Q. The time period of a satellite of earth is 5h. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become
A) 10 h
B) 18 h
C) 40 h
D) 20 h
Explanation
(c) According to Kepler’s law
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T^{2} \propto r^{3}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5^{2} \propto r^{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
$and\ \ \ \ \ \ \ \ \ \ \left( T^{'} \right)^{2} \propto (4r)^{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
From Eqs.(i) and (ii), we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{25}{\left( T^{'} \right)^{2}} = \frac{r^{3}}{64r^{3}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = \sqrt{1600} = 40h\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.