Gravitation Question 10
Question 10 - 29 January - Shift 2
The time period of a satellite of earth is 24 hours. If the separation between the earth and the satellite is decreased to one fourth of the previous value, then its new time period will become.
(1) 4 hours
(2) 6 hours
(3) 12 hours
(4) 3 hours
Show Answer
Answer: (4)
Solution:
Formula: Law of periods
$T^{2} \propto R^{3}$
$\frac{T_1^{2}}{T_2^{2}}=\frac{R_1^{3}}{R_2^{3}} \Rightarrow(\frac{T_1}{T_2})^{2}=(\frac{R}{\frac{R}{4}})^{3}$
$\therefore \quad \frac{T_1^{2}}{T_2^{2}}=64$
$\therefore \quad T_2^{2}=\frac{T_1^{2}}{64}$
$\therefore \quad T_2=\frac{24}{8}=3$
