A planet moves around the sun in nearly circular orbit. Its period of revolution 'T' depends upon. (i) radius 'r' or orbit, (ii) mass 'm' of the sun and. (iii) The gravitational constant G show dimensionally that `T^(2) prop r^(3)`.

Explanation

Let `T=Kr^(a)M^(b)G^(c )" …(1)"`
Where `K=a` dimensionless constant
dimensions of the various quantities are [T] `=T,[r]=L, [M]=M`
`[G]=(Fr^(2))/(m_(1)m_(2))=(MLT^(-1)L^(2))/(MM)`
`=M^(-1)L^(3)T^(-2)`
Substituting these dimensionally in equation (1) we get,
`[T]=[L]^(a)[M]^(b)[M^(-1)L^(3)T^(-2)]^(c )`
`M^(0)L^(0)T^(1)=M^(b-c)=M^(b-c)=M^(b-a)L^(a+3c)T^(-2c)`
Equating the dimensions of M, L and T, we get `b-c=0, a+3c=0,-2c=1`
on solving, `a=(3)/(2), b=-(1)/(2), c=-(1)/(2)`
`therefore T=Kr^(3//2)M^(-1//2)G^(-1//2)`
or `T^(2)=(K^(2)R^(3))/(MG)rArr therefore T^(2) prop r^(3)`