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Physics Question on Centripetal forces

Suppose the gravitational force varies inversely as the nth power of distance. Then, the time period of a planet in circular orbit of radius R around the sun will be proportional to

A

R(n+12)R^{\left(\frac{n+1}{2}\right)}

B

R(n−12)R^{\left(\frac{n-1}{2}\right)}

C

RnR^n

D

R(n−22)R^{\left(\frac{n-2}{2}\right)}

Answer

R(n+12)R^{\left(\frac{n+1}{2}\right)}

Explanation

Solution

The necessary centripetal force required for a planet to move round the sun
== gravitational force exerted on it
i.e., mv2R=GMemRn\frac{mv^{2}}{R} = \frac{GM_{e} m}{R^{n}}
v=(GMeRn−1)1/2v = \left(\frac{GM_{e}}{R^{n-1}}\right)^{1/2}
Now, T=2πRv=2πR×(Rn−1GMe)1/2T = \frac{2\pi R}{v} = 2\pi R \times \left(\frac{R^{n-1}}{GM_{e}}\right)^{1/2}
⇒2π(R2×Rn−1GMe)1/2\Rightarrow 2\pi \left(\frac{R^{2} \times R^{n-1}}{GM_{e}}\right)^{1/2}
T=2π(R(n+1)/2(GMe)1/2)T = 2\pi\left(\frac{R^{\left(n+1\right)/2}}{\left(GM_{e}\right)^{1/2}}\right)
∴T∝R(n+1)/2\therefore T \propto R^{\left(n+1\right)/2}