Question 6: Deduce the dimensions of gravitational constant.
Newton’s law of Universal Gravitation is
Here G is the gravitational constant. We have to find the dimensions of G. Rearranging the above equation
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.