Question

Two bodies of masses $M_1$ and $M_2$ are placed at a distance $R$ apart. Then at the position where the gravitational field due to them is zero, the gravitational potential is

• A. $-\frac{G\sqrt{M_1}}{R}$
• B. $-\frac{G\sqrt{M_2}}{R}$
• C. $-\frac{G(\sqrt{M_1}+\sqrt{M_2}{)}^2}{R}$
• D. $-\frac{G(\sqrt{M_1}-\sqrt{M_2}{)}^2}{R}$

Answer 1

The correct option is C $-\frac{G(\sqrt{M_1}+\sqrt{M_2}{)}^2}{R}$

Let point $\text{P}$ be the null point, which is at a distance $x$ from $m$.


The field of $M_1$, $E_{M_1}$ is cancelled by the field of $M_2$, $E_{M_2}$ at this point. Thus,

$E_{M_1}=E_{M_2}$

Substituting the values,

$\Rightarrow \frac{G{M}_1}{x^2}=\frac{G{M}_2}{(R-x{)}^2}$

$\Rightarrow \frac{\sqrt{M_1}}{x}=\frac{\sqrt{M_2}}{R-x}$

$\Rightarrow \sqrt{M_1}R-\sqrt{M_1}x=\sqrt{M_2}x$

$\Rightarrow x=\frac{\sqrt{M_1}R}{\sqrt{M_1}+\sqrt{M_2}}$

Now, the potential at point $\text{P}$ will be,

$V_P=V_{M_1}+V_{M_2}$

$\Rightarrow V_P=-\frac{G{M}_1}{x}+(-\frac{G{M}_2}{R-x})$

Substituting the value of $x$,

$\Rightarrow V_P=-\frac{G{M}_1}{\sqrt{M_1}R}(\sqrt{M_1}+\sqrt{M_2})-\frac{G{M}_2}{\sqrt{M_2}R}(\sqrt{M_1}+\sqrt{M_2})$

$\Rightarrow V_P=-\frac{G\sqrt{M_1}}{R}(\sqrt{M_1}+\sqrt{M_2})-\frac{G\sqrt{M_2}}{R}(\sqrt{M_1}+\sqrt{M_2})$

$\Rightarrow V_P=-(\sqrt{M_1}+\sqrt{M_2})(\sqrt{M_1}+\sqrt{M_2})\frac{G}{R}$

$\Rightarrow V_P=-{(\sqrt{M_1}+\sqrt{M_2})}^2\frac{G}{R}$

Hence, option (c) is the correct answer.

Key Concept: The gravitational potential due to a point mass is given by $$V=\frac{-GM}{r}$$ Why this question: The gravitational potential at a point is equal to the sum of gravitational potentials due to all masses in the vicinity. Also gravitational potential is a scalar quantity and hence algebraic sum is sufficient.