Assume that the force of gravitation `F prop 1/(r^(n))` . Then show that the orbital speed in a circular orbit of radius r is proportional to ` 1/(r^((n-1)//2))` , while its period T is proportional to ` r^((n+1)//2)`

Explanation

Kepler's first law: The orbit of a planet is an ellipse with the Sun at one of the foci.
The adjoining figure shows the elliptical orbit of a planet revolving around the Sun. S denotes the position of the Sun.
Kepler's second law : The line joining the planet and the Sun Sweeps equal areas in equal intervals of time.
Area ASB=area CSD=area ESF.
Kepler's third law :
The square of the period of revolution of a plant around the sun is directly proportional to the cube of the mean distance of the planets from the Sun.
Thus , if r is the average distance of the planet from the Sun and T is its period of revolution , then,
`T^(2) prop r^(3), ` i.e. `(T^(2))/(r^(3))` =constant =K
For simplicity we shall assume the orbit to be a circle. In the adjoining figure, S denotes the poistion of the Sun, P denotes the position of a planet at a given instant and r denotes the radius of the orbit (`-=` the distance of the planet from the Sun ). Here, the speed of the planet is uniform. It is
`v=("circumference of the circle")/("period of revolution of the planet")=(2pir)/(T)`
If m is the mass of the planet , the centripetal force exerted on the planet by the Sun (`-=` gravitational force), `F=(mv^(2))/(r)`
`therefore F=(m(2pir//T)^(2))/(r)=(4pi^(2)mr^(2))/(T^(2)r) =(4pi^(2)mr)/(T^(2))`
Accordign to Kepler's third law,
`T^(2)=Kr^(3)`
`therefore F=(4^(2)mr)/(Kr^(3))=(4pi^(2)m)/(K)((1)/(r^(2)))`
Thus, `F prop (1)/(r^(2))` as `(4pi^(2)m)/(K)` is constant in a particular case.