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Sidreal Year: 20 Minutes?

  1. Dec 9, 2009 #1
    I've recently become aware of a supposed well known fact, that actually seems quite wrong. It is said in many places that the difference between the Tropical Year and the Sidereal Year is about 20 minutes. Here are some examples on Wikipedia, but I've seen this in astronomy and science textbooks:


    http://en.wikipedia.org/wiki/Axial_precession_(astronomy)#Effects"

    and

    http://en.wikipedia.org/wiki/Sidereal_year" [Broken]

    But if this were to actually happen, if we were to really have 20 extra minutes when the sun reaches the same place at the vernal equinox for the completion of a tropical year, then we would be cycling through the zodiac 5 full degrees each year and not 50 arcseconds.

    I believe the incorrect equation is being used. The difference in the tropical and sidereal year is not 20 minutes. This is clearly not happening. Precession through the zodiac at the vernal equinox would be only 72 years(because 20 minutes covers 5 full degrees of the sky) if this were happening. The difference is only 50 arcseconds per year, which in real time is only 3.3 seconds.

    To say this: "One sidereal year is roughly equal to 1 + 1/26000 or 1.0000385 tropical years." is not correct. It's 1 year plus 1/2600 of the angle of the sky, which is 50 arcseconds of angle. We've been lumping in the precession measure with the measure for the spin of the earth in a year. They are independent movements. The stars do not shift 20 minutes(5 degrees) compared to the sun at vernal equinox every year.

    This is very interesting!

    Mark
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 9, 2009 #2

    mgb_phys

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    The 20minutes difference is over a year.
    So the angle shift is 360/(365.25*24*(20/60)) = 0.0137deg = 50"
     
  4. Dec 9, 2009 #3
    Hi,

    Thank you for responding.

    To include the 365.25 spins of the earth in the equation is not correct. Precession moves independent of the spin of the earth.

    If we start at say 0 degrees with sun overhead on the vernal equinox, every year when the sun is in the same place we have 0 degrees angle again. To line up with the stars we need 50 more arcseconds. This 50 arcseonds is observable.

    So you took 50/1296000 (all the arcseconds possible). But why did you then multiply this number by all the time in one year? The precession wobble occurs independently of all the 365.25 spins of the year.

    We can't have 20 minutes of extra time and only 50 arcseconds of extra angle.

    Mark
     
  5. Dec 9, 2009 #4
    Something to ponder:

    360 degree shift = 1 day of real time

    180 degree shift = 12 hours of real time

    90 degree shift = 6 hours of real time

    45 degree shift = 3 hours of real time

    15 degree shift = 1 hour of real time

    10 degree shift = 40 minutes of real time

    5 degree shift = 20 minutes of real time

    1 degree shift = 4 minutes of real time

    1 arcminute shift = 4 seconds of real time

    1 arc second = 1/15 of a second of real time

    50 arc seconds = 3.3 seconds of real time

    The shift is observable and it's 50 arcseconds. Hipparchus recorded this way back when. It can't be anything other than 3.3 seconds just like a 180 degree shift can't be anything other than 12 hours. And 12 hours can't be anything other than a 180 degree shift......unless the earth is spinning much faster. And it certainly is not.

    The argument that the 50 arc seconds shifts over the whole year does not work. An arc second is a measure of time because the earth rotates at a constant speed. The 24 hour day is constant. So is the 50 arcsecond period. It's 3.3 seconds.

    The measure of the spin of the earth and the movement of precession must be separated.

    Mark
     
    Last edited: Dec 9, 2009
  6. Dec 9, 2009 #5
    Say we shift 50" over a whole year.

    50" is 50/3600 = 1/72nd of a degree.

    1deg of Earth's orbital circumference is roughly 924,375,700km/360 = 2,567,710.28 km (per degree of circumference).

    1/72nd of this distance is 2,567,710.28km/72 = 35662.64 km (per arcsecond of circumference).

    With an orbital velocity of roughly 30 km/s, we cover this distance in:

    35662.64km/(30km/s) = 1188.75 seconds = 19.81 minutes.

