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)

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
$M^{\prime }=M-\frac{M}{8}=\frac{7}{8}M$
Therefore gravitational field due to the left over part of the sphere
$=\frac{G{M}^{\prime }}{X^2}=\frac{7}{8}\frac{GM}{x^2}$