Question

Find the unit and dimensional formula of the universal gravitational constant

Answer 1

Step1: Calculating the dimensional formula of force

Acceleration is the rate of change of velocity with time so, we can say that $a=\frac{v}{t}$

Here velocity is divided by time again we will substitute the dimensional formulas of velocity band time to get the dimensional formula for acceleration

$$\begin{gathered} a=\left[\frac{L{T}^{-1}}{T}\right] \\ a=[L{T}^{-2}] \end{gathered}$$

Force is defined as the mass time of acceleration so,

$$\begin{gathered} Force=Mass\times Acceleartion \\ F=\left[M\right]\times \left[L{T}^{-2}\right] \\ Dimensionalformulaofforce=\left[M^1{L}^1{T}^{-2}\right]..........equation\left(i\right) \end{gathered}$$

Step 2: Calculating dimensional formula of universal gravitational constant

The gravitation constant is the proportionality constant that was used in Newton's law of gravitation.

$$\begin{gathered} G=\frac{F\times r^2}{m_1{m}_2} \\ G=F\times r^2\times {\left(m_1{m}_2\right)}^{-1}......equation\left(ii\right) \end{gathered}$$

On substituting equation ($i$) in equation ($ii$) we get

$$\begin{gathered} G=\left[M^1{L}^1{T}^{-2}\right]\times \left[L^2\right]\times {\left[M\right]}^{-2}As,Mass=\left[M\right] \\ G=\left[M^{-1}{L}^3{T}^{-2}\right] \end{gathered}$$

Hence, the unit of universal gravitational constant is $N{m}^2/k{g}^2$ and the dimensional of universal gravitational constant is $\left[M^{-1}{L}^3{T}^{-2}\right]$.