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

From the given expression, V=−rGM​+r2A​, 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 r2A​ 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: c2G2M2​.

Comparing the dimensions of c2G2M2​ ([M L2T−2[M L2T−2) with the dimensions of A ([M L2T−2[M L2T−2), we find that A=c2G2M2​.

Hence, from dimensional analysis, A is equal to c2G2M2​.

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

Answer 2

From the given expression, \(V = \frac{GM}{r}\) + \(\frac{A}{r^2}\), we can determine the dimensions of the constant A using dimensional analysis.

The dimensions of gravitational potential V are [M L²T^-2], where M represents mass, L represents length, and T represents time.

The dimensions of the term GM/r are also [M L²T^-2].

The dimensions of the term A/r² must be [M L²T^-2].

By equating the dimensions of both sides, we get: [M L²T^-2] = [M L²T^-2].

This implies that the dimensions of constant A are [M L⁴T^-2].

Since the gravitational constant G has dimensions [M^-1 L³T^-2], mass M has dimensions [M], and the speed of light c has dimensions [L T^-1], we can form a quantity c²G²M² with dimensions [M L⁴T^-2].

Comparing the dimensions of c²G²M² with the dimensions of A, we find that A = c²G²M².

Hence, from dimensional analysis, A is equal to c²G²M².

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