    This is a rough calculation but should explain how 20 min is reckoned from 50". It is you who is using the spin of the Earth to get 5deg out of 20min, i.e. 5deg of Earth's rotational period, not Earth's orbital period. In other words we may say Earth rotates about 1 degree for every 10 arcseconds of orbit. Which makes sense given we have roughly as many degrees in our orbit as days.
     
    Last edited: Dec 9, 2009
  7. Dec 9, 2009 #6
    Thank you blkqi. I think your post may help us to highlight the answer!

    You are leaving out the 365.25 spins of the earth in 924,375,700km orbital circumference! That makes the degrees 365.25(plus 1) times longer!

    Divide your answer by the 365 spins of the earth.

    Very good. Thank you blkqi!

    Mark
     
  8. Dec 9, 2009 #7
    Let's do some rough calculations to find out how many degrees there are in a year.

    365 days in a year
    x
    each day has 360 degrees of celestial sky passing over
    = Each tropical year has 131400 degrees of celestial sky coverage which equals 7884000 arcminutes which equals 473040000 arcseconds.

    In other words, 365 x the number of arcseconds in a full circle(1296000 ) =473040000 arcseconds.



    Each day, just like it has 360 degrees, has 24 hours, as I showed above.
    365 days x 24 hours = 8760 hours = 525600 minutes(yes, just like the song) = 31536000 seconds.

    Independent of this is the observable movement of the celestial stars, called precession. The stars move 50'' each as opposed to the sun on the equinox compared with the previous year. This is independent of the spinning earth. The earth is spinning much much more in angle than precession is advancing it in angle. Precession is only tilting the celestial sky 50 arcseconds per year. The spinning earth is advancing the celestial sky 473040000 arcseconds per year!


    50''/473040000''

    This ratio of total celestial sky advanced to the total arcseconds in a year of celestial sky passed will give us the ratio we need. Multiply this by the total number of seconds in a year to find the value of real-time seconds in 50 arcseconds:

    50''/473040000'' x 31536000sec = 3.3 seconds!

    Mark
     
  9. Dec 9, 2009 #8

    D H

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    You are tilting at windmills in disputing known facts, Mark. You would better serve yourself if you tried to understand this rather than debating it.
     
  10. Dec 9, 2009 #9
    I'm asking that we be honest with ourselves and with the facts, rather than relying on a number that someone else came up with. If we say, 'you're disputing known facts', then we're not being honest with ourselves. Look at the actual facts and you will see something different. It's too easy to just brush it off as "known facts".

    These are some real facts:

    Fact: There are 360 degrees in one circle.

    Fact: There are 1296000 arcseconds in one full circle.

    Fact: There are 1296000 arcseconds in one full day.

    Fact: There are, give or take, 473040000 arcseconds in one year.

    Fact: The observed celestial sphere movement as opposed to the observed solar movement is 50 arcseconds.

    Fact: 24 hours is roughly the same as 360 degrees of celestial sphere movement.

    Fact: 24 hours is roughly the same as 1296000 arcseconds.

    Fact: 1296000 arcseconds is equivalent to 24 hours of real time.

    Fact: 648000 arcseconds is equivalent to 12 hours of real time

    Fact: 54000 arcseconds is equivalent to 1 hour of real time

    Fact: 900 arcseconds is equivalent to 1 minute of real time.

    Fact: 15 arcseconds is euqivalent to 1 second of real time.

    Fact: 50 arcseconds is equivalent to 3.3 seconds of real time.

    Fact: The tropical year is roughly 365 days of real time.

    Fact: The tropical year is roughly 525600 minutes of real time.

    Fact: The tropical year is roughly 31536000 seconds of real time.

    Fact: The tropical year is 473040000 arcseconds.

    Fact: Precession is 50 arcseconds movement of the ecliptic as opposed to the sun every year.

    Fact: The sidereal year is 473040000 arcseconds + 50 arcseconds.

    I challenge you, in a friendly way, to dispute any one of these facts.

    Mark
     
    Last edited: Dec 9, 2009
  11. Dec 9, 2009 #10

    D H

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    You are confusing the rotation of the Earth about its axis with its orbit about the Earth. This is an observed fact. Why are you disputing it?
     
  12. Dec 9, 2009 #11
    DH,

    50 arcseconds of displacement between the sun and the background stars occurs every year. Only 50 arcseconds. Observable by anyone who takes the time. 50 arcseconds is 3.3 seconds of real time.

