Center of Mass between the Earth and Sun

  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:

  2. jcsd
  3. 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. Doc Al

    Staff: Mentor


    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. (M1X1+M2X2....MnXn)/(M1+M2)

    I'm guessing I made it too complicated?
  6. Doc Al

    Staff: Mentor

    That's all there is to it.
  7. So are all the Ms unknown, or do I use the 84.5 as a mass
  8. Doc Al

    Staff: Mentor

    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. 84.5m?
  10. Doc Al

    Staff: Mentor

    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. Gokul43201

    Gokul43201 11,044
    Staff Emeritus
    Science Advisor
    Gold Member

    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. so my equation looks like this

  13. Doc Al

    Staff: Mentor

    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. Oh, now I'm really confused as to how to write this equation.
  15. Doc Al

    Staff: Mentor

    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. dynamicsolo

    dynamicsolo 1,656
    Homework Helper

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


    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. Doc Al

    Staff: Mentor

    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. so the practice problems

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

    So this is my new idea:

  20. Doc Al

    Staff: Mentor

    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. 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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