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Question about Gravity and ratio of attraction

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Gerald stands on the roof looking up at the full moon, which has a mass of 7.39 x 10^22 kg. At this moment, Gerald is 3.77 x 10^8 m from the center of the moon. If the mass of the earth is 5.96 x 10^24 kg, and the radius of the earth is 6.39 x 10^6 m, what is the ratio of (Gerald's attraction to the moon) to (Gerald's attraction to the earth)?


    2. Relevant equations
    F=G[itex]\frac{M1*M2}{r^2}[/itex]

    3. The attempt at a solution
    So the first thing I know I need to solve for is the radius of the moon. I wasn't sure how to go about this so I decided that R[itex]_{Gerald from moon}[/itex] - R[itex]_{Earth}[/itex] = R[itex]_{Moon}[/itex] I feel like that's wrong, but I got an answer of 3.71x10[itex]^{8}[/itex].

    Next I tried to use the gravity formula (posted above) to find the "F" values for the moon and the earth. I got:
    F[itex]_{moon}[/itex] = 7.927x10[itex]^{28}[/itex] and
    F[itex]_{earth}[/itex] = 4.597x10[itex]^{30}[/itex]

    Then I just divided the moon value by the earth value to get a ratio. The answer wasn't correct, and I feel like it has to do with the first part. Problem is, I have no idea where to go. Some help would be appreciated.
     
  2. jcsd
  3. Oct 21, 2012 #2
    There is no reason to do this.
     
  4. Oct 21, 2012 #3
    How far is Gerald from the center of the moon? How far is Gerald from the center of the Earth?
     
  5. Oct 21, 2012 #4
    Well, now that you mention it, it seems like I was finding the ratio of attraction between the Earth and the moon...

    Those values were given, so I'm assuming that's what I'm supposed to be using. Alright, I think I get this. I shouldn't need to worry about Gerald's mass right?
     
  6. Oct 21, 2012 #5
    It would cancel, yes.
     
  7. Oct 21, 2012 #6
    After taking another stab at the problem, I got another wrong answer. What I was trying to do was find Fm and Fe (through the formula posted). I used the distance that Gerald was from the moon (I used that as the radius value) to find Fm, then I found Fe and divided Fm/Fe. Could you explain to me what I'm doing wrong?
     
  8. Oct 21, 2012 #7
    Would you mind posting the solution you have in terms of variables (i.e. no numbers)?
     
  9. Oct 21, 2012 #8
    Sure. I have:

    Ratio = Fm/Fe

    Thus:
    Ratio = G*M[itex]_{M}[/itex]*M[itex]_{E}[/itex]*[itex]\frac{1}{rm^2}[/itex]/G*M[itex]_{M}[/itex]*M[itex]_{E}[/itex]*[itex]\frac{1}{re^2}[/itex].
    So the G and M's cancel out and I'm left with [itex]\frac{1}{rm^2}[/itex]/[itex]\frac{1}{re^2}[/itex]

    That equals [itex]\frac{re^2}{rm^2}[/itex]

    Where rm = Radius of Gerald from the moon
    and re = Radius of the Earth given in the problem.
     
  10. Oct 21, 2012 #9
    You are using the wrong values for the masses. Think about how you would set up the problem if the question simple asked what is Gerald's attraction to the Earth.
     
  11. Oct 21, 2012 #10
    There's your problem. You have to recognize that you're finding two separate forces: one between Earth and Gerald, and one between the Moon and Gerald. Thus, the masses need to be Earth's and Gerald's, and the Moon's and Gerald's. The masses of the Earth and the Moon do not cancel out of the ratio; only Gerald's mass will cancel.
     
  12. Oct 21, 2012 #11
    You should put Gerald's mass in and cancel it out yourself. I think that it will help you to work through the problem without making mistakes if you do it that way.
     
  13. Oct 21, 2012 #12
    Oh ok! So I redid the problem and I'm left with M[itex]_{m}[/itex]*[itex]\frac{1}{rm^2}[/itex]/M[itex]_{e}[/itex]*[itex]\frac{1}{re^2}[/itex] (where rm is Gerald's distance from the moon). I'm assuming that's my real answer.
     
  14. Oct 21, 2012 #13
    Looks kosher, yes.
     
  15. Oct 21, 2012 #14

    BruceW

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    Homework Helper

    yep, nice work
     
  16. Oct 21, 2012 #15
    Alrighty then, thank you all!
     
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