The escape velocity v of a body depends upon (i) the acceleration due to gravity of the planet and (ii) the radius of the planet R. Establish dimensionally the relationship between v, g and R.

Explanation

Let `v=Kg^(a)R^(b)`
Where K = a dimensionless constant Putting the dimensions,
`LT^(-1)=[LT^(-2)]^(a)[L]^(b)=L^(a+b)T^(-2a)`
Equating the powers of L and T,
`a+b=1, -2a=-1`
`therefore a=(1)/(2),b=(1)/(2)`
Hence `v=Kg^((1)/(2))R^((1)/(2))=Ksqrt(gR)`