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Earth and moon where does gravity cancel out

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    where does the gravity cancel out between the earth and moon?


    2. Relevant equations
    s=3.84405x10^8m
    earth m = 5.98x10^24kg
    moon m =7.35x10^22kg
    G = 6.67x10^-11N m^2/kg^2
    mass of object between = 1kg for simplicity
    F = G x m / r^2
    r = ?

    F earth = F moon
    G x earth / r^2 = G x moon / (s-r)^2
    3. The attempt at a solution
    I get so far with that and ive ended up with:
    earth = (moon x r^2) / (s-r)^2

    not sure if its right and how do i isolate r?
     
    Last edited: Oct 23, 2012
  2. jcsd
  3. Oct 23, 2012 #2
    Let's call the distance away from Earth r, and the total distance between the Earth and moon R.

    A mass in between the Earth and moon will experience a gravitational force from both.

    Fearth = GMem/r2

    Fmoon = GMmm/(R-r)^2

    For the net force to be zero the two above forces must equal each other.
     
  4. Oct 23, 2012 #3
    i understand that but i dont know how to solve for r (R = s in my part)
     
  5. Oct 23, 2012 #4
    You need to use simple algebra to isolate r.

    start with multiplying both sides by the rights side denominator. then you need to multiply out Earth and divide everything by r2. By doing this you should be able to get a single r2 term. then just isolate r.
     
  6. Oct 23, 2012 #5
    I worked it out, and it's definitely doable but the algebra is a little more intense than it probably needs to be. I thought of an easier way to do it. Let R be the total distance between the moon and the earth.

    The distance away you are from the earth must be some percentage of total R. So if you're halfway of the total distance then .5R , if you're a quarter then .25R, etc. So let's say you're distance from earth is aR where a is some number <1

    The distance from the moon must then be 1-a

    So we can say this.

    GM/aR = Gm/(1-a)R here I called M the mass of earth and m the mass of the moon.

    Now you can just solve for a, that is significantly easier. Multiply a by the total distance and that's how far from earth you need to be (well the center of the earth)
     
  7. Oct 23, 2012 #6
    thats a good way to do it! thanks for that.

    do post the intense algebra? :)
     
  8. Oct 23, 2012 #7
    You cancel G and cross multiply to get

    (R-r)2/r2 = m/M

    Foil out R-r

    (R2-2Rr+r2)/r2 = m/M

    You cross multiply the m/M to bring it to the left and r2 to bring it to the right.

    Distribute the new M/m term on the left. Now subtract r2 so it's back on the left. Give r2 a common denominator wit the rest of the terms. Now you can factor out r2

    The end result will be a constant plus r times some constant plus r2 times some constant = 0

    So you can plug that into the quadratic formula.
     
  9. Oct 23, 2012 #8

    haruspex

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    Oops.
     
    Last edited: Oct 23, 2012
  10. Oct 23, 2012 #9

    I'm not sure what you mean? What does being inside the earth have to do with this? You're in space located in between the moon and the earth.
     
  11. Oct 23, 2012 #10

    haruspex

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    Oops, was thinking of a different question entirely. Sorry for the noise.
     
  12. Oct 23, 2012 #11
    So it is impossible without quadratic formula.. guess I have some reading to do thanks for all your help!
     
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