Initial potential energy of the body at the surface of the earth is given by:
Ui=−ReGMm
Final potential energy of the body at a height h=2Re from the surface of the earth is:
Uf=−Re+hGMm=−Re+2ReGMm=−3ReGMm
The increase in potential energy is:
ΔU=Uf−Ui=−3ReGMm−(−ReGMm)
ΔU=ReGMm(1−31)=3Re2GMm
Acceleration due to gravity at the surface of the earth is g=Re2GM, which gives GM=gRe2.
Substituting this value, we get:
ΔU=3Re2(gRe2)m=32mgRe
Answer: 32mgRe