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where fs is the frequency of the source and fo is the observed frequency. Again, the top sign is associated with objects moving towards each other and the bottom sign associated with moving away. Suppose a space probe emits a radio signal at a frequency of 4.500000 MHz (Megahertz). It travels away from Earth with a speed of 2.000×104 m/s. What is the frequency of the signal when it is detected on Earth? (Note: give 7 significant digits)

Respuesta :

Answer:

Observed frequency is 4.499700 MHz

Explanation:

Doppler effect is defined as the change in the frequency or wavelength of the wave as the source or/and observer are moving away or towards each other.

Given :

Frequency of the source, [tex]f_{s}[/tex] = 4.500000 MHz

Speed of the source, v = 2.00 x 10⁴ m/s

Consider f₀ be the observed frequency.

Since, the source is moving away from the observer, so the relation of relativistic Doppler Effect is:

[tex]f_{0} =f_{s} \sqrt{\frac{c-v}{c+v} }[/tex]

Here c is speed of light and its value is 3 x 10⁸ m/s.

Substitute the suitable values in the above equation.

[tex]f_{0} =4.5 \sqrt{\frac{3\times10^{8} -2\times10^{4}}{3\times10^{8}+2\times10^{4}} }[/tex]

f₀ = 4.499700 MHz

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