NEETPhysicsGravitationMCQ+4 / −1
Two bodies of mass and are placed at a distance . The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( gravitational constant) :
- A
- B
- C
- D
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Correct answer: B

Position of Neutral point (Zero Gravitational Field)
$$r_{1}=\frac{\sqrt{m_{1}} R}{\sqrt{m_{1}}+\sqrt{m_{2}}}=\frac{\sqrt{m} R}{\sqrt{m}+\sqrt{9 m}}=\frac{R}{4}$$
$$\mathrm{r}_{2}=\mathrm{R}-\mathrm{R} / 4=3 \mathrm{R} / 4$$
Now Gravitational potential at point $\mathrm{P}$
$$ \begin{aligned} \& V_{P}=-\frac{G M}{R / 4}-\frac{9(\mathrm{GM})}{3 \mathrm{R} / 4} \\ \& =\frac{-16 \mathrm{GM}}{\mathrm{R}} \end{aligned} $$
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