A projectile is projected with velocity k{v_e | vedclass

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Question

(C) According to the law of conservation of energy,the total energy at the surface of the Earth is equal to the total energy at the maximum height $h$

Vedclass · Accepted Answer

$\frac{R}{1 - k^2}$

Vedclass · Answer

(C) According to the law of conservation of energy,the total energy at the surface of the Earth is equal to the total energy at the maximum height $h$ from the surface.Total energy at surface = $-\frac{GMm}{R} + \frac{1}{2}m(kv_e)^2$Total energy at maximum height = $-\frac{GMm}{R+h} + 0$Since $v_e = \sqrt{\frac{2GM}{R}}$,we have $v_e^2 = \frac{2GM}{R}$.Equating the energies: $-\frac{GMm}{R} + \frac{1}{2}m k^2 \left(\frac{2GM}{R}\right) = -\frac{GMm}{R+h}$Dividing by $GMm$: $-\frac{1}{R} + \frac{k^2}{R} = -\frac{1}{R+h}$$\frac{k^2 - 1}{R} = -\frac{1}{R+h} \implies \frac{1 - k^2}{R} = \frac{1}{R+h}$$R+h = \frac{R}{1 - k^2}$Thus,the distance from the center of the Earth is $r = R+h = \frac{R}{1 - k^2}$.

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