A spherical part of radius R/2 is excavated f | vedclass

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(B) Let the original mass of the asteroid be $M$ and its radius be $R$. The gravitational field at the surface is $g = \frac{GM}{R^2}$.When a spherica

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$\frac{GM}{2R^2}$

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(B) Let the original mass of the asteroid be $M$ and its radius be $R$. The gravitational field at the surface is $g = \frac{GM}{R^2}$.When a spherical part of radius $r = R/2$ is excavated,its mass $M'$ is given by $M' = \rho \cdot V' = \rho \cdot \frac{4}{3}\pi (R/2)^3 = \frac{1}{8} \rho \cdot \frac{4}{3}\pi R^3 = \frac{M}{8}$.The gravitational field at the point on the surface just above the excavation is the vector sum of the field due to the original sphere and the field due to the removed part (treated as a negative mass).The field due to the original sphere at the surface is $g_1 = \frac{GM}{R^2}$ (directed towards the center).The field due to the removed sphere at its own surface is $g_2 = \frac{GM'}{r^2} = \frac{G(M/8)}{(R/2)^2} = \frac{GM/8}{R^2/4} = \frac{GM}{2R^2}$ (directed away from the center of the excavation,i.e.,outwards).The net gravitational acceleration is $g_{net} = g_1 - g_2 = \frac{GM}{R^2} - \frac{GM}{2R^2} = \frac{GM}{2R^2}$.

$A$ spherical part of radius $R/2$ is excavated from an asteroid of mass $M$ and radius $R$ as shown in the figure. The gravitational acceleration at a point on the surface of the asteroid just above the excavation is

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