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(B) Let the point where the gravitational field intensity is zero be at a distance $x$ from mass $m$. The gravitational field intensity $E$ is given b

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the point where the gravitational field intensity is zero be at a distance $x$ from mass $m$. The gravitational field intensity $E$ is given by $E = \frac{Gm}{x^2} - \frac{GM}{(d-x)^2} = 0$.This implies $\frac{m}{x^2} = \frac{M}{(d-x)^2}$,so $\frac{\sqrt{m}}{x} = \frac{\sqrt{M}}{d-x}$.Solving for $x$,we get $x = \frac{\sqrt{m}}{\sqrt{m} + \sqrt{M}} d$ and $d-x = \frac{\sqrt{M}}{\sqrt{m} + \sqrt{M}} d$.The gravitational potential $V$ at this point is the sum of potentials due to both masses: $V = - \frac{Gm}{x} - \frac{GM}{d-x}$.Substituting the values of $x$ and $d-x$: $V = - \frac{Gm}{\frac{\sqrt{m}}{\sqrt{m} + \sqrt{M}} d} - \frac{GM}{\frac{\sqrt{M}}{\sqrt{m} + \sqrt{M}} d}$.$V = - \frac{G}{d} [\sqrt{m}(\sqrt{m} + \sqrt{M}) + \sqrt{M}(\sqrt{m} + \sqrt{M})]$.$V = - \frac{G}{d} (\sqrt{m} + \sqrt{M})^2$.

Two bodies of masses $m$ and $M$ are placed at a distance $d$ from each other. What is the gravitational potential $V$ at the point where the gravitational field intensity is zero?

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