Question

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

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

Answer 1

Answer: (D)

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

Let m and M be the masses.

d is the distance between body

Let x be the distance from the mass 'm'

at gravitational field is zero.

Net Force at x,F=$\frac{Gm}{x^2}-\frac{GM}{(d-x{)}^2}$

at x, F=0

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

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

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

$x\times (\sqrt{M}+\sqrt{m})=\sqrt{m}\times d$

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

Potential at point x, V=$\frac{-Gm}{x}-\frac{GM}{d-x}$

Put x in V, we get

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

V=$\frac{-G\sqrt{m}}{\frac{d}{\sqrt{M}+\sqrt{m}}}-\frac{GM}{\left(\frac{\sqrt{M}d}{\sqrt{M}+\sqrt{m}}\right)}$

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

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