Question

The light from a nearby star is observed to be red-shifted by 10% Is the star approaching or receding from the earth? How fast is it moving?

Answer 1

3.149×10^7 m/s

Step-by-step explanation:

the frequency is shifted to lower value hence the star is receding the Earth.

`f=f_0\sqrt{(1-\beta)/(1+\beta) }`................1

f= 0.9 f_0

putting the value in 1 we get

`0.9f_0=f_0\sqrt{(1-\beta)/(1+\beta) }`.

`0.9=\sqrt{(1-\beta)/(1+\beta) }`.

on solving we get

β= 0.1049

now

`(v)/(c) = \beta`

v= 0.1049×3×10^8= 3.149×10^7 m/s