Question

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

• A. $\frac{6Gm}{r}$
• B. $\frac{-9Gm}{r}$
• C. Zero
• D. $\frac{-4Gm}{r}$

Answer 1

Answer: (B)

$\frac{-9Gm}{r}$

Answer 2

Let the two masses be $m$ and $4m$ placed at a distance $r$. Let the point where the gravitational field is zero be at a distance $x$ from mass $m$. Then its distance from mass $4m$ is $(r-x)$. For the gravitational field to be zero, the point must lie between the two masses.

The gravitational field due to mass $m$ at this point is $E_1 = \frac{Gm}{x^2}$. The gravitational field due to mass $4m$ at this point is $E_2 = \frac{G(4m)}{(r-x)^2}$.

For the net gravitational field to be zero, their magnitudes must be equal: $\frac{Gm}{x^2} = \frac{G(4m)}{(r-x)^2}$ $\frac{1}{x^2} = \frac{4}{(r-x)^2}$ Taking the square root of both sides: $\frac{1}{x} = \frac{2}{r-x} \quad \text{(since } x \text{ and } r-x \text{ are positive)}$ $r-x = 2x$ $r = 3x$ $x = \frac{r}{3}$ The distance of the point from mass $4m$ is $r-x = r - \frac{r}{3} = \frac{2r}{3}$.

The gravitational potential at this point is the sum of the potentials due to each mass: $V = V_m + V_{4m}$ $V = -\frac{Gm}{x} - \frac{G(4m)}{r-x}$ Substitute the values of $x$ and $r-x$: $V = -\frac{Gm}{r/3} - \frac{G(4m)}{2r/3}$ $V = -\frac{3Gm}{r} - \frac{4Gm \times 3}{2r}$ $V = -\frac{3Gm}{r} - \frac{12Gm}{2r}$ $V = -\frac{3Gm}{r} - \frac{6Gm}{r}$ $V = -\frac{9Gm}{r}$