Question

Two bodies of masses $m$ and $M$ are placed atdistance $d$ apart. The gravitational potential ($V$) at the position where the gravitational field due to them is zero is $V$.

• A. $V=-\frac{G}{d}(m+M)$
• B. $V=-\frac{G}{d}$
• C. $V=-\frac{GM}{d}$
• D. $V=-\frac{G}{d}(\sqrt{m}+\sqrt{M}{)}^2$

Answer 1

The correct option is D $V=-\frac{G}{d}(\sqrt{m}+\sqrt{M}{)}^2$

Gravitational field at a distance r, $E=\frac{GM}{r^2}$

Let at point O net gravitational field intensity is zero.

$\frac{Gm}{x^2}-\frac{GM}{(d-x{)}^2}=0$

$\frac{(d-x{)}^2}{x^2}=\frac{M}{m}$

$[\frac{d}{x}-1{]}^2=\frac{M}{m}$

$⟹x=\frac{d\sqrt{m}}{\sqrt{m}+\sqrt{M}}$ and $d-x=\frac{d\sqrt{M}}{\sqrt{m}+\sqrt{M}}$

Now potential at point O, $V=(-\frac{Gm}{x})+(-\frac{GM}{d-x})$

$V=-\frac{Gm}{\frac{d\sqrt{m}}{\sqrt{M}+\sqrt{m}}}-\frac{GM}{\frac{d\sqrt{M}}{\sqrt{M}+\sqrt{m}}}$

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

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