Question 6: Deduce the dimensions of gravitational constant.

## Answer

Newtonâ€™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 r^{2} = [L][L] = [L^{2}] —– (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}][L^{2}]/[M^{2}] = [ML^{3}T^{-2}][M^{-2}] = [ML^{3}T^{-2}M^{-2}] = [M^{-1}L^{3}T^{-2}]

OR

G = [M^{-1}L^{3}T^{-2}]

This is the required expression.

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