Question #234773
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
1
Expert's answer
2021-09-08T16:13:50-0400

Given:

v=0.03cv=-0.03c

f0=6107Hzf_0=6*10^7\:\rm Hz


The Doppler shift of a frequency

f=f01+v/c1v/cf=f_0\sqrt{\frac{1+v/c}{1-v/c}}

So

f=6.0010710.031+0.03=5.82107Hzf=6.00*10^7\sqrt{\frac{1-0.03}{1+0.03}}=5.82*10^7\:\rm Hz


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