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(A) The gravitational potential $V$ inside the Earth at a distance $r$ from the center is given by $V = -\frac{GM}{2R^3}(3R^2 - r^2)$.At the surface $

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$\sqrt{\frac{3GM}{4R}}$

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(A) The gravitational potential $V$ inside the Earth at a distance $r$ from the center is given by $V = -\frac{GM}{2R^3}(3R^2 - r^2)$.At the surface $(r = R)$,the potential is $V_s = -\frac{GM}{R}$.At a distance $r = \frac{R}{2}$ from the center,the potential is $V_p = -\frac{GM}{2R^3}(3R^2 - (\frac{R}{2})^2) = -\frac{GM}{2R^3}(3R^2 - \frac{R^2}{4}) = -\frac{GM}{2R^3}(\frac{11R^2}{4}) = -\frac{11GM}{8R}$.Using the law of conservation of energy between the surface and the point at $\frac{R}{2}$:$K_i + U_i = K_f + U_f$$0 + m V_s = \frac{1}{2}mv^2 + m V_p$$m(-\frac{GM}{R}) = \frac{1}{2}mv^2 + m(-\frac{11GM}{8R})$$-\frac{GM}{R} + \frac{11GM}{8R} = \frac{1}{2}v^2$$\frac{3GM}{8R} = \frac{1}{2}v^2$$v^2 = \frac{3GM}{4R}$$v = \sqrt{\frac{3GM}{4R}}$

$A$ tunnel is dug across the diameter of the Earth. $A$ ball is released from the surface of the Earth into the tunnel. The velocity of the ball when it is at a distance $\frac{R}{2}$ from the center of the Earth is (where $R$ = radius of the Earth and $M$ = mass of the Earth):

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