Dimension of mass in gravitational constant

Question

Professor PC Thomas Classes · Answer

$F = G \frac{m_1 m_2}{r^2} \implies G = \frac{Fr^2}{m_1 m_2}$. Dimension of $G = \frac{[\mathrm{MLT}^{-2}][\mathrm{L}^2]}{[\mathrm{M}^2]} = [\mathrm{M}^{-1}\mathrm{L}^3\mathrm{T}^{-2}]$. The dimension of mass is $-1$.