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(B) The escape velocity of a planet is given by $v_e = \sqrt{\frac{2GM}{R}}$.Using the principle of conservation of energy,the total energy at the sur

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

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(B) The escape velocity of a planet is given by $v_e = \sqrt{\frac{2GM}{R}}$.Using the principle of conservation of energy,the total energy at the surface equals the total energy at the center.The gravitational potential at the surface is $V_S = -\frac{GM}{R}$.The gravitational potential at the center is $V_C = -\frac{3GM}{2R}$.Loss in potential energy = Gain in kinetic energy:$m(V_S - V_C) = \frac{1}{2}mv^2$$m\left(-\frac{GM}{R} - (-\frac{3GM}{2R})\right) = \frac{1}{2}mv^2$$m\left(\frac{3GM}{2R} - \frac{2GM}{2R}\right) = \frac{1}{2}mv^2$$\frac{GM}{2R} = \frac{1}{2}v^2$$v^2 = \frac{GM}{R}$Since $v_e^2 = \frac{2GM}{R}$,we have $\frac{GM}{R} = \frac{v_e^2}{2}$.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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