Q. An earth satellite emits a radio signal of frequency $10^{8}$ Hz. An observer on the ground detects beats between the received signal and a local signal also of frequency $10^{8}$ Hz. At a particular moment, the beat frequency is 2400 Hz. What is the component of satellite velocity directed towards earth at this moment?
A) $1080\ ms^{- 1}$
B) $1800\ ms^{- 1}$
C) $3600\ ms^{- 1}$
D) $7200\ ms^{- 1}$
Explanation
(d) Here, $v - v^{'} = \mathrm{\Delta}v = 2400\ Hz$
As $\mathrm{\Delta}v = \frac{vv}{c}$
or $v = \frac{c\mathrm{\Delta}v}{v} = \frac{3 \times 10^{8} \times 2400}{10^{8}} = 7200\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.