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You've entered the Great Space Race. Your engines are hearty enough to keep you in second place. While racing, the person in front of you begins to have engine troubles and turns on his emergency lights that emit at a frequency of 5.990 1014 Hz. If the person in front of you is traveling 2640 km/s faster than you when he turns on his lights, what is the frequency of the emergency lights that you observe when it reaches you in your spaceship? (Enter your answer to four significant figures.)

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Answer

The frequency of the emergency lights that you observe when it reaches you in your spaceship =  5.94 × 10 ¹⁴ Hz

Explanation:

Using Relativistic Doppler's effect equation;

[tex]\frac{f_s}{f_o}= \sqrt {\frac{1+ \beta}{1-\beta} }[/tex]

where;

[tex]\beta =[/tex] v/c

[tex]\beta =[/tex]  2640 / 3×10⁵

[tex]\beta =[/tex] 0.0088

Rearranging the equation for frequency observed;

we have:

f_o = f_s [ ( √1 + [tex]\beta[/tex])/ (√1 - [tex]\beta[/tex]) ]⁻¹

f_o = (5.99 × 10 ¹⁴ Hz)   [ ( √1 + 0.0088)/ (√1 - 0.0088) ]⁻¹

f_o = 5.94 × 10 ¹⁴ Hz

Therefore, The frequency of the emergency lights that you observe when it reaches you in your spaceship =  5.94 × 10 ¹⁴ Hz

The frequency of the emergency lights that you observe when it reaches you in your spaceship is  5.94 × 10 ¹⁴ Hz.

Relativistic Doppler's effect equation:

[tex]\frac{f_s}{f_o}= \sqrt {\frac{1+ \beta}{1-\beta} }[/tex]

where;

[tex]\beta = v/c[/tex]

[tex]\beta = 2640 / 3*10^5\\\\\beta = 0.0088[/tex]

On rearranging:

[tex]f_o = f_s [ ( \sqrt{1+ \beta)} / (\sqrt{1 - \beta} ) ]^{-1}\\\\f_o = (5.99 *10^{14} Hz) [ ( \sqrt{1 + 0.0088)} / (\sqrt{1- 0.0088)} ]]^{-1}\\\\\f_o = 5.94 * 10^{14}Hz[/tex]

Therefore, the frequency of the emergency lights that you observe when it reaches you in your spaceship =  5.94 × 10 ¹⁴ Hz.

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