Two bodies of masses m\: and \: 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

  • Option 1)

    zero

  • Option 2)

    -\frac{4Gm}{r}

  • Option 3)

    -\frac{6Gm}{r}

  • Option 4)

    -\frac{9Gm}{r}

 

Answers (1)
S Sabhrant Ambastha

As we discussed in

Gravitational field due to Point mass -

F=\frac{GmM}{r^{2}}

I= \frac{F}{m}= \frac{GMm}{r^{2}m}

\therefore I= \frac{GM}{r^{2}}

\dpi{100} G \rightarrow Gravitational\: constant

M\rightarrow mass\: of\: earth

 

 

- wherein

As the distance (r) of test mass from point (M) Increases I decreases.

I\propto \frac{1}{r^{2}}

I= 0\: at\: \left ( r= \infty \right )

 

 \frac{Gm}{r^{2}}=\frac{G(4m)}{(r-x)^{2}}

\left [ \frac{x}{\left ( r-x \right )} \right ]^2=\frac{1}{4}

\Rightarrow x=\frac{r}{3}

U=\frac{-Gm}{x}-\frac{G\left ( 4m \right )}{\left ( r-x \right )}

=\frac{-Gm}{\left ( \frac{r}{3} \right )}-\frac{G\left ( 4m \right )}{\left ( r-\frac{r}{3} \right )}

= -\frac{3Gm}{r}-\frac{3G\left ( 4m \right )}{2r}

= -\frac{9Gm}{2r}

 


Option 1)

zero

Incorrect option

Option 2)

-\frac{4Gm}{r}

Incorrect option

Option 3)

-\frac{6Gm}{r}

Incorrect option

Option 4)

-\frac{9Gm}{r}

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