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(A) The given equation is $V = -\frac{GM}{r} + \frac{A}{r^2}$.According to the principle of homogeneity of dimensions, the dimensions of each term on

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

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(A) The given equation is $V = -\frac{GM}{r} + \frac{A}{r^2}$.According to the principle of homogeneity of dimensions, the dimensions of each term on both sides must be the same.Therefore, the dimension of $\frac{A}{r^2}$ must be equal to the dimension of $\frac{GM}{r}$.$[V] = [\frac{GM}{r}] = [\frac{A}{r^2}]$From this, we get $[A] = [\frac{GM}{r}] 	imes [r^2] = [GM] 	imes [r]$.We know that the gravitational potential $V$ has the dimensions of energy per unit mass, which is $[L^2 T^{-2}]$.Also, $\frac{GM}{r}$ represents the gravitational potential, so $[\frac{GM}{r}] = [L^2 T^{-2}]$.Since the velocity of light $c$ has dimensions $[L T^{-1}]$, then $c^2$ has dimensions $[L^2 T^{-2}]$.Thus, $[\frac{GM}{r}] = [c^2]$.Substituting $[r] = \frac{[GM]}{[c^2]}$ into the expression for $[A]$:$[A] = [GM] 	imes \frac{[GM]}{[c^2]} = \frac{G^2 M^2}{c^2}$.

$A$ 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:

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