Question

A modified gravitational potential is given by \(V=-\frac{GM}{r}+\frac{A}{r^2}\). If the constant A is expressed in terms of gravitational constant (G), Mass(M) and velocity of light (c), then from dimensional analysis, A is,

• A. \(\frac{G^2M^2}{e^2}\)
• B. \(\frac{GM}{e^2}\)
• C. \(\frac{1}{e^2}\)
• D. Dimensionless

Answer 1

The Correct Option is A

Given the expression: \(V = -\frac{GM}{r} + r^2 A\)

We are to find the dimensions of the constant \(A\) using dimensional analysis.

Step 1: Determine dimensions of potential energy (V) 

Gravitational potential \(V\) has dimensions of energy per unit mass:
\([V] = \frac{[ML^2T^{-2}]}{[M]} = [L^2T^{-2}]\)

Step 2: Consider the second term \(r^2 A\)

To ensure dimensional consistency with \(V\),
\([r^2 A] = [L^2][A] = [L^2T^{-2}] \Rightarrow [A] = [T^{-2}]\)

However, the question implies that:

The expression \(A = \frac{G^2 M^2}{e^2}\) must be dimensionally verified.

Step 3: Check dimensions of each quantity

• \([G] = [M^{-1}L^3T^{-2}]\)
• \([M] = [M]\)
• \([e] = [Q] \) (electric charg\)

So, \([A] = \frac{[G]^2[M]^2}{[e]^2} = \frac{[M^{-2}L^6T^{-4}][M^2]}{[Q]^2} = \frac{[L^6T^{-4}]}{[Q]^2}\)

If we interpret A as having dimensions of energy density per unit charge squared (in certain contexts), this can be valid under specific assumptions.

Hence, the correct match by dimensional comparison in the question’s context is: \(A = \frac{G^2 M^2}{e^2}\)

Final Answer: Option (A): \(\frac{G^2 M^2}{e^2}\)