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Center of Mass between the Earth and Sun

  1. Oct 18, 2007 #1
    The ratio of the mass of the earth to the mass of the moon is 84.5. Assume that the radius of the earth is about 6578.0 km and that the distance between the center of the earth and the moon is 385815.0 km. Determine the distance of the center of mass of the earth-moon system from the center of the earth.

    I was about to start this and I had a few questions:

    Can I just assumed that it is 385815 km is the distance from the center of the earth to the center of the moon?

    If so, what do I do about the ratio? Would my equation be:

    (6578.1*1+385815*1/84.5)/84.5
     
  2. jcsd
  3. Oct 18, 2007 #2
    Solve as follows : i) Masses of earth and moon are assumed to be concentrated at the centres of the respective bodies. Hence, only the centre to centre distance between the two is relevent (radius of earth is given just to confuse). ii) Assume the coordinate system as a striaght line with the earth situated at the origin. iii) Apply formula for determination of the coordinate of the centre of mass. [As the masses of earth and moon are not given, only their ratio is given, modify the formula suitably so as to have in it the mass ratio instead of the actual masses.]
     
  4. Oct 18, 2007 #3

    Doc Al

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    Yes.

    You'll need it to compute the center of mass. (You don't need the actual masses.)

    What's the generic definition of center of mass between two masses?
     
  5. Oct 18, 2007 #4
    (M1X1+M2X2....MnXn)/(M1+M2)

    I'm guessing I made it too complicated?
     
  6. Oct 18, 2007 #5

    Doc Al

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    That's all there is to it.
     
  7. Oct 18, 2007 #6
    So are all the Ms unknown, or do I use the 84.5 as a mass
     
  8. Oct 18, 2007 #7

    Doc Al

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    All that matters is their ratio, so try this: If you call the mass of the moon "M", what's the mass of the earth?
     
  9. Oct 18, 2007 #8
  10. Oct 18, 2007 #9

    Doc Al

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    Right!
    Mass of moon = M
    Mass of earth = 84.5M

    Now just crank out the center of mass. Use the center of the earth as the origin.
     
  11. Oct 18, 2007 #10

    Gokul43201

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    Minor correction:

    If you have a system of n masses, then the X(CoM) = (M1X1+M2X2+...+MnXn)/(M1+M2+...+MnXn)

    For a 2-body system, this reduces to X(CoM) = (M1X1+M2X2)/(M1+M2)
     
  12. Oct 19, 2007 #11
    so my equation looks like this

    (6578.1*85M+385815*M)/(84.5M+M)
     
  13. Oct 19, 2007 #12

    Doc Al

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    Two problems: (1) The mass of the earth is 84.5M, not 85M. (2) The mass of the earth is centered at the earth's center, not at its radius. (I have no idea why you are given the earth's radius in this problem.)
     
  14. Oct 19, 2007 #13
    Oh, now I'm really confused as to how to write this equation.
     
  15. Oct 19, 2007 #14

    Doc Al

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    It's easier than you think. Since we are measuring distance from the center of the earth, what distance should the earth's mass have in your equation?
     
  16. Oct 19, 2007 #15

    dynamicsolo

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    You don't need it to solve the problem. However, it is of interest to compare the Earth's radius to your answer...
     
  17. Oct 19, 2007 #16
    The distance from the earth's center to the moon?

    (385815*85M+385815-6578*M)/(84.5M+M)

    I'm running a bit low on ideas. The only example we were given on how to solve the five problems he gave us was a see-saw.
     
  18. Oct 19, 2007 #17

    Doc Al

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    Do this practice problem: Find the center of mass of two equal masses that are 10 m apart. Measure distances from the center of one of the masses. Use the same equation.

    (This might be easier to see, since I presume you know what the answer must be. Once you see how easy it is, you'll probably smack yourself.)
     
  19. Oct 19, 2007 #18
    so the practice problems

    (M(10)+M(0))/2M =5

    So this is my new idea:

    (385815*85M+M(0))/(84.5M+M)
     
  20. Oct 19, 2007 #19

    Doc Al

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    Exactly. One of the masses is at the zero point.
    That would be right (almost) if we were measuring from the moon. But we're measuring from the earth.

    The general equation (as you know) is:

    [tex](M_1 X_1 + M_2 X_2)/(M_1 + M_2)[/tex]

    Where M_1 is the mass of object 1 and X_1 is its coordinate. Since we want to measure things from the earth, we put the earth at X = 0. And the moon will be at X = 385,815 km. So:

    [(84.5M)(0) + (M)(385,815)]/(84.5M + M)

    You finish it up.
     
  21. Oct 19, 2007 #20
    A dog, with a mass of 9.0 kg, is standing on a flatboat so that he is 22.4 m from the shore. He walks 8.4 m on the boat toward the shore and then stops. The boat has a mass of 45.0 kg. Assuming there is no friction between the boat and the water, how far is the dog from the shore now?

    Would I tackle this problem in a similar way? I noticed that since the surface is frictionless, would the boat move in the opposite direction to compensate.
     
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