Q. A satellite in launched in a circular orbit of radius $R$ around the earth. A second satellite is launch in to an orbit of radius 1.01R . The period of second satellite is longer than the first one (approximately) by
A) 1.5 %
B) 0.5%
C) 3%
D) 1%
Explanation
(a) From Kepler’s law, $T^{2} \propto R^{3}$
$or\ \ \ \ \ \ \left( \frac{T_{2}}{T_{1}} \right)^{2} = \left( \frac{R_{2}}{R_{1}} \right)^{3} = \left( \frac{1.01\ R}{R} \right)^{3} = (1 + 0.01)^{3}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \frac{T_{2}}{T_{1}} = (1 + 0.01)^{3/2} = 1 + \left( \frac{3}{2} \times 0.01 \right)$
$or\ \ \ \frac{T_{2} - T_{1}}{T_{1}} = \frac{1.5}{100} = 1.5\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.