Physics - Gravitation Question with Solution | TestHub

Let the point where the gravitational field is zero be at a distance $x$ from the mass $m$. The distance from the mass $M$ will be $d-x$.

Equating the magnitudes of the gravitational fields due to both masses at this point:

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

Taking the square root of both sides:

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

Solving for $x$:

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

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

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

The distance from the mass $M$ is:

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

The gravitational potential $V$ at this point is the sum of the potentials due to both masses:

$V=-\frac{Gm}{x}-\frac{GM}{d-x}$

Substituting the values of $x$ and $d-x$:

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

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

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

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

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