Question

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):

• A. $\frac{5}{6} \frac{GM}{x^{2}}$
• B. $\frac{8}{9} \frac{GM}{x^{2}}$
• C. $\frac{7}{8} \frac{GM}{x^{2}}$
• D. $\frac{6}{7} \frac{GM}{x^{2}}$

Answer 1

Answer: (C)

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

Answer 2

Let mass of smaller sphere (which has to be removed) is m
Radius $=\frac{R}{2}$ (from figure)
$\frac{M}{\frac{4}{3}\pi R^{3}}= \frac{m}{\frac{4}{3}\pi\left(\frac{R}{2}\right)^{3}}$
$\Rightarrow m = \frac{M}{8}$
Mass of the left over part of the sphere
$M' =M - \frac{M}{8} = \frac{7}{8}M$
Therefore gravitational field due to the left over part of the sphere
$= \frac{GM'}{X^{2}} = \frac{7}{8} \frac{GM}{x^{2}}$