An Earth satellite S has Orbit radius four 4 that of communication satellite C. The period of revolution of S will be.
A. 32 days
B. 18 days
C. 8 days
D. 9 days

Hint: Use the formula of Kepler’s law. For this question students know about Kepler’s third Law. Use the concept of elliptical motion of orbit.

Complete step by step solution:
Kepler’s Third Law – The Law of Periods:
Shorter the orbit of the planet around the sun, shorter the time taken to complete one revolution. According to Kepler’s law of periods, the square of the time period of revolution (of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis).
 \[{{T}^{2}}\propto {{R}^{3}}\]
For this question student know about Kepler’s third Law that of the square of time period in proportional to the cube of the semi major areas of the orbit \[{{T}^{2}} \propto {{R}^{3}}\]
Where \[T\Rightarrow Time\ period\]
 \[R\Rightarrow Semi\ major\ areas\]
In bounded motion, the particle has negative total energy (E<0) and has two or more extreme points where the total energy is always equal to the potential energy of the particle i.e the kinetic energy of the particle becomes zero.
For eccentricity 0≤ e <1, E<0 implies the body has bounded motion. A circular orbit has eccentricity e = 0 and elliptical orbit has eccentricity e < 1.
In unbounded motion, the particle has positive total energy (E>0) and has a single extreme point where the total energy is always equal to the potential energy of the particle i.e the kinetic energy of the particle becomes zero.
For eccentricity e ≥ 1, E > 0 implies the body has unbounded motion. Parabolic orbit has eccentricity e = 1 and Hyperbolic path has eccentricity e>1.
From Kepler’s Third law
 \[{{T}^{2}} \times {{R}^{3}}\]
 \[\Rightarrow {{\left( \dfrac{T_c}{T_c} \right)}^{2}}\pm {{\left( \dfrac{R_s}{R_c} \right)}^{3}}\]
 \[R_s = 4R_c \]
 \[\Rightarrow \dfrac{T_s}{T_c}={{\left( \dfrac{4Rc}{Rc} \right)}^{{}^{3}/{}_{2}}}\]
 \[\Rightarrow \dfrac{T_s}{T_c}={{\left( 4 \right)}^{{}^{3}/{}_{2}}}
\therefore \dfrac{T_s}{T_c}=8\]
for \[T_c=1\] day
$\therefore$ \[Ts=8\] days

Option (C) is the correct answer.

Note: Students should have a clear understanding of Kepler’s Third Law. Students should also consider that \[T_c=1\] day. Don’t confuse between the Kepler’s laws.