The escape velocity from a planet is V_e. A t | vedclass

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(A) The potential energy at the surface of the planet is $U_s = -\frac{GMm}{R}$.The potential energy at the center of the planet is $U_c = -\frac{3GMm

Vedclass · Accepted Answer

$\frac{V_e}{\sqrt{2}}$

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(A) The potential energy at the surface of the planet is $U_s = -\frac{GMm}{R}$.The potential energy at the center of the planet is $U_c = -\frac{3GMm}{2R}$.By the law of conservation of energy,the total energy at the surface equals the total energy at the center:$-\frac{GMm}{R} + 0 = -\frac{3GMm}{2R} + \frac{1}{2}mv^2$.Rearranging the terms,we get $\frac{1}{2}mv^2 = \frac{3GMm}{2R} - \frac{GMm}{R} = \frac{GMm}{2R}$.Thus,$v^2 = \frac{GM}{R}$.Since the escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$,we have $V_e^2 = \frac{2GM}{R}$,which implies $\frac{GM}{R} = \frac{V_e^2}{2}$.Substituting this into the expression for $v^2$,we get $v^2 = \frac{V_e^2}{2}$,so $v = \frac{V_e}{\sqrt{2}}$.

The escape velocity from a planet is $V_e$. $A$ tunnel is dug along the diameter of the planet and a small body is dropped into it. The speed of the body at the centre of the planet will be

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