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 = k/R sqrt(r^3/g)`. where k is a dimensionless constant and g is acceleration due to gravity.

Answer 1

By Kepler's third law,

`T^2 ∝ r^3` ⇒ `T ∝ r^(3/2)`

We know that T is a function of R and g.

Let `T ∝ r^(3/2) R^a g^b`

⇒ `T = kr^(3/2) R^a g^b`  ......(i)

Where k is a dimensionless constant of proportionality.

Substituting the dimensions of each term in equation (i), we get

`[M^0L^0T] = k[L]^(3/2) [L]^a [LT^-2]^b`

= `k[L^(a+ b + 3/2 T^-2b)]`

On comparing the powers of same terms, we get

`a + b + 3/2` = 0  ......(ii)

`- a2b` = 1 ⇒ b = `- 1/2`  ......(iii)

From equation (ii), we get

`a - 1/2 + 3/2` = 0 ⇒ a = – 1

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

`T = kr^(3/2) R^-1 g^(-1/2)`

⇒ `T = k/R sqrt(r^3/g)`