Question

If the unit of length be doubled then the numerical value of the universal gravitation constant G will become (with respect to present value)

• A. Double
• B. Half
• C. 8 times
• D. 1/8 times

Answer 1

Answer: (D)

1/8 times

Answer 2

The universal gravitation constant $G$ has dimensions $[G] = [M^{-1} L^3 T^{-2}]$. Let the original units of mass, length, and time be $u_M, u_L, u_T$ respectively. Let the numerical value of $G$ in this system be $N$. So, $G = N \ u_M^{-1} u_L^3 u_T^{-2}$.

If the unit of length is doubled, the new unit of length becomes $u'_L = 2 u_L$. The units of mass and time remain unchanged: $u'_M = u_M$ and $u'_T = u_T$. Let the new numerical value of $G$ be $N'$. Then, $G = N' \ u'_M{}^{-1} (u'_L)^3 u'_T{}^{-2}$. Substituting the new units: $G = N' \ u_M^{-1} (2 u_L)^3 u_T^{-2}$ $G = N' \ u_M^{-1} (8 u_L^3) u_T^{-2}$ $G = 8 N' \ u_M^{-1} u_L^3 u_T^{-2}$.

Equating the two expressions for $G$: $N \ u_M^{-1} u_L^3 u_T^{-2} = 8 N' \ u_M^{-1} u_L^3 u_T^{-2}$ $N = 8 N'$ $N' = \frac{N}{8}$.

Thus, the numerical value of $G$ becomes $\frac{1}{8}$ times its present value.