A planet is rotating round a heavy star along a circular path of radius R with time T. if the gravitational force of attraction between planet and star is proportional to `R^(-5/2)` , then `T^2 prop R^x`. What is the value of x ?

Explanation

Gravitational force `=kR^(-5//2)` [k=constant]
If the mass of the planet is m and its velocity is v, then centripetal force `=(mv^2)/R`.
So, `(mv^2)/R=kr^(-5//2) or, v=sqrt(k/mR^(-3//2))`
Time period,
`T=(2pi R)/v =2pi sqrt(m/k)sqrt(R^2R^(3//2)) = "constant" xxsqrt(R^(7//2))`
`therefore T^2prop R^(7//2)`
So, value of `x=7/2`.