Question

What is the dimensional formula of gravitational constant?

$\text{A}\text{. }\left[ M{{L}^{2}}{{T}^{-2}} \right]$
$\text{B}\text{. }\left[ M{{L}^{1}}{{T}^{-1}} \right]$
$\text{C}\text{. }\left[ {{M}^{-1}}{{L}^{3}}{{T}^{-2}} \right]$
D. none of these

Answer 1

Hint: When we write an equation of any physical quantity, the left-hand-side and the right-hand-side of an equation have the same dimensional formula because the dimensional formula of a physical quantity can never change. Use this fact on the equation of force gravity.

Formula used:

$F=G\dfrac{{{M}_{1}}{{M}_{2}}}{{{r}^{2}}}$
$\text{Force = mass }\\!\\!\times\\!\\!\text{ acceleration}$
$\text{acceleration = }\dfrac{\text{change in velocity}}{\text{time}}$
$\text{velocity = }\dfrac{\text{change in distance}}{\text{time }\\!\\!\times\\!\\!\text{ time}}$

Complete step by step answer:
Gravitational constant is a universal proportionality constant used in the equation to show the relationship of the force of attraction between two bodies with their masses and the distance between them i.e. $F=G\dfrac{{{M}_{1}}{{M}_{2}}}{{{r}^{2}}}$ …..(1) ,where M1 and M2 are the masses of the bodies and r is the distance between them.

We can write equation (1) as $G=\dfrac{F{{r}^{2}}}{{{M}_{1}}{{M}_{2}}}$ …... (2).

Although we directly do not know the dimensional formula of the gravitational constant but we do know the dimensional formula of force, mass, and distance. The left-hand-side and the right-hand-side of an equation have the same dimensional formula. That means in equation (2) $G$ will have the same dimensional formula as the right-hand-side expression. So, now calculate dimensional formula of $\dfrac{F{{r}^{2}}}{{{M}_{1}}{{M}_{2}}}$ to get the dimensional formula for $G$.

We know the dimensional formula of mass and distance. Let us calculate the dimensional formula of force.

$\text{Force = mass }\\!\\!\times\\!\\!\text{ acceleration}$

$\Rightarrow \text{Force = mass }\\!\\!\times\\!\\!\text{ }\dfrac{\text{change in velocity}}{\text{time}}$

$\Rightarrow \text{Force = mass }\\!\\!\times\\!\\!\text{ }\dfrac{\text{change in distance}}{\text{time }\\!\\!\times\\!\\!\text{ time}}$

So, $[F]=[ML{{T}^{(-2)}}]$ (dimensional formula is always written inside square brackets),

where M, L, T are the dimensional formula of mass length and time respectively.

Therefore,

$[G]=\dfrac{[F][{{r}^{2}}]}{[{{M}_{1}}][{{M}_{2}}]}=\dfrac{[ML{{T}^{(-2)}}][{{L}^{2}}]}{[M][M]}=[{{M}^{-1}}{{L}^{3}}{{T}^{(-2)}}]$

Therefore, the dimensional formula of the gravitational constant ($G$) is $[{{M}^{-1}}{{L}^{3}}{{T}^{(-2)}}]$.

Hence, the correct option is C.

Now, we can also calculate the unit of the gravitational constant. The unit of $G$ will be $\dfrac{{{m}^{3}}{{s}^{(-2)}}}{kg}$.

Note: We can calculate the dimensional formula of unknown quantities with the help of known quantities. We can even calculate the unit of the unknown quantity with the same method.