A metal sphere has radius $R$ and mass $M$ . A spherical hollow of diameter $R$ is made in this sphere such that its surface passes through the centre of the metal sphere and touches the outside surface of the metal sphere. A unit mass is placed at a distance from the centre of metal sphere. The gravitational field at that point is

Question

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Answer 1

Let $ρ$ be the mass density.
mass of metal sphere $M = ρ . \frac{4}{3} π R^{3}$
mass of hollow sphere , $m = ρ . \frac{4}{3} π (R / 2)^{3} = M / 8$
The field at that point $=$ field due to metal sphere $-$ field due to hollow sphere.
$f = \frac{G M}{a ^{2}} - \frac{G m}{( a - \frac{R}{2} ) ^{2}} = \frac{G M}{a ^{2}} - \frac{G M}{8 ( a - \frac{R}{2} ) ^{2}} = \frac{G M}{a ^{2}} - \frac{G M}{8 a ^{2} ( 1 - \frac{R}{2 a} ) ^{2}} = \frac{G M}{a ^{2}} (1 - \frac{1}{8 ( a - \frac{R}{2 a} ) ^{2}})$

Answer 2

Let

ρ

be the mass density.
mass of metal sphere

M = ρ . \frac{4}{3} π R^{3}

mass of hollow sphere ,

m = ρ . \frac{4}{3} π (R / 2)^{3} = M / 8

The field at that point

=

field due to metal sphere

-

field due to hollow sphere.

f = \frac{G M}{a ^{2}} - \frac{G m}{( a - \frac{R}{2} ) ^{2}} = \frac{G M}{a ^{2}} - \frac{G M}{8 ( a - \frac{R}{2} ) ^{2}} = \frac{G M}{a ^{2}} - \frac{G M}{8 a ^{2} ( 1 - \frac{R}{2 a} ) ^{2}} = \frac{G M}{a ^{2}} (1 - \frac{1}{8 ( a - \frac{R}{2 a} ) ^{2}})