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Astronomy : Velocity of approach of the Sun's equator

  1. Oct 2, 2016 #1
    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.

    Relevant equations
    Circular motion formula v = (2*pi*r)/period

    Frequency = 1/period
    wavelength = c/frequency

    The attempt at a solution
    Convert period into seconds
    P = 2 160 000 sec

    Convert radius of Sun into m
    r = 7*10^7 m

    v = (2*pi*7*10^7m)/2 160 000 s
    = 2.04 m/s

    frequency = 1/period
    f = 1/ 2 160 000

    wavelength = c/f
    = 6.48*10^14 m

    I'm not really sure what the question is asking beyond this
  2. jcsd
  3. Oct 2, 2016 #2


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    How many meters are there in 1 km? You also neef to recheck your other computations because you are randomly getting the powers of 10 wrong.

    No, this is not the frequency of the light. It is the frequency of rotation. The actual frequency of the light is irrelevant for the question as it asks you for a ratio for the Doppler shift of the light frequency.
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