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: Understanding the Concept:
Gravitational potential ($V$) at a specific point in space is defined as the amount of work done per unit mass to bring a small test mass from infinity to that point. Crucially, gravitational potential is a scalar quantity, meaning it has magnitude and sign but possesses no spatial direction. According to the Principle of Superposition, when dealing with a configuration of multiple point masses, the total net gravitational potential at any point is simply the direct algebraic sum of the individual scalar potentials produced by each mass independently.

Step 2: Key Formula or Approach:
1. Potential due to a Point Mass: At a distance $r$ away from a point mass $m$, its scalar gravitational potential is given by: $$ V = -\frac{Gm}{r} $$ 2. Total Superposition Potential: Since all four point masses are identical ($m$) and arranged symmetrically at the corners of a square, they sit at an identical straight-line distance ($r$) from the square's geometric center. Therefore: $$ V_{\text{net}} = V_1 + V_2 + V_3 + V_4 = 4 \times \left(-\frac{Gm}{r}\right) = -\frac{4Gm}{r} $$ 3. Diagonal Geometry of a Square: For a square possessing a side length $a$, the total length of its diagonal line is given by $d = \sqrt{a^2 + a^2} = \sqrt{2}a$. The distance from any corner vertex to the central intersection point is exactly half of this full diagonal length: $$ r = \frac{d}{2} = \frac{\sqrt{2}a}{2} = \frac{a}{\sqrt{2}} $$

Step 3: Detailed Explanation:
Let us substitute our calculated geometric distance $r = \frac{a}{\sqrt{2}}$ directly into our total superposition potential expression: $$ V_{\text{net}} = -\frac{4Gm}{\left(\frac{a}{\sqrt{2}}\right)} $$ To simplify this fraction, move the term $\sqrt{2}$ from the denominator up into the primary numerator: $$ V_{\text{net}} = -4\sqrt{2} \frac{Gm}{a} $$ This algebraic solution matches option (B). Note that because gravitational forces are always attractive, gravitational potential configurations always carry a negative scalar value, representing a bound stable system.

Step 4: Final Answer:
The gravitational potential at the center of the square is $-4\sqrt{2} Gm / a$.