Question

If speed of light, acceleration due to gravity and pressure are taken as fundamental quantity, then what are the dimensions of gravitational constant

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{Let}\mathrm{us}\mathrm{assume}, \\ \mathrm{G}={\mathrm{c}}^{\mathrm{x}}{\mathrm{g}}^{\mathrm{y}}{\mathrm{P}}^{\mathrm{z}} \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{G}=\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right] \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{c}=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right] \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{g}=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-2}\right] \\ \mathrm{Dimensional}\mathrm{formula}\mathrm{of}\mathrm{P}=\left[{\mathrm{M}}^1{\mathrm{L}}^{-1}{\mathrm{T}}^{-2}\right] \\ \left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]={\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]}^{\mathrm{x}}{\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]}^{\mathrm{y}}{\left[{\mathrm{M}}^1{\mathrm{L}}^{-1}{\mathrm{T}}^{-2}\right]}^{\mathrm{z}} \\ \left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]=\left[{\mathrm{M}}^{0+0+\mathrm{z}}{\mathrm{L}}^{\mathrm{x}+\mathrm{y}-\mathrm{z}}{\mathrm{T}}^{-\mathrm{x}-2\mathrm{y}-2\mathrm{z}}\right] \\ \mathrm{Comparing}\mathrm{the}\mathrm{coefficients}, \\ \mathrm{z}=-1.........\left(1\right) \\ \mathrm{x}+\mathrm{y}-\mathrm{z}=3.........\left(2\right) \\ -\mathrm{x}-2\mathrm{y}-2\mathrm{z}=-2............\left(3\right) \\ \mathrm{Substitute}\mathrm{equation}\left(1\right)\mathrm{in}\left(2\right)\&\left(3\right), \\ \mathrm{x}+\mathrm{y}=2........\left(4\right) \\ -\mathrm{x}-2\mathrm{y}=-4 \\ \mathrm{or},\mathrm{x}+2\mathrm{y}=4........\left(5\right) \\ \mathrm{from}\left(4\right)\&\left(5\right),\mathrm{we}\mathrm{get}, \\ \mathrm{y}=2 \\ \mathrm{x}=0 \\ \mathrm{Hence},\mathrm{G}={\mathrm{c}}^0{\mathrm{g}}^2{\mathrm{P}}^{-1} \end{gathered}$$