Question

The sun has a radius of 6.955 x 108 km and a rotational period of 25.38 days. What is the difference is there in the peak of the observed spectrum between light received from the approaching side of the Sun and from the receding side of the sun? Observations from the middle of the sun place the spectral peak at 480nm.

Answer 1

Answer:

Difference in wavelengths from the receding and approaching parts is $6.4\,nm$

Explanation:

We have given radius of sun $R=6.955\times 10^8m$

And rotational period = 25.38 days

A source is moving away from us at a speed v. Frequency of the radiation at the source is$f_s.$

Observed frequency is $f_o=f_s\sqrt{\frac{1-\beta}{1+\beta}} where \beta=\frac{v}{c} .$

For wavelengths above formula is$\lambda_o=\lambda_s\sqrt{\frac{1+\beta}{1-\beta}}$

If a source is moving towards us, substitute velocity with negative sign.

Speed of receding part of sun or approaching part of sun $v=\frac{2\pi{R}}{T}=\frac{(2\pi)(6.955\times10^{8})}{25.38\times86400}=1.99\times10^{3}\,m/s$

Wavelength of radiation from receding part of sun $\lambda_o=\lambda_s\sqrt{\frac{1+\beta}{1-\beta}}=(480\times10^{-9})\sqrt{\frac{1+0.00664745}{1-0.00664745}}=483.2\,nm$

Wavelength of radiation from approaching part of sun $\lambda_o=\lambda_s\sqrt{\frac{1+\beta}{1-\beta}}=(480\times10^{-9})\sqrt{\frac{1-0.00664745}{1+0.00664745}}=476.8\,nm$

Difference in wavelengths from the receding and approaching parts is $6.4\,nm$