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Earth's orbit

  1. Oct 9, 2008 #1
    To help me you will need tho know Excel VBA very well.

    Earth has an elliptical orbit around the sun. JPL's website has the data to calculate this orbit:
    In the spreadsheet the parameters are calculated from widely used equations and some of my making. The major discrepencies are in how to calculate the distance between the Earth and the Sun,r. By my calculations:
    r<1AU and a minimum @True Anomaly=0
    r>1AU and a maximum @True Anomaly=180
    This is not the case when directly calculated from the table(In Green). My calculations of what they should be are in yellow.

    Any suggestions would be deeply appreciated!

    Attached Files:

  2. jcsd
  3. Oct 9, 2008 #2

    D H

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    Your equation for calculating the eccentric anomaly is incorrect:
    Code (Text):

         E1 = E0 - (E0 - e(k) * 180 / Pi * Math.Sin(E0) - MA(k)) / (1 - e(k) * Math.Cos(E0))
       Loop While E1 = E0
       EA(k) = E1
    Your input values for the true anomaly (column M) are completely wrong. True anomaly is an increasing function of time, not decreasing, and it is not uniform in time.
  4. Oct 9, 2008 #3
    If I change the values of the True Anomaly to increase by starting with 210,240,270,300,330,0,30,60,90,120,150,180 that would be increasing and will still prove my point.
    I got the Eccentric Anomaly from this website:
    If I may, does anyone have a better equation for Eccentric Anomally?
  5. Oct 9, 2008 #4
    Here is the corrected True Anomaly with increasing degrees in E and M. The True, Mean and Eccentric Anomaly increase each year by about 360 degrees making a wrap around function. I think that the True Anomaly v=0 should start @Jun 21 the Summer Solstice not at v=168.0600083 as calculated from the table. I think the calculations of M and E from the table (in Green) are wrong.

    Attached Files:

  6. Oct 9, 2008 #5

    D H

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    What makes you think the initial value is 210? (Hint: It isn't.) What makes you think the true anomaly increases by 30 degrees per month? (Hint: It doesn't.)

    What is your point, exactly? That you posted this in the "Beyond the Standard Model" section makes me think your point might be some crackpot theory. I hope that that is not the case.

    There are two things wrong with your Newton's method implementation. (1) You are treating the mean anomaly as if it were in radians. (2) Your loop condition is wrong. As written, the loop will only execute once.
  7. Oct 9, 2008 #6

    D H

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    You're wrong. In 2008, for example, perihelion occurred on January 3. True anomaly does not increase by exactly 30 degrees per month, for several reasons.
    1. No month is 365.25/12 days long.
    2. You are confusing the tropical year (solstice to solstice) and the sidereal year (apsis to apsis). The sidereal year is 20 minutes and 24 seconds longer than the tropical year.
    3. True anomaly is not a uniform function of time. Think of it in terms of a comet's orbit. Near apohelion, the true anomaly changes very little from day to day, while near perihelion, the true anomaly changes very rapidly.
  8. Oct 9, 2008 #7
    My equations (in Yellow) are just my way of trying to understand, PLEASE DISREGARD. The main equations (in Green) are what need to be correct then comes understanding. Are these equations correct?
    'Argument of the Perihelion
    PA(k) = LP(k) - LAN(k)
    'Mean Longitude of Earth at a particular time
    ML(k) = .Cells(7, 5) + .Cells(8, 5) * JT(k)
    ML(k) = ML(k) - 360 * Int(ML(k) / 360)
    'Mean Anomaly of Earth at a particular time
    MA(k) = ML(k) - LP(k)
    MA(k) = MA(k) - 360 * Int(MA(k) / 360)
    'Eccentric Anomaly of Earth at a particular time from above website
    EA(k) = MA(k) + 180 / Pi * e(k) * Math.Sin(MA(k) * Pi / 180) * (1 + e(k) * Math.Cos(MA(k) * Pi / 180))
    EA(k) = EA(k) - 360 * Int(EA(k) / 360)
    Last edited: Oct 9, 2008
  9. Oct 9, 2008 #8
    So in the winter, say Jan 3, the earth is closer to the sun just tilted away from the Northern Hemisphere.
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