Menu Close

Question 6: Deduce the dimensions of gravitational constant.


Newton’s law of Universal Gravitation is

gravitational force

Here G is the gravitational constant. We have to find the dimensions of G. Rearranging the above equation
gravitational constant

Now a consistent equation has same dimensions on both sides. So we find the dimensions on RHS.

Since F = ma. And dimensions of mass = [M]

Acceleration is measured in ms-2, therefore, dimensions of acceleration = [LT-2]

Therefore, dimensions of F = [M][LT-2] —– (i)

Dimensions for r2 = [L][L] = [L2] —– (ii)

Dimensions of mass = [M] —– (iii)

Now substitute the dimensions from (i), (ii) and (iii) in the above equation for G, we get

Dimensions of G = [M][LT-2][L2]/[M2] = [ML3T-2][M-2] = [ML3T-2M-2] = [M-1L3T-2]

G = [M-1L3T-2]
This is the required expression.


  1. Pingback:conceptual-questions-of-measurement-chapter-1-physics-11 – msa

  2. Pingback:dimension-of-kinetic energy … msa – msa

  3. Pingback:Dimensional consistency of Einstein-formula … msa – msa

Leave a Reply

Your email address will not be published. Required fields are marked *