An asteroid, initially at rest, falls towards a planet from a distance of 4R from the center of the planet, where R is the radius of the planet. If v_e is the escape velocity from the surface of the planet, what will be

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Correct answer: C. 3 2 v_e

Escape velocity from the surface of the planet is given by

v_e = \sqrt{\dfrac{2GM}{R}}

.

By conservation of mechanical energy between the initial position and the surface of the planet:

K_i + U_i = K_f + U_f

0 - \dfrac{GMm}{4R} = \dfrac{1}{2}mv^2 - \dfrac{GMm}{R}

\dfrac{1}{2}mv^2 = \dfrac{GMm}{R} - \dfrac{GMm}{4R}

\dfrac{1}{2}mv^2 = \dfrac{3GMm}{4R}

v^2 = \dfrac{3GM}{2R}

Since

\dfrac{GM}{R} = \dfrac{v_e^2}{2}

, substituting this into the equation for

v^2

gives:

v^2 = \dfrac{3}{2} \left( \dfrac{v_e^2}{2} \right) = \dfrac{3}{4} v_e^2

v = \dfrac{\sqrt{3}}{2} v_e

Answer:

\dfrac{\sqrt{3}}{2} v_e

An asteroid, initially at rest, falls towards a planet from a...

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