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Gravitational attraction problem

  1. May 2, 2007 #1

    A particle is projected vertically upward from the earth's surface with initial speed Vo. Prove that the maximun height H reached above the earth's surface is H= Vo^2R/(2gR-Vo^2)

    3. The attempt at a solution
    R is the earth's radius. If the gravitational attraction at the earth's surface is mg, then the attraction at some height r above the surface will be mgR^2/(R+r)^2. The attraction can be integrated from r = 0 to r = H to get the total work involved, which must equal the kinetic energy mVo^2/2. HOWever, when we integral from
    r=0 to r=H of mgR^2/(R+r)^2 and set it equal to mVo^2/2,
    H= does not give the value Vo^2R/(2gR-Vo^2) ??
  2. jcsd
  3. May 2, 2007 #2


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    Gold Member

    It's easier to use
    final potential energy = initial kinetic energy.
  4. May 2, 2007 #3
    If we do it that way, it gives us...

    mv^2(1/2) = mgh

    h= v^2*(1/2)/g

    and it still doesnt satisfy the answer :eek:
  5. May 2, 2007 #4


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    Gold Member

    You're assuming the potential is the same at R and R+h.
    But they stand in the ratio R/(R+h)

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