An artificial satellite is revolving around a planet of mass M and radius R, in a circular orbit of radius r. From Kelper'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 usnig dimensional analysis, that `T = (k)/(R ) sqrt((r^3)/(g)),` Where k is a dimensionless constant and g is acceleration due to gravity.

Explanation

By kepler's third law, `T^(2)proptor^(3)RightarrowTproptor^(3//2)`
We know that T is a function of R and g.
`"Let"" "T propto r^(3//2)R^(a)g^(b)`
`Rightarrow T=kr^(3//2)R^(a)g^(b)" "...(i)`
where, k is a dimensionless constant of proportionality.
Substituting the dimensions of each term in Eq.(i), we get
`[M^(0)L^(0)T]=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)`
`-2b=1Rightarrowb=-1//2" "...(iii)`
From Eq. (ii), we get
`a-1//2+3//2=0Rightarrowa=-1`
Substituting the values of a and b in Eq. (i), we get
`T=kr^(3//2)R^(-1)g^(-1//2)`
`Rightarrow T=k/Rsqrt((r^(3))/g)`