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Step-by-Step Solution
Step 1: Identify the Scenario
We have two masses, one of mass $m$ and the other of mass $4m$, separated by a distance $r$. We need to find the gravitational potential at a point $P$ on the line joining these two masses where the resultant gravitational field is zero.
Step 2: Condition for the Gravitational Field to be Zero
Let the distance of point $P$ from mass $m$ be $x$. Then, its distance from the other mass ($4m$) would be $(r - x)$. The gravitational field due to a mass $M$ at a distance $d$ is given by $G M / d^2$ (directed toward the mass). At $P$, the net gravitational field becomes zero, implying the magnitudes of fields due to the two masses cancel each other:
$$
\frac{G m}{x^2} \;=\; \frac{4G m}{(r - x)^2}.
$$
Step 3: Solve for x
Canceling $Gm$ on both sides and rearranging, we get:
$$
\frac{1}{x^2} \;=\;\frac{4}{(r - x)^2}
\quad\Longrightarrow\quad
4\,x^2 \;=\;(r - x)^2.
$$
Taking the square root on both sides leads to (considering the physically valid root):
$$
2x \;=\; r - x
\quad\Longrightarrow\quad
x \;=\; \frac{r}{3}.
$$
Step 4: Substituting x into the Potential Formula
The gravitational potential $V$ at a distance $d$ from a mass $M$ is given by $- \tfrac{G M}{d}$. Since point $P$ is at distance $x=\tfrac{r}{3}$ from $m$ and distance $r - x = r - \tfrac{r}{3} = \tfrac{2r}{3}$ from $4m$, we calculate:
$$
V \;=\; - \frac{G m}{\frac{r}{3}} \;-\; \frac{G\,(4m)}{\frac{2r}{3}}.
$$
Simplify each term:
$$
- \frac{G m}{\frac{r}{3}}
\;=\;
- \frac{G m \times 3}{r}
\;=\;
-\frac{3G m}{r},
$$
$$
- \frac{4G m}{\frac{2r}{3}}
\;=\;
- \frac{4G m \times 3}{2r}
\;=\;\;
- \frac{12G m}{2r}
\;=\;\;
- \frac{6G m}{r}.
$$
Adding these two contributions gives:
$$
V
\;=\;
-\frac{3G m}{r}
\;-\;
\frac{6G m}{r}
\;=\;
-\frac{9G m}{r}.
$$
Step 5: Final Result
The gravitational potential at the point $P$ where the gravitational field is zero is:
$$
\boxed{ - \frac{9G m}{r} }.
$$
