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,the 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 a dimensionless constant and $g$ is the acceleration due to gravity.

(A) According to Kepler's third law,$T^2 \propto r^3$,which implies $T \propto r^{3/2}$.
We assume $T$ is a function of $r$,$R$,and $g$ such that $T = k r^{3/2} R^a g^b$,where $k$ is a dimensionless constant.
Substituting the dimensions of each term: $[T] = [L]^{3/2} [L]^a [LT^{-2}]^b$.
$[M^0 L^0 T^1] = [L^{3/2 + a + b} T^{-2b}]$.
Comparing the powers of $T$: $1 = -2b \Rightarrow b = -1/2$.
Comparing the powers of $L$: $3/2 + a + b = 0 \Rightarrow 3/2 + a - 1/2 = 0 \Rightarrow a + 1 = 0 \Rightarrow a = -1$.
Substituting $a$ and $b$ back into the equation: $T = k r^{3/2} R^{-1} g^{-1/2}$.
Therefore,$T = \frac{k}{R} \sqrt{\frac{r^3}{g}}$.