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Measuring the Distance to an Exploded Star

  1. Nov 11, 2006 #1
    The gas cloud known as the Crab Nebula can be seen with
    even a small telescope. It is the remnant of a supernova, a
    cataclysmic explosion of a star. The explosion was seen
    on the earth on July 4, 1054 AD. The streamers in the
    figure glow with the characteristic red color of heated
    hydrogen gas. In the laboratory on earth, heated hydrogen
    produces red light with frequency 4.568 × 1014 Hz; the red
    light received from the streamers in the Crab Nebula
    pointed at the earth has a frequency of 4.586 × 1014 Hz.

    a) Assume that the speed of the center of the nebula relative to the earth is negligible.
    Estimate the speed with which the outer edges of the Crab Nebula are expanding.
    b) Assuming that the expansion has been constant since the supernova explosion, estimate the
    average radius of the Crab Nebula in year 2006. (Immediately after the explosion, the size of the
    Crab Nebula was neligible.) Give your answer in light years.
    The angular size of the Crab Nebula, as seen from the earth, is about 4 arc minutes by 6 arc
    minutes, for an average of 5 arc minutes. (1 arc minute = 1/60 of a degree)
    c) Estimate the distance (in light years) to the Crab Nebula.
    d) Estimate in what year BC the supernova explosion actually took place.
    A light-year (ly) is the distance traveled by light in one year.
    You may find it helpful to use the the first terms of the binomial expansion: (1+ε)^p ≅ 1 + pε

    Hi, I'm not too sute how to approach this question but I think it has soething to do with the Doppler effect for light. If this is the case I would employ the following formula:

    f_r = sqrt((c-v/c+v))(f_s)

    where f_r is the frequency measured by the reciever and f_s the frequency of the source. What confuses me here is that both the frequencies are the same so I'm not sure how to set up the problem.

    But, since they say to assume that the speed of the center of the nebula relative to the earth is negligible, could I write the formula as follows?

    f_r = (1 + v/c)^(-1/2)

    4.586 × 1014 Hz = (1 + v/(3.0 x 10^8))^-1/2

    before I start solving for v, could you guys give me any hints on whether my setup is correct or not?

    thank you !!
    Last edited: Nov 11, 2006
  2. jcsd
  3. Nov 12, 2006 #2


    User Avatar

    Staff: Mentor

    The frequencies are different.

    One needs to know the distance as a function of reshift, or vice versa.

    The angle subtended by an arc is given by [itex]\theta[/itex]=s/d, where s is the arc length and d is the distance, and the angle is in radians. For small angles, the arc length is very nearly the linaer dimension (shortest distance between two points).
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