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Potential energy of a ball falling

  1. Dec 26, 2006 #1
    need help plz urgent

    1. The problem statement, all variables and given/known data
    An object is released from rest at an altitude h above the surface of the Earth. (a) Show that its speed at a distance r from the Earth’s center, where RE < r < RE + h, is given by
    v = sqrt(2GME (1/r -1/ (RE + h) )

    (b) Assume the release altitude is 500 km. perform the integral:
    (Delta) t = (integral from i to f) dt = - (integral from i to f) dr/v

    to find the time of fall as the object moves from the release point to the Earth’s surface. The negative sign appears because the object is moving opposite to the radial direction, so its speed is v = -dr / dt. Perform the integral numerically.

    3. The attempt at a solution

    The Potential Energy at a disance h above earth's surface (and therefore h+Re above earth's center) = -GMm/(Re+h).

    Potential at a distance r from center = -GMm/r
    Change in PE = (-GMm/(Re+h)) - -(GMm/r) = GMm/r - GMm/(Re+h)

    change in PE = Gain in KE = 1/2mv2

    Solving for v

    getting confused....../
  2. jcsd
  3. Dec 26, 2006 #2


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    Staff Emeritus
    Science Advisor

    I presume here, ME is mass of the earth

    Your method is correct. Where are you having problems solving for v? Set the KE equal to the PE (note that M shuld be ME in the notation in the question) and you more or less have the answer!

    For the second bit, plug v into the integral.
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