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An alien civilization occupies a planet circling a brown dwarf, several light-years away. The plane of the planet's orbit is perpendicular to a line from the brown dwarf to the Sun, so the planet is at nearly a fixed position relative to the Sun. The extraterrestrials have come to love broadcasts of MacGyver, on television channel 2, at carrier frequency 150. MHz. Their line of sight to us is in the plane of the Earth's orbit. Find the difference between the highest and lowest frequencies they receive due to the Earth's orbital motion around the Sun. (Assume the Earth has a mass of 1.99 1030 kg, and has a circular orbit around the Sun of radius 1.496 1011 m.)

Sagot :

Answer:

f'₁ = 149.985 10⁶ Hz,    f'₂ = 150.0149 10⁶ Hz

Explanation:

We can solve this problem using the relativistic doppler effect

          f’= f₀  [tex]\sqrt{ \frac{1- \beta }{1+ \beta } }[/tex]

          β = v/c

where the velocity is positive for the emitter and the receiver are moving away

In this case we must calculate the speed of the Earth, they tell us to consider the orbit as circular

          v = w r

the angular velocity is related to the period

          w = 2π / T

we substitute

           v = 2π r / T

Let's reduce the period to the SI system

           T = 1 year (365day / 1 year) (24h / 1day) (3600s / 1h) = 3.15 10⁷ s

         

we substitute

           v = 2π 1.496 10¹¹ / 3.145 10⁷

           v = 2.989 10⁴ m / s

We have two maximum possibilities when the planets approach and when the planets move away

the change of the emitted frequency

            β = 2.989 10⁴/3 10⁸

            β = 0.99625 10⁻⁴

we substitute in the equation to find the two frequencies

               f’= f₀  [tex]\sqrt{ \frac{1- \beta }{1+ \beta } }[/tex]

planets move away

             f’₁ = 150 10⁶ [tex]\sqrt{ \frac{1 - 0.99625 \ 10^{-4}}{1 + 0.99625 \ 10^{-4}} }[/tex]

             f’₁ = 150 10⁶ [tex]\sqrt{ \frac{0.999900}{1.000099625} }[/tex]

             f’₁ = 1.50 10⁶ 0.999900

             f'₁ = 149.985 10⁶ Hz

planets approaching

             f’₂ = 150 10⁶ [tex]\sqrt{ \frac{1 + 0.99625 \ 10^{-4}}{1 - 0.99625 \ 10^{-4}} }[/tex]

             f’₂ = 150 10⁶   1.0000996

             f'₂ = 150.0149 10⁶ Hz

we see that the changes in the frequencies are extremely small

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