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  • An astronaut is approaching the moon. He sends out a radio signal of frequency <img alt="5 \times 10^9 \mathrm{~Hz}" src="https://entrancecorner.oncodecogs.com/gif.latex?5%20%5Ctimes%2010%5E9%

An astronaut is approaching the moon. He sends out a radio signal of frequency 5 \times 10^9 \mathrm{~Hz} and finds that the frequency shift in the echo received is 10^3 \mathrm{~Hz}. Then the speed of approach is 

Option: 1

40m/s


Option: 2

120m/s


Option: 3

30m/s


Option: 4

50m/s


Answers (1)

best_answer

Actual frequency of  radio signal = 5 \times 10^9 \mathrm{~Hz}=n

frequency n_m=n \sqrt{\frac{1+v / c}{1-v / c}}    ....(1)

frequency of echo n_a=n_m \sqrt{\frac{1+v / c}{1-v / c}}   ....(2)

from (1) and (2) \frac{n_a}{n}=\frac{1+v / c}{1-v / c}=\frac{c+v}{c-v}

V=c \cdot \frac{n_a-n}{n_a+n}=3 \times 10^8 \times \frac{10^3}{10^3+10^{10}}=30 \mathrm{~m} / \mathrm{s}

 

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Rishabh

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