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(A) Let the mass of the particle be $m$,the mass of the planet be $M$,and its radius be $R$. Taking the gravitational potential at a large distance (i

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$\sqrt{1.5} v_e$

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(A) Let the mass of the particle be $m$,the mass of the planet be $M$,and its radius be $R$. Taking the gravitational potential at a large distance (infinity) as zero,the total energy of the particle at infinity is $E_i = 0$. The gravitational potential at the centre of the planet is $V_c = -\frac{3GM}{2R}$. By the law of conservation of energy,the total energy at infinity equals the total energy at the centre of the planet: $K_i + U_i = K_c + U_c$ $0 + 0 = \frac{1}{2}mv^2 + m(-\frac{3GM}{2R})$ $\frac{1}{2}mv^2 = \frac{3GMm}{2R}$ $v^2 = \frac{3GM}{R}$ $v = \sqrt{\frac{3GM}{R}}$. We know that the escape velocity at the surface of the planet is $v_e = \sqrt{\frac{2GM}{R}}$,which implies $\frac{GM}{R} = \frac{v_e^2}{2}$. Substituting this into the expression for $v$: $v = \sqrt{3 \cdot \frac{v_e^2}{2}} = \sqrt{1.5} v_e$.

$A$ particle released at a large distance from a planet reaches the planet only under gravitational attraction and passes through a smooth tunnel through its centre. If $v_e$ is the escape velocity of a body at the planet,then the particle's speed at the centre of the planet is

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