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

Answer: (A)

$\frac{G^2M^2}{e^2}$

Answer 2

From the given expression, V =− rGM ​+r 2 A ​, we can determine the dimensions of the constant A using dimensional analysis.

The dimensions of gravitational potential V are [M L2T−2[M L2T−2, where MM represents mass, LL represents length, and TT represents time.

The dimensions of the left-hand side GM ​ are [M L2T−2[M L2T−2.

The dimensions of the right-hand side r 2 A ​ are [M L2T−2[M L2T−2.

By equating the dimensions of both sides, we have: [M L2T−2[M L2T−2 = [M L2T−2[M L2T−2 + [M L2T−2[M L2T−2.

Simplifying, we get: [M L2T−2[M L2T−2 = [M L2T−2[M L2T−2.

This implies that the dimensions of constant A are [M L2T−2[M L2T−2.

Since gravitational constant G has dimensions [M−1 L3T−2[M−1 L3T−2, mass M has dimensions [M][M], and velocity of light c has dimensions L T−1L T−1, we can form a dimensionless quantity as: c 2 G 2 M 2​.

Comparing the dimensions of c 2 G 2 M 2​ ([M L2T−2[M L2T−2) with the dimensions of A ([M L2T−2[M L2T−2), we find that A =c 2 G 2 M 2​.

Hence, from dimensional analysis, A is equal to c 2 G 2 M 2​.

The correct option is(A): $\frac{G^2M^2}{e^2}$