A star moves away from Earth at a speed of 0. | vedclass

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Question

(B) For a source moving away from the observer at a relativistic speed $v$,the observed frequency $
u'$ is given by the relativistic Doppler effect f

Vedclass · Accepted Answer

$2.0$

Vedclass · Answer

(B) For a source moving away from the observer at a relativistic speed $v$,the observed frequency $
u'$ is given by the relativistic Doppler effect formula:$
u' = 
u \sqrt{\frac{1 - \beta}{1 + \beta}}$,where $\beta = \frac{v}{c}$.Given $
u = 6 	imes 10^{14} 	ext{ Hz}$ and $v = 0.8 c$,we have $\beta = 0.8$.Substituting the values:$
u' = 6 	imes 10^{14} \sqrt{\frac{1 - 0.8}{1 + 0.8}}$$
u' = 6 	imes 10^{14} \sqrt{\frac{0.2}{1.8}}$$
u' = 6 	imes 10^{14} \sqrt{\frac{1}{9}}$$
u' = 6 	imes 10^{14} 	imes \frac{1}{3} = 2 	imes 10^{14} 	ext{ Hz}$.Thus,the observed frequency is $2 	imes 10^{14} 	ext{ Hz}$.

$A$ star moves away from Earth at a speed of $0.8 c$ while emitting light of frequency $6 \times 10^{14} \text{ Hz}$. What frequency will be observed on the Earth (in units of $10^{14} \text{ Hz}$)? ($c$ is the speed of light)

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