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An object of mass m is lifted from the surface of the Earth to a height equal to n times the Earth's radius R. Using g for the surface gravitational acceleration, what is the increase in its gravitational potential energy?

1. (n/(n + 1)) m g R
2. ((n - 1)/n) m g R
3. (n/(n - 1)) m g R
4. (n + 1/n) m g R

Correct answer: (n/(n + 1)) m g R

Solution

U(r) = -GMm/r. Initial: U_i = -GMm/R. Final at r = (n+1)R: U_f = -GMm/((n+1)R). Change dU = U_f - U_i = -GMm/((n+1)R) + GMm/R = (GMm/R)(1 - 1/(n+1)) = (GMm/R)(n/(n+1)). Since g = GM/R², GMm/R = mgR. So dU = (n/(n+1)) m g R.

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