An artificial satellite is revolving around a planet of mass $M$ and radius $R$ in a circular orbit of radius $r$. From Kepler’s third law about the period of a satellite around a common central body, square of the period of revolution $T$ is proportional to the cube of the radius of the orbit $r$. Show using dimensional analysis that $T\, = \,\frac{k}{R}\sqrt {\frac{{{r^3}}}{g}} $, where $k$ is dimensionless constant and $g$ is acceleration due to gravity.

According to Kepler's third law,

$\mathrm{T}^{2} \propto r^{3} \Rightarrow \mathrm{T} \propto r^{\frac{3}{2}}$

We know that $\mathrm{T}$ is a function of $\mathrm{R}$ and $\mathrm{g}$,

Let, $\quad \mathrm{T} \propto r^{\frac{3}{2}} \mathrm{R}^{a} g^{b}$

$\therefore \mathrm{T}=k r^{\frac{3}{2}} \mathrm{R}^{a} g^{b}$

where $k$ is a dimensionless constant of proportionality. Substituting the dimensions of each term in equ. $(i)$, we get

$\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}\right] =[\mathrm{L}]^{\frac{3}{2}}[\mathrm{~L}]^{a}\left[\mathrm{LT}^{-2}\right]^{b}$

$=\left[\mathrm{L}^{a+b+\frac{3}{2}} \mathrm{~T}^{-2 b}\right]$

By comparing the powers,

$a+2 b+\frac{3}{2}=0$

$\quad=-2 b=1 \Rightarrow b=-\frac{1}{2}$

Using equ. $(ii)$,

$a-\frac{1}{2}+\frac{3}{2}=0 \Rightarrow a=-1$

By substituting the values of $a$ and $b$ in equ. $(i)$,

$\mathrm{T} =k r^{\frac{3}{2}} \mathrm{R}^{-1} g^{-\frac{1}{2}}$

$\therefore \mathrm{T} =\frac{k}{\mathrm{R}} \sqrt{\frac{r^{3}}{g}}$