Question

the radii of circular orbit of two sattelilites around the earthare in the ratio of 1:4 , then ratio of the respective periods of revolution?
A) 1:4
B) 4:1
C) 1;8
D) 8:1
​

Answer 1

Given :

• Ratio of radii of circular orbits = 1 : 4

To find :

Ratio of time periods of the revolution.

Solution :

We know that ,

According to the Kepler's third law of Planetary Motion , the square of the Period of revolution of a planet is directly proportional to the cube of the radius of the planets orbit i.e,

$\boxed{\bf{T^{2} \propto R^{3}}}$

Where :

• T = Time period of revolution.
• R = Radius of the planet.

In other case , if two planets have a period of revolution of $\bf{T_{1}}$ and $\bf{T_{2}}$ , and the average radii of their planet as , $\bf{R_{1}}$ and $\bf{R_{2}}$, then ;

$\boxed{\bf{\dfrac{T_{1}^{2}}{R_{2}^{3}} = \dfrac{T_{1}^{2}}{R_{2}^{3}}}}$

From the above equation , we get :

$\bf{\dfrac{T_{1}^{2}}{T_{2}^{2}} = \dfrac{R_{1}^{3}}{R_{2}^{3}}}$

Let the radius of the two planets be 1x and 4x. Now Substituting it in the equation, we get :

$:\implies \bf{\dfrac{T_{1}^{2}}{T_{2}^{2}} = \dfrac{(1x)^{3}}{(4x)^{3}}}$

$:\implies \bf{\dfrac{T_{1}^{2}}{T_{2}^{2}} = \dfrac{x^{3}}{64x^{3}}}$

$:\implies \bf{\dfrac{T_{1}^{2}}{T_{2}^{2}} = \dfrac{\not{x^{3}}}{64\not{x^{3}}}}$

$:\implies \bf{\dfrac{T_{1}^{2}}{T_{2}^{2}} = \dfrac{1}{64}}$

$:\implies \bf{\bigg(\dfrac{T_{1}}{T_{2}}\bigg)^{2} = \dfrac{1}{64}}$

$:\implies \bf{\dfrac{T_{1}}{T_{2}} = \sqrt{\dfrac{1}{64}}}$

$:\implies \bf{\dfrac{T_{1}}{T_{2}} = \dfrac{1}{8}}$

$:\implies \bf{T_{1} : T_{2} = 1 : 8}$

$\boxed{\therefore \bf{T_{1} : T_{2} = 1 : 8}}$

Hence the ratio of time periods of of the revolution of the planet around the sun is 1 : 8.