Question

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 a dimensionless constant and g is acceleration due to gravity?

Answer 1

Kepler’s third law According to Kepler's third law, $T^2\alpha a^3$ i.e., square of time period $\left(T^2\right)$ of a satellite revolving around a planet, is proportional to the cube of the radius of the orbit $\left(a^3\right)$

As we know Kepler's third law, $T^2\alpha r^3\Rightarrow T\alpha r^{3/2}$ Also, T depends on R and g. Let $T\alpha r^{3/2}{g}^a{R}^b$

Then,$T=k{r}^{3/2}{g}^a{r}^b$

where, k is a dimensionless constant of proportionality. By putting the dimensions of various quantities on both the sides, we get

$\left[M^0{L}^{0}T\right]=\left[L{]}^{3/2}\right[L{T}^{-2}{]}^a\left[L{]}^b\right[M^0{L}^{0}T]=[M^0{L}^{a+b+3/2}{T}^{-2a}]$

On comparing the dimensions of both sides, we get

$a+b+\frac{3}{2}=0\dots \cdots .\left(i\right)$
$-2a=1\dots \cdots .\left(ii\right)$

$a=-\frac{1}{2}$
From Eq. (ii), we get

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

$b=-1$

Substituting the values of a and b in Eq. (i), we get

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

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

Final Answer: $T=\frac{k}{R}\sqrt{\frac{r^3}{g}}$