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 dimensionless RV g constant and g is acceleration due to gravity.

Explanation

According to Kepler.s third law
`T^(2) prop r^(2) rArr T prop r^((3)/(2))`
We know that T is a function of R and G,
Let `T prop 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 equ. (i) we get
`[M^(0)L^(0)T]=[L]^((3)/(2))[L]^(a)+[LT^(-2)]^(b)`
`=[L^(a+b+(3)/(2))T^(-2b)]`
By comparing the powers
`a+2b+(3)/(2)=0 " "...(ii)`
`=-2b=1 rArr b=-(1)/(2) " "...(iii)`
Using equ. (ii)
`a-(1)/(2)+(3)/(2)=0 rArr a=-1`
By substituting the values of a and b in equ (i)
`T=kr^((3)/(2))R^(-1)g^((1)/(2))`
`:.T=(k)/(R) sqrt((r^(3))/(g))`