An artificial satellite is revolving round a planet of mass `M` and radius `R` in a circular orbit of radius `'r'` . If its period of revolution `T` obeys Kepler's law i.e. `T^(2)propr^(3).` What is relation for the period of revolution in terms of `R,r` and `"g"` the accelerations due to gravity on the planet ?

Options

A. `T=kr^(3/2)R^(1/2)g^(1/2)`
B. `T=kr^(3)Rg^(1/2)`
C. `T=kr^(3/2)R^(-1)g^(1/2)`
D. `T=kr^(3/2)R^(2)g`

Answer

C

Explanation

Here `T^(2)propr^(3)` ( Kepler.s law )
`:.Tpropr^(3//2)...(2)`
Also let `Tpropg^(a)R^(b)`
`:.Tpropr^(3//2)g^(a)R^(b)`
Putting the dimensions
`M^(0)L^(0)T^(1)=k[L]^(3//2)[L^(1)T^(-2)]^(a)[L^(1)]^(b)`
`=kL^(3//2+a+b)T^(-2a)`
Comparing the dimension
`3//2+a+b=0`
`3//2-(1)/(2)+b=0`
`:.b=-1`
`-2a=1`
`a=-(1)/(2)`
`:.T=kr^(3//2)g^(-1//2)R`
`=(k)/(R)sqrt((r^(3))/(g))`
Hence `(c )` is correct.