The gravitational field,due to the 'left over | vedclass

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(C) Let the mass of the original uniform sphere be $M$ and its radius be $R$. The mass of the smaller sphere (which is removed) is $m$. From the figur

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$\frac{7}{8} \frac{GM}{x^2}$

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(C) Let the mass of the original uniform sphere be $M$ and its radius be $R$. The mass of the smaller sphere (which is removed) is $m$. From the figure,the radius of the removed sphere is $r = R/2$. Assuming the density of the sphere is $\rho$,we have: $\rho = \frac{M}{\frac{4}{3}\pi R^3} = \frac{m}{\frac{4}{3}\pi (R/2)^3}$ $\Rightarrow m = M \cdot \left(\frac{R/2}{R}\right)^3 = \frac{M}{8}$ Mass of the left over part of the sphere is: $M' = M - m = M - \frac{M}{8} = \frac{7}{8}M$ For a point $P$ at a very large distance $x$ from the center of the original sphere,the gravitational field $E$ due to the remaining mass $M'$ is given by: $E = \frac{GM'}{x^2} = \frac{G(7/8)M}{x^2} = \frac{7}{8} \frac{GM}{x^2}$

The gravitational field,due to the 'left over part' of a uniform sphere (from which a part as shown,has been 'removed out'),at a very far off point,$P$,located as shown,would be (nearly)

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