    Try to imagine the earth going around the sun without rotating , but still precessing the same. It would still be 50 arcseconds. And we would still see the same change on the vernal equinox. The background star(assuming we were on the night side of a non-spinning earth) change would look the same even if the earth was not spinning, but still precessing the same.

    You are not disputing any of my facts, you are just saying that I'm wrong. That won't work.

    Mark : )
     
  13. Dec 9, 2009 #12
    Perhaps the most important thing to realize is that the spinning of the earth is a different action than the movement of precession.

    20 minutes(1200 seconds of precession) distributed over every day of the year means 3.3 seconds(1200/365) of real time every single day. Likewise it means 50 arcseconds of change every single day. It is obvious that this is not happening. The stars are not adjusting 50 arcseconds every day compared to the sun. It just is not happening.
     
  14. Dec 9, 2009 #13

    ideasrule

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    So the Sun takes 20 extra minutes to reach the equinox from the previous equinox than it does to reach a certain star field from the previous time it was in the star field. The Sun moves 360/365 degrees per day relative to the stars, which is 0.0137 degrees per 20 minutes. At that rate, it would take 26 000 years to move 360 degrees. That's the correct precession period of the Earth.
     
  15. Dec 9, 2009 #14
    Very good discussion.

    This is confusing.

    .0137 degrees is 50 arcseconds. The sky never takes more or less than 3.3 real time seconds to travel 50 arcseconds. Likewise, the sky never takes more or less than about 20 minutes to travel 5 full degrees.

    The sun "moves" in the sky 5 degrees per 20 minutes.

    The sun moves 15 degrees per hour.

    The sun moves 90 degrees per 6 hours.

    The celestial sky moves 50 arcseconds every year due to precession and nothing else.

    If it moved 20 real time minutes over the course of the whole year, the whole year would net 5 full degrees change in the celestial sky. It doesn't matter if you mark it as 3.3 real seconds/50 arcseconds per day or 20 real minutes/5 degrees per year.

    An arcsecond, arcminute and degree are units of time just as seconds, minutes and hours are units of time because the spin of the earth is a constant speed.
     
  16. Dec 9, 2009 #15

    ideasrule

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    Then try to understand it! Earth is orbiting the Sun, so the Sun appears to move relative to the stars.
     
  17. Dec 9, 2009 #16
    Yes. And the sun comes back to the same position after one tropical year. It takes an additional 50 arcseconds(at the half year it would be 25 arcseconds) to line up with the same background stars. 50 arcseconds is 3.3 seconds of real time. It's just what it is. You can't dispute that.

    Why have I separated the tropical year and the precession measure? Because they are completely different motions. The tropical year doesn't multiply by 50 arcseconds which is what is causing the error. The 50 arcseconds occur independently of the spin.
    This concept is very important. Precession movement occurs independently of rotation movement.

    20 minutes is 5 degrees of celestial sky movement. I challenge you to disprove this.
     
  18. Dec 9, 2009 #17

    D H

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    You are STILL confusing the rotation of the Earth with its orbit.
     
  19. Dec 9, 2009 #18
    We shall see : )




    I want to highlight blkqi's post because it is illustrious of the issue.

    Blkqi came up with 20 minutes for his answer. He measured the orbital circumference of the earth. But 50 arcseconds should not represent 50 arcseconds of the orbital cricumference, it represents 50 arcseconds of the celestial sphere. Hipparchus was not viewing a 50 arcsecond change of the orbital cricumference in a year. He could not have. He was viewing 1/365 of that.

    Blkqi got 20 minutes because he didn't factor in the approximate 365 spins of the earth in this orbital period. Again, the 50 arcseconds are of the celestial sphere moving, not the 50 arcseconds of the orbital circumference of the earth around the sun.

    Blkqi says:
    circumference of the orbital. But the earth is spinning 365 times in this orbital period. We must divide our answer by 365.

    Any clear thinking, reasonable person will see this. We must view the facts from a clean slate, without any presumptions. It's all there.
     
    Last edited: Dec 9, 2009
  20. Dec 10, 2009 #19

    jtbell

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    No, you need to re-think this. Thread closed.
     
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