Question

. An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit or radius r, from Keplers Third law about 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.

Answer 1

According to Kepler's third law,$T^2\infty a^3$ i.e., square if time period $\left(T^2\right)$ of a satellite revolving around a planet, is proportional to the cube of the radius of the orbit $\left(a^3\right)$
We have to apply Kelper's third lar,
$T^2\infty r^3\Rightarrow T\infty r^{3/2}$
Also, T depends on R and g.
Let $T\infty r^{3/2}{g}^a{R}^b$
$\Rightarrow T=k{r}^{3/2}{R}^a{g}^b$...............(i)
where k is a dimensionless constant of proportionality.
Writing the dimensions of various quantities on both the sides,we get
$\left[M^0{L}^{0}T\right]=\left[L{]}^{3/2}\right[L{T}^{-2}{]}^a[L{]}^b$
$=\left[M^0{L}^{a+b+3/2T^{-2a}}\right]$
On comparing the dimensions of both sides, we get
$a+b+\frac{3}{2}=0$...............(ii)
$-2a=1\Rightarrow a=-1/2$........(iii)
From eq. (ii), we get
$b-\frac{1}{2}+\frac{3}{2}=0\Rightarrow b=-1$
Substituting the values of a and b in eq. (i)

we get

$T=k{r}^{3/2}{R}^{-1}{g}^{-1/2}$

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