How much additional energy is needed to fire a 7.2 × 10^2 kg weather monitor on the earth’s surface to a height of 120 km above the earth?
(rE = 6.38 × 10^6 m, ME = 5.98 × 10^24 kg) Assume that the weather monitor rises to that height, stops, and falls back to earth.

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asked by grey
  1. PE = - G m Me/R
    where G = 6.67 * 10^-11
    Look at change in potential energy at R = 6.38*10^6
    and at 6.38*10^6 + .120 *10^6 meters

    I bet it would not be far off to use m g h where g = 9.81 m/s^2 :) because 120 km is not far compared to Rearth

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    posted by Damon

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