# The sun's rotational period is 25 days at the equator. Given that the radius of the sun is 700,000 km, calculate the max velocity of approach or recession of the Sun's equator as viewed from Earth....

The sun's rotational period is 25 days at the equator. Given that the radius of the sun is 700,000 km, calculate the max velocity of approach or recession of the Sun's equator as viewed from Earth. Find the max change in wavelength of a spectral line due to the rotation and express it as a percentage of the rest wavelength of the line.

*print*Print*list*Cite

### 1 Answer

The problem provides the rotational period and the radius, hence, you may easily find out the velocity, such that:

`v = (2pi*R)/T`

You need to take in consideration the Earth's radius, measured from equator to it's center, that is `6.37*10^6` m. You need to remember that in 24 hours there are 86400 seconds, hence, the period T = 86400*25

`v = (2*3.14*(700*10^3 + 6.37*10^6))/(86400*25)`

`v = (2*3.14*1337*10^3)/(864*10^2*25)`

`v = 3887.1 m/s`

Since the frequency is inverse of period, yields:

`nu = 1/T => nu = 1/(86400*25)`

You need to evaluate the wavelength, using the following formula, such that:

`lambda = c/(nu) => lambda = (3*10^8)/(1/(86400*25))`