Solution:
At the surface, \(U_s = -\frac{GMm}{R_e} = -mgR_e\). At a height \(h = R_e\), the distance from the center is \(r = 2R_e\). Thus, \(U_h = -\frac{GMm}{2R_e} = \frac{-mgR_e}{2}\).
The gravitational potential energy of a body of mass \(m\) at the earth's surface is \(-mgR_e\). Its gravitational potential energy at a height \(R_e\) from the earth's surface will be (Here \(R_e\) is the radius of the earth)
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