The time period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius 4R is:

A) 4T
B) T/4
C) 8T
D) T/8

Solution:
We are given that the time period of a satellite in a circular orbit of radius RR is TT. We need to find the time period of another satellite in a circular orbit of radius 4R4R.
The formula for the time period TT of a satellite in a circular orbit around a planet is given by Kepler's Third Law:
T2∝R3T^2 \propto R^3
Where:
• TT is the time period of the satellite.
• RR is the radius of the orbit.
This means:
T12T22=R13R23\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}
Let the time period of the second satellite be T2T_2, and its orbital radius be R2=4RR_2 = 4R. For the first satellite, the time period is T1=TT_1 = T, and the orbital radius is R1=RR_1 = R.
Substitute these into the formula:
T2T22=R3(4R)3\frac{T^2}{T_2^2} = \frac{R^3}{(4R)^3}
Simplifying the equation:
T2T22=R364R3\frac{T^2}{T_2^2} = \frac{R^3}{64R^3} T2T22=164\frac{T^2}{T_2^2} = \frac{1}{64}
Taking the square root of both sides:
TT2=18\frac{T}{T_2} = \frac{1}{8}
Thus:
T2=8TT_2 = 8T
Answer:
The time period of the satellite in the orbit of radius 4R4R is 8T.
Thus, the correct answer is: C. 8T.