Astronomy Question 2

(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