Question

SI unit of gravitational constant is:
(A) $ N{m^2}k{g^2} $
(B) $ Nmk{g^2} $
(C) $ {N^2}mk{g^2} $
(D) $ N{m^2}k{g^{ - 2}} $

Answer 1

Hint : The gravitational constant is the constant of proportionality of Newton's universal gravitational law. Newton's universal gravitational law states that the force of attraction between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Formula used: In this solution we will be using the following formula;
 $ F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}} $ , $ F $ is the gravitational force of attraction between two masses $ {m_1} $ and $ {m_2} $ , $ r $ is the distance between their centres, and $ G $ is the gravitational constant.

Complete step by step answer
The gravitational constant is considered a fundamental constant of the universe. It is the constant of proportionality possessed by Newton’s law of the universal gravitation which states that the force of attraction between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between their centres. This is given mathematically as
 $ F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}} $ , $ F $ is the gravitational force of attraction between two masses $ {m_1} $ and $ {m_2} $ , $ r $ is the distance between their centres, and $ G $ is the gravitational constant.
Hence, to find the unit, we can replace the quantities with dimension, we can write them as
 $ \left[ F \right] = \left[ G \right]\dfrac{{\left[ {{m_1}} \right]\left[ {{m_2}} \right]}}{{{{\left[ r \right]}^2}}} $ where the square brackets signifies dimension of the various terms.
We have $ \left[ F \right] = \left[ G \right]\dfrac{{{M^2}}}{{{L^2}}} $ by making $ \left[ G \right] $ subject of formula
 $ \left[ G \right] = \left[ F \right]\dfrac{{{L^2}}}{{{M^2}}} $
Hence, by replacing the dimension with units we have
 $ {G_u} = N{m^2}k{g^{ - 2}} $
Hence, the correct option is D.

Note
Alternatively, we could directly use units, to solve the problem as in:
 $ N = {G_u}\dfrac{{kg \times kg}}{{{m^2}}} $ where $ {G_u} $ stands for the unit of $ G $
Hence, $ {G_u} = N\dfrac{{{m^2}}}{{k{g^2}}} = N{m^2}k{g^{ - 2}} $
It is easier and faster to deal with units directly. However, it is often considered unscientific and when we are told to perform a dimensional analysis, the dimensions should be used not units, even when the final answer is given in its unit. We can do this here because it’s a multiple choice question.