# Astronomy Question 2

Astronomy Question 2

(a) Using the formula (wavelength)x(frequency)=c, prove that

(original frequency - observed frequency) / observed frequency = v/c

for a Doppler shifted source of EM wave.

(b) Look up the frequency of the H-alpha line.

(c) The observed frequency of the H-alpha line of a star is 4.51x1014Hz. What is its radial speed relative to us?

(d) An artificial satellite is orbiting around the Sun in an elliptical orbit. Will it attain its maximum orbital speed when it is farthest away from the Sun?

(e) The distance between its perihelion and aphelion is 10A.U. What is its orbital period in term of years?

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(a) Such proof can be found in every standard physics textbook. A brief proof is given below:

Let v be the velocity of a light source

c be the speed of light

f is the frequency of light emitted, T be the period between emission of two wavefronts (i.e. T =1/f)

Consider a time interval of T second, the source moves forward by a distance = v, hence the distance between a wavefront emitted now with one emitted T sec before w' = w-v.T, where w is the wavelgth of the light

w' = c/f - v/f , since c = f.w

thus, w'f = (c-v)

But w' = c/f', where f' is the apparent frequency

cf/f' = c-v

after arragement, you would get (f'-f")/f = v/s

(b) The wavelength of H alpha line is about 656 nm

hence frequency = 3x10^8/656 x 10^-9 Hz = 4.57 x 10^14 Hz

(c) Using the formula in part (a)

(5.57-4.51) x 10^14/4.57x10^14 = v/3x10^8

solve for v

(d) No. When it is farthest from the sun, the potential enegy is maximum, the kinetic energy and hencethe orbital speed is minimum.

(e) The difference between the perihelion and aphhelion

= 2a.e

where a is the simi-major axis of the elliptical orbit, and e is the ecentricity of the orbit

hence, 2a.e = 10

By Kepler's Third Law, T^2 = a^3, i.e. a = T^(2/3)

hence, 2.[T^(2/3)].e = 10

T^(2/3) = 10/2.e

T = [5/e]^(3/2)

You need to know the ecentricity of the elliptical orbit in order to calculate T