1.

Imagine a light planet revolving around a very massive star in a circular orbit of radius 'r' with period of revolution T. On what power of 'r' will the square of time period depend on the gravitational force of attraction between the planet and the star is proportional to r^(-5//2)

Answer»

SOLUTION :The gravitaitonal force provides necessary centripetal force
`(MV^(2))/(r) = (K)/(r^(5//2)) rarr V^(2) = (K)/(m r^(3//2))`
So that `T = (2pir)/(V) = 2PI rsqrt((MR^(3//2))/(K))`
So `T^(2) prop r^(7//2)`


Discussion

No Comment Found