Question

Four point masses each of mass $m$ are placed at the four corners of a square of side length $a$. The gravitational potential at the center of the square is:

• A. $-4Gm / a$
• B. $-4\sqrt{2} Gm / a$
• C. $-2\sqrt{2} Gm / a$
• D. Zero Correct Answer: (B) $-4\sqrt{2} Gm / a$

Hint:

Be very careful not to confuse Gravitational Potential (a scalar field) with Gravitational Field Intensity (a vector field). At the center of a symmetric square layout: - The vector fields pull in opposite directions and completely cancel out, making the field intensity Zero. - The scalar potentials do not cancel out; instead, their negative values pile up constructively, giving a total potential of $\mathbf{-4\sqrt{2}\frac{Gm}{a}}$.

Answer 1

The Correct Option is B

Step 1: Remember potential is a plain number.
Gravitational potential has no direction, so I just add up the four contributions. Each corner mass is the same distance from the centre, so I find that distance first.

Step 2: Find the corner to centre distance.
The diagonal of the square is $\sqrt{2}\,a$, and the centre sits halfway along it: \[ r = \frac{\sqrt{2}\,a}{2} = \frac{a}{\sqrt{2}} \]

Step 3: Add the four equal potentials.
Each mass gives $-\dfrac{Gm}{r}$, and there are four of them: \[ V = 4\times\left(-\frac{Gm}{r}\right) = -\frac{4Gm}{a/\sqrt{2}} \]

Step 4: Tidy the fraction.
Dividing by $a/\sqrt{2}$ is the same as multiplying by $\sqrt{2}/a$: \[ V = -\frac{4\sqrt{2}\,Gm}{a} \]
So the potential at the centre is $-4\sqrt{2}\,Gm/a$, which is option (B).
\[ \boxed{-\frac{4\sqrt{2}\,Gm}{a}} \]