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

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(B) Let the mass of the asteroid be $M$ and its radius be $R$. The density $\rho$ is given by $\rho = M / (\frac{4}{3} \pi R^3)$.When a spherical hole

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$GM/2R^2$

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(B) Let the mass of the asteroid be $M$ and its radius be $R$. The density $\rho$ is given by $\rho = M / (\frac{4}{3} \pi R^3)$.When a spherical hole of radius $r = R/2$ is excavated,the mass of the removed part is $m = \rho \cdot (\frac{4}{3} \pi r^3) = \rho \cdot (\frac{4}{3} \pi (R/2)^3) = \rho \cdot (\frac{4}{3} \pi R^3) / 8 = M/8$.The gravitational acceleration at the point $P$ (just above the hole) due to the complete asteroid is $g_1 = GM/R^2$ directed towards the center.The gravitational acceleration at point $P$ due to the removed sphere (hole) is $g_2 = G(M/8) / (R/2)^2 = (GM/8) / (R^2/4) = GM / 2R^2$ directed away from the center of the hole (towards the center of the asteroid).Since the hole is removed,the net gravitational acceleration $g_{net}$ is the vector sum of the acceleration due to the complete sphere and the negative mass of the hole: $g_{net} = g_1 - g_2 = GM/R^2 - GM/2R^2 = GM/2R^2$.

$A$ spherical hole 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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