Q. A satellite is launched into a circular orbit of radius $R$ around the earth. A second satellite is launched into an orbit of radius $1.03R$. The time period of revolution of the second satellite is larger than the first one approximately by
A) $9\%$
B) $3\%$
C) $4.5\%$
D) $2.5\%$
Explanation
[Solution by Dr. Abubacker] $T\propto r^{3/2}$, so $\frac{T_2}{T_1}=(1.03)^{3/2}\approx1+1.5(0.03)=1.045$, an increase of about $4.5\%$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.