formula?

    I assume it does, so then v = [tex]\sqrt{GM/r}[/tex]

    or

    v = [tex]\sqrt{4/3*G*pi*rho*r^2}[/tex] which has me a little worried, since the problem has a note that says we shouldn't have to do any arithmetic at all to solve it.
     
    Last edited: Mar 14, 2010
  7. Mar 14, 2010 #6

    Doc Al

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    Note that you are asked to find the period, so express v in terms of the period.

    It's not obvious to me how you can get a numerical answer without doing any arithmetic. (Perhaps I just need more coffee.) You can certainly express the answer in terms of ρ.
     
  8. Mar 14, 2010 #7

    ehild

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    Compare it with an object orbiting just above the surface of the Earth.

    ehild
     
  9. Mar 14, 2010 #8
    Doc Al: So I could do it with v*omega instead of v2/r

    ehild: I was thinking something like that, but how..?
     
  10. Mar 14, 2010 #9

    ideasrule

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    Don't be discouraged; keep going. How do you get the period from the speed? Remember that this is a circular orbit, so you can easily calculate the distance that the satellite travels. Once you get an expression for the period, you'll realize why you don't need any arithmetic to solve the problem.
     
  11. Mar 14, 2010 #10

    ideasrule

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    What's v*omega? Anyhow, you don't need to bring omega into the picture.
     
  12. Mar 14, 2010 #11

    Doc Al

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    Don't bother using ω²r, if that's what you're thinking. Just replace v with distance per period.
     
  13. Mar 14, 2010 #12

    ehild

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    Just write out the equations acceleration due to gravity = centripetal acceleration , both for Earth and for the rock. Find out how the angular velocities ω are related. Do not forget that the densities are equal. And you know the minimum speed of a satellite which can orbit around the Earth .


    ehild
     
    Last edited: Mar 14, 2010
  14. Mar 14, 2010 #13
    Since it's circular, the period is T = 2*pi*r/v, but that is not making it immediately obvious why I shouldn't have to do arithmetic.
     
  15. Mar 14, 2010 #14

    Doc Al

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    That's certainly true, but you still need some arithmetic to figure out the period.
     
  16. Mar 14, 2010 #15

    Doc Al

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    Good. Express v in terms of T and use that in your early equation. Then you can solve for the period.
    Unless they want you to express the answer symbolically, you'll need a little arithmetic.
     
  17. Mar 14, 2010 #16
    You are almost there. Carry out the isolation for T in general terms and you will observe that the radius term cancels out. Thus it really doesn't matter how big the rock is at all. If the density of the rock is the same as the earth's, then the orbital period at the surface would also be the same. If you knew this orbital period for the earth, then you would not have to do any arithmetic (although I doubt anybody would know).
     
  18. Mar 14, 2010 #17
    Actually, I do know the orbital period on earth's surface, since it is in the packet this question is in. It is given as 5069 seconds. But, the problem is that the answer I'm getting is not 5069.

    So, after the algebra, you get:

    T^2 = (2pi)/(4/3*G*rho)

    solving for T yields: 11709.99

    Did I do something wrong?
     
  19. Mar 14, 2010 #18

    Doc Al

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    Ah, so that's why you don't need any arithmetic! :wink:

    That's not quite right. Redo your algebra.

    But the main thing to realize is that the period is independent of the radius.
     
  20. Mar 14, 2010 #19
    Hmm, I am supposed to be solving:

    GMm/r2 = m * v2/r

    plugging in and simplifying to:

    G(4/3*π*r*ρ)= (2πr/T)2/r

    right?

    Because that simplifies all the way to: T^2 = (3pi)/(G*ρ) which doesn't give 5069.
     
  21. Mar 14, 2010 #20

    Doc Al

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    Good.
    What did you use for G and ρ? (Show how you calculated ρ.)
     
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