The escape velocity for a planet is v_e. A tu | vedclass

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(B) The gravitational potential at the surface of the planet is $V_s = -\frac{GM}{R}$.The gravitational potential at the center of the planet is $V_c

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$\frac{v_e}{\sqrt{2}}$

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(B) The gravitational potential at the surface of the planet is $V_s = -\frac{GM}{R}$.The gravitational potential at the center of the planet is $V_c = -\frac{3GM}{2R}$.Using the law of conservation of energy,the kinetic energy gained by the body equals the loss in potential energy: $\frac{1}{2}mv^2 = m(V_s - V_c)$.Substituting the values: $\frac{1}{2}mv^2 = m\left(-\frac{GM}{R} - (-\frac{3GM}{2R})\right)$.$\frac{1}{2}v^2 = \frac{3GM}{2R} - \frac{GM}{R} = \frac{GM}{2R}$.$v^2 = \frac{GM}{R}$.Since the escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$,we have $v_e^2 = \frac{2GM}{R}$,which implies $\frac{GM}{R} = \frac{v_e^2}{2}$.Therefore,$v^2 = \frac{v_e^2}{2}$,which gives $v = \frac{v_e}{\sqrt{2}}$.

The escape velocity for a planet is $v_e$. $A$ tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet,its speed will be

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