Answer to Question #234773 in Astronomy | Astrophysics for Sridhar

Question

A spaceship is moving away from Earth with speed 0.03c where c is speed of light in air.The spaceship emits signal of frequency 6×10^(7)Hz. Frequency of signal observed by the observer on Earth will be

Answer 1

Given:

$v=-0.03c$

$f_0=6*10^7\:\rm Hz$

The Doppler shift of a frequency

$$f=f_0\sqrt{\frac{1+v/c}{1-v/c}}$$

So

$$f=6.00*10^7\sqrt{\frac{1-0.03}{1+0.03}}=5.82*10^7\:\rm Hz$$