Question

let the period of revolution of a planet at a distance R from a star be T. Prove that if it was a diatance of 2R from the star, it's revolution will be √8T​

Answer 1

When a planet of mass m revolves along an orbit of radius R, it experiences centripetal force which is necessary provided by the force of gravitation.

$\sf{\longrightarrow \dfrac{mv^2}{R}=\dfrac{GMm}{R^2}}$

$\sf{\longrightarrow v=\sqrt{\dfrac{GM}{R}}}$

The time period of the planet is,

$\sf{\longrightarrow T=\dfrac{2\pi R}{v}}$

$\sf{\longrightarrow T=\dfrac{2\pi R}{\sqrt{\dfrac{GM}{R}}}}$

$\sf{\longrightarrow T=2\pi R\sqrt{\dfrac{R}{GM}}}$

$\sf{\longrightarrow T=2\pi\sqrt{\dfrac{R^3}{GM}}}$

Then we get,

$\sf{\longrightarrow T\propto\sqrt{R^3}}$

$\sf{\longrightarrow T\propto R^\frac{3}{2}}$

Therefore,

$\sf{\longrightarrow \dfrac{T_2}{T_1}=\dfrac{(R_2)^\frac{3}{2}}{(R_1)^{\frac{3}{2}}}}$

$\sf{\longrightarrow T_2=T_1\times\dfrac{(R_2)^\frac{3}{2}}{(R_1)^{\frac{3}{2}}}}$

Here,

• $\sf{R_1=R}$

• $\sf{R_2=2R}$

• $\sf{T_1=T}$

Then,

$\sf{\longrightarrow T_2=T\times\dfrac{(2R)^\frac{3}{2}}{R^{\frac{3}{2}}}}$

$\sf{\longrightarrow T_2=T\times\dfrac{2^\frac{3}{2}R^\frac{3}{2}}{R^{\frac{3}{2}}}}$

$\sf{\longrightarrow T_2=2^\frac{3}{2}\,T}$

$\sf{\longrightarrow\underline{\underline{T_2=\sqrt8\,T}}}$

Hence the Proof!