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Homework Help: Solar/Sidereal day/SpiralArm/Orbital Speed With Radio Telescope

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Hello, I have 48-hour data from a radio telescope on Earth, and my goal is to answer these four questions:

    1) Calculate the length of a solar day.
    2) Calculate the length of a sidereal day
    3) Explain how to locate a spiral arm on one of your plots and how to determine
    its velocity, and then find the velocities of all spiral arms present.
    4) Calculate the orbital speed of the Earth

    I was given a picture and massive data tables(which I plotted into graphs) The first three correlates with the picture, and the last is an intensity of hydrogen vs time:

    Picture: http://docs.google.com/fileview?id=...jQtNzgzZS00ODFkLWI5OTItZDYwNjFkMzQzY2Ez&hl=en

    Graphs: http://docs.google.com/Doc?docid=0AQ0MqLuFEwp7ZGYyaDdmN3FfMGdxa2tjYnc5&hl=en

    The picture data set given is actually three dimensional, with frequency in MHz on the x-axis, time in modified julian date (MJD) on the y-axis, and power in arbitrary units on the z-axis (represented by the greyscale). Taking a look at the pcture, which represents two days of observing. Some of the key features of this image.

    Thick White Bands
    The telescope is set at an appropriate angle such that the Sun transits through the beam twice in a 48-hour period. The Sun is the most radio-loud object in the sky, and when it passes through the telescope’s beam it appears as an extremely bright object at all frequencies in the data set, resulting in the large white bands you see.

    S-shape of the Hydrogen Observations
    With the exception of the hydrogen detected from the spiral arms, all other detections of hydrogen are from the local region of space. This material has the same rotation velocity about the galactic centre as the Earth. The result would be a Doppler shift of zero except for the fact that the Earth is orbiting the Sun. This orbital speed causes a Doppler shift in the frequency of the hydrogen, and, as the Earth’s daily rotation about its own axis carries the antenna beam around the sky, we see the orbital speed as the sinusoidal function present in the data.

    Broad Diffuse White Object in the Centre of the Data
    In addition to the Sun, the galactic plane also passes through the telescope’s beam. The galaxy is not a disk of evenly distributed hydrogen rotating around a centre, but rather a series of spiral arms. Along any given line of sight, arms at different distances will have different radial velocities as seen from the Sun, so the hydrogen emission from the different arms is Doppler shifted by different amounts and shows up at different frequencies. This creates an emission feature, with the hydrogen distributed over a large frequency range (e.g. seen at ≈ 54459.25 MJD). The Doppler shifts can be explicitly seen in the centerspectrum graph.



    2. Relevant equations

    There will be several, unfortunately I don't know yet

    3. The attempt at a solution

    My main issue here is interpreting the information, I know the Sun is the most radio loud, but how do I figure length of solar day from the data? Can I do that with an intensity of hydrogen and time graph? I know length of a sidereal day means relative to the stars, but how go by figuring that out? I'm not even sure how to do the last two..Any clues would be appreciated.
     
    Last edited: Mar 25, 2010
  2. jcsd
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