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 = k/R √r3/g where k is a dimensionless constant and g is acceleration due to gravity.

Home » 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 = k/R √r3/g where k is a dimensionless constant and g is acceleration due to gravity.

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 = k/R √r3/g where k is a dimensionless constant and g is acceleration due to gravity.

Kepler’s third law states that,

$T^{2} \propto a^{3}$ 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 $\left(a^{3}\right)$ .

On applying Kepler’s third law,

$T^{2} \propto r^{3} \Rightarrow T \propto r^{3 / 2}$

Also, $T$ depends on $R$ and $g$ .

Let $T \propto r^{3 / 2} g^{a} R^{b}$
$\Rightarrow \quad T=k r^{3 / 2} R^{a} g^{b}$ …..(i)

where,
$k=$ dimensionless constant.

On writing the dimensions of various quantities on both the sides, we get

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