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

The correct option is C $\frac{7}{8}\frac{GM}{x^2}$

Mass of small part $=$ mass density of big sphere $\times$ volume of small part

or, $m=\frac{M}{\frac{4}{3}{\pi R}^3}\times \frac{4}{3}\pi {\left(\frac{R}{2}\right)}^3=\frac{M}{8}$

Gravitational field due to rest part $=$ Gravitational field due to big sphere $-$ Gravitational of small part

$=\frac{GM}{x^2}-\frac{GM}{{(\frac{R}{2}+x)}^2}$

$=\frac{GM}{x^2}-\frac{GM}{8{(\frac{R}{2}+x)}^2}$

$=GM[\frac{1}{x^2}-\frac{1}{8{(\frac{R}{2}+x)}^2}]$

For very far point $X>>R/2$

So, $(R/2+X)=x$

Gravitational field due to rest part

$=GM[\frac{1}{x^2}-\frac{1}{8x^2}]=\frac{7GM}{{8x}^2}$