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(A) According to the law of conservation of energy:$(K.E + P.E)_{	ext{surface}} = (K.E + P.E)_{	ext{max height}}$Since the velocity at maximum heigh

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$\frac{n^2 R}{(1 - n^2)}$

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(A) According to the law of conservation of energy:$(K.E + P.E)_{	ext{surface}} = (K.E + P.E)_{	ext{max height}}$Since the velocity at maximum height is $0$,we have:$-\frac{GMm}{R} + \frac{1}{2} m(n V_e)^2 = -\frac{GMm}{R + h} + 0$We know that the escape velocity $V_e = \sqrt{\frac{2GM}{R}}$,so $V_e^2 = \frac{2GM}{R}$.Substituting this into the equation:$\frac{1}{2} m n^2 \left( \frac{2GM}{R} ight) = GMm \left( \frac{1}{R} - \frac{1}{R + h} ight)$$\frac{n^2 GMm}{R} = GMm \left( \frac{R + h - R}{R(R + h)} ight)$$\frac{n^2}{R} = \frac{h}{R(R + h)}$$n^2 = \frac{h}{R + h}$$n^2(R + h) = h$$n^2 R + n^2 h = h$$n^2 R = h(1 - n^2)$$h = \frac{n^2 R}{1 - n^2}$

$A$ body is projected vertically upward from the surface of the earth of radius $R$ into space with velocity $n V_e$ (where $n < 1$ and $V_e$ is the escape velocity). The maximum height from the surface of the earth to which the body can reach is

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