Question

An earth satellite S has an orbit radius which is 4 times that of a communication satellite C.the period of revolution of S is

Answer 1

Answer:

8Tc

Explanation:

Satellite of earth = S (Given)

Satellite of communication = C (Given)

Radius = 4 times of the satellite C (Given)

As per Kepler’s third law, the square of the period of any planet is proportional to the cube of the semimajor axis of its orbit. This phenomena captures the relationship between the distance of planets from the Sun, and their totalorbital periods.

Thus, (Ts/Tc)² = (Rs/Rc)³

= (Ts/Tc)² = ( 4Rc/Rc)²

Ts/Tc = √64

Ts/Tc = 8

Therefore, the period of revolution of S is 8Tc.

Answer 2

An earth satellite of the orbit radius is 4 times of a ‘communication satellite C’, then ‘period of revolution’ of S is 8 days.

Given:  

Radius of the earth satellite is 4 times of a ‘communication satellite’.

To find:

The period of revolution of S is ?

Explanation:

As per Kepler’s third law, the “square of the period of any planet” is proportional to the “cube of the semi-major axis” of its orbit.  

Hence:

$\begin{aligned}\left(\frac{T_{s}}{T_{c}}\right)^{2} &=\left(\frac{R_{s}}{R_{c}}\right)^{3} \\\left(\frac{T_{s}}{T_{c}}\right)^{2} &=\left(\frac{4 R_{c}}{R_{c}}\right)^{3} \\\left(\frac{T_{s}}{T_{c}}\right) &=\sqrt{64} \\\left(\frac{T_{s}}{T_{c}}\right) &=8 \\ T_{s} &=8 T_{c} \end{aligned}$