The speed of distant stars and planet has been estimated from measurements of the wavelengths of the spectrum lines which they emit. Suppose a star or planet is moving with a velocity v away from the earth and emits light of wavelength Ʌ. If the frequency of the vibrations is *f* cycles per second, then *f* waves are emitted in one second, where *C=f* Ʌ and c is the velocity of light in *vacuo. *Owing to the velocity v, the *f* waves occupy a distance (c + v). Thus the apparent wavelength Ʌ’ to an observer on the earth in line with the star’s motions is

Ʌ =

**. ^{.}. **Ʌ’

**–**Ʌ = “shift” in wavelength =………….……(i)

And hence fractional change in wavelength = ……………(ii)

From (i), it follows that Ʌ’ is greater than Ʌ when the star or planet is moving away from the earth, that is, there is a ‘’shift’’ or displacement towards the red. The position of a particular wavelength in the spectrum of the star is compared with that obtained in the laboratory, and the difference in the wavelength, Ʌ’ – Ʌ is measured. From (i), knowing Ʌ and c, the velocity v can be calculated.

If the star is moving towards the earth with a velocity u, the apparent wavelength is given by

=

**. ^{.}. **Ʌ- =

Since is less than Ʌ, there is a displacement towards the blue in this case.

In measuring the speed of a star, a photograph of its spectrum is taken. The spectral lines are then compared with the same lines obtained by photographing in the laboratory an arc or spark spectrum of an element present in the star. If the former are displaced towards the red, the star is approaching the earth. By this method the velocities of the stars have been found to be between about 10km s^{– 1} and 300km s^{-1}. The Doppler effect has also been used to measure the speed of rotation of the sun, photograph are taken of the east and west edges of the sun; in the sun, and also absorption lines due to elements such as iron vaporized atmosphere. When the two photographs are put together so that the oxygen lines coincide, the iron lines in the two photographs are displaced relative to each other. In one case the edge of the sun approaches the earth, and in the other the opposite edge recedes from the earth. Measurements show a rotational speed of about 2 km s^{-1}.

** Implications Of Doppler Effect On Radio Waves**

A radio wave is an electromagnetic wave, like light, and travels with the same velocity, c, in free space of 3.0 x 10^{5} km s^{-1}. The Doppler Effect with radio waves can be utilised for finding the speed of aeroplanes and satellites.

As an illustration, suppose an aircraft C sends out two radio beams at a frequency of 10^{10} Hz; one in a forward direction, and the other in a backward direction, each beam being inclined downward at an angle of 30^{} to the horizontal. A Doppler effect is obtained when the radio waves are scattered at the ground at A, B, and when the returning waves to C are combined; a beat frequency equal to their difference is measured. Suppose the beat frequency is 3 x 10^{4} Hz

If the velocity of the aircraft C is v, the velocity of radio waves is c and the frequency of the emitted beams is *f*, the apparent frequency *f’* of the waves reaching A is given by

*F’* =

Where ϴ is 30^{}. The frequency *f _{1}* of the wave received back at C from A is given by

*F _{1}* =

Where V^{’ }is the velocity of the wave relative to C and Ʌ’ is the wavelength of the waves reaching C. since V’ c – v cos ϴ and Ʌ’ = c/f’,

**. ^{.}. **……………..(ii)

From (i). similarly, the frequency f_{2} of the waves received back at c from B is given by

**. ^{.}.**

*f*= . . . ………..(iii)

_{2}**. ^{.}. **beat frequency at C =

*f*

_{2 }–*f*=

_{1 }Now C = 3 x 10^{5} km s^{-1}, ϴ = 30^{}, *f _{2}* = 3 x 10

^{4}H

_{z},

*f*= 10

^{10 }H

_{2}, and v

^{2}cos

^{2}ϴ is negligible compared with c

^{2}

^{ }**. ^{.}. ** 3 x 10

^{4}= = =

**. ^{.}. **

*v*=

= 0. 26 km s^{-1}

= 936 km h^{-1 } (approx.)

The speed of the aircraft relative to the ground is thus nearly 940 km h^{-1}

^{ }

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