First slide

Gravitaional potential at a point

Question

Two bodies of mass m and M are placed a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V. Then :

Difficult

A

$V = - \frac{G}{d} (M + m)$
B

$V = - \frac{G m}{d}$
C

$V = - \frac{G M}{d}$
D

$V = - \frac{G}{d} {(\sqrt{m} + \sqrt{M})}^{2}$

Solution

$d i s t f r o m m w h e r e f i e l d i s z e r o i s x w h i c h s a t i s f y t h e c o n d i t i o n \frac{G m}{x^{2}} = \frac{G M}{{(d - x)}^{2}} s o l v i n g f o r x w e g e t x = \frac{d \sqrt{m}}{\sqrt{m} + \sqrt{M}} a n d (d - x) i s \frac{d \sqrt{M}}{\sqrt{m} + \sqrt{M}} t h e n p o t e n t i a l t h e r e V = - \frac{G m}{x} - \frac{G M}{d - x} s u b s t i t u t i n g t h e v a l u e s w e g e t V$

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