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Gravitational pull with the earth and moon

  1. Aug 17, 2012 #1
    1. The problem statement, all variables and given/known data

    The mass of the moon is 7.35x10^22 kg. At some point between Earth and the Moon, the force of Earth's gravitational attraction on an object is cancelled by the Moon's force of gravitational attraction. If the distance between the earth and the moon (center to center) is 3.84x10^5 km, calculate where this will occur, relative to the Earth.




    3. The attempt at a solution
    ok now this is from another thread, and i completely understand how she got everyting but the 81! could someone please explain to me how she got the 81.. thts the only part im lost in..

    We assume an object is at the L1 point (the point where the gravitational fields of the two objects cancels out). We will call this object X

    r is the radius between the Earth and the Moon
    We will call
    r1
    the point from the center of the earth to x and
    r2
    the point from x to the center of the moon
    Therefore
    (1)
    r=r1+r2


    Using this assumption of the object X we get..

    Fg=Gm1xr21
    where
    m1
    is the mass of the earth, and x is the mass of object X
    Fg=Gm2xr22
    where
    m2
    is the mass of the moon, and x is the mass of object X

    Since we know that, at the point where the object X is situated, the gravitiational pull from both the Earth and the Moon will be equal, we can equate the above two equations getting the following:
    (2)
    Gm1xr21=Gm2xr22


    Now, we obtain a ratio between
    m1
    and
    m2
    so we can express
    m1
    in terms of
    m2


    m1m2=81
    (Note: I rounded to 81 just to make typing it up here easier...)
    Therefore,
    m1=81m2


    Now, going back to equation (2), G and x will cancel out and we replace
    r1
    with
    r−r2
    (from equation (1)) leaving us with:
    81m2(r−r2)2=m2r22


    m2
    will cancel out:
    81(r−r2)2=1r22


    81=(r−r2)2r22


    i take the square root of both sides, and put the denominator on the left side
    9r2=r−r2


    10r2=r


    r2=r/10


    r2=3.84∗104


    Since
    r1=r−r2


    r1=3.84∗105−3.84∗104


    r1=3.456∗105
     
  2. jcsd
  3. Aug 17, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Alicia,

    your post is very hard to read. Use the x2 button for a square and x2 for an index.

    So distance between the Earth and the object at point L1 is r1, the distance between the Moon and the object is r2, the force from the Earth is Gm1x/r12, the force from the Moon is Gm2x/r22, they are equal, so m1r22=m2r12.


    You can find the mas of Earth in any table about Earth. Google "mass of Earth" if you do not have it on your Physics book. It is 5.9736˙1024 kg. You see that the ratio of the masses of Earth and Moon is about 81.


    ehild
     
  4. Aug 17, 2012 #3

    Thank you very much and sorry for my horrible computer skills!! I'll try that next time !
     
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