Doppler Shift and Astronomy

[risingabove]

Homework Statement

The Sun rotates with a period of 24.7 days and has a radius of 7.00 x 10⁸ m. For a terrestrial observer, Calculate the resultant Doppler shift of light of wavelength 500 nm which is emitted from the solar equator at :

i)each side of the disc,
ii)the center of the solar disc

The speed of light, c = 3.00 x 10⁸ m/s

Homework Equations

wavelength shift Δλ = v/c λ

v : speed of of the source
c: speed of light
λ : wavelength of source

speed = distance/ time

24.7 days = 2134080 s

500 nm = 5.0 x 10^-7 m

Circumference of a circle = 2∏r = 4.4 x 10⁹ m

The Attempt at a Solution

i) I first found the speed of the source, v

speed = 4.4 x 10⁹ m / 2134080 s
= 2060.9 m/s

using the equation for wavelength shift, plug in values

Δλ = (2060.9 / 3.00 x 10⁸ m/s ) 5.0 x 10^-7 m

= 3.43 x 10^-12 m
= 0.00343 nm

ii) i assumed that λ = 0

therefore Δλ = (328 / 3.00 x 10⁸ m/s ) 0
= 0 m

.. not sure if my workings are correct ...thanks for any help

Homework Statement

Homework Equations

The Attempt at a Solution

 

[rude man]

ii) i assumed that λ = 0

therefore Δλ = (328 / 3.00 x 10⁸ m/s ) 0
= 0 m

Why λ = 0 at the center?

Where did 328 come from? This is roughly the average speed of the cg's of the Earth from the Sun. But if you include that speed you have to include it in part (i) also. So the doppler shift from one side of the sun's disc will be different than that from the other and will depend on the time of year.