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Question

(C) According to the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$ from the surface.

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 equals the total energy at the maximum height $h$ from the surface.Initial energy at surface: $E_i = -\frac{GMm}{R} + \frac{1}{2}m(k{v_e})^2$Final energy at maximum height: $E_f = -\frac{GMm}{R+h}$Since ${v_e} = \sqrt{\frac{2GM}{R}}$,we have ${v_e}^2 = \frac{2GM}{R}$.Setting $E_i = E_f$:$-\frac{GMm}{R} + \frac{1}{2}m k^2 \left(\frac{2GM}{R}\right) = -\frac{GMm}{R+h}$$-\frac{GMm}{R} + \frac{GMm k^2}{R} = -\frac{GMm}{R+h}$$-\frac{1}{R}(1 - k^2) = -\frac{1}{R+h}$$R+h = \frac{R}{1-k^2}$Thus,the maximum distance from the center of the earth is $r = R+h = \frac{R}{1-k^2}$.

$A$ body is projected vertically upwards from the surface of the earth with a speed of $k{v_e}$,where $k < 1$ and ${v_e}$ is the escape velocity of the earth. What is the maximum height from the center of the earth that the body will reach? (Given: $R$ is the radius of the earth)

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