Exact time Earth completing its orbit around the Sun...

[robotoidhuman]

Hi... I'm New here and this is my first question...

 

I'm not sure if here the proper sub-forum to ask this question...

 

2013 is almost over, every year in the edge of the year, I'm alwys curious about...

 

According to the Gregorian calendar, the calendar system which internationally the most widely accepted and used in civil calendar.

 

In Gregorian Calendar, 1 Year = 365 days, 5 hours, 49 minutes, 12 seconds. By that, can be defined as: 365 days, 5 hours, 49 minutes, 12 seconds is the exact time Earth will finish 939,889,368 km of its orbit around the Sun.

Or, in other word, by that time Earth will completely Finished and begins new orbital period around the Sun.

 

2012 is a Leap year, Earth finished its orbit within exactly 366 days.

2013 is not a leap, Earth will finish its orbit within 365,25 days.

 

Here's the question, When will the exact time Earth will completes its orbit and begins new one again?

 

In my assumption, The new orbital period will begins at Jan 1, 2014 GMT +6.

My calculation is simply, 365 (days) + 1 (day) / 0.25 (day).

 

Is it correct? Where could i get explanation bout this? could somebody please help me explain to this?

 

Thank you...

Regards

 

PS: - Sorry for my poor English, I'm not physic student, I'm just interested in Physics and astronomy, so I can't explain it very clearly. Hope you all get my point.

Edited December 30, 2013 by robotoidhuman

[swansont]

If you started the clock at midnight on Dec 31 2012, then the orbit will be complete on Jan 1 at 5:49:12 2014, according to your numbers. However, according to wikipedia, the sidereal year is 365.256363 solar days. 0.2563666 * 24 = 6.1528. That's 6 hours, 9 minutes, 10 seconds, which means 6:09:10 on Jan 1.

[Janus]

If you started the clock at midnight on Dec 31 2012, then the orbit will be complete on Jan 1 at 5:49:12 2014, according to your numbers. However, according to wikipedia, the sidereal year is 365.256363 solar days. 0.2563666 * 24 = 6.1528. That's 6 hours, 9 minutes, 10 seconds, which means 6:09:10 on Jan 1.

As I pointed out to this poster on another forum where he asked the same question, The sidereal year given is a mean value that changes from orbit to orbit. The orbit starting at midnight on Dec 31 2012 would be ~0.86 min longer than the mean value.

[swansont]

As I pointed out to this poster on another forum where he asked the same question, The sidereal year given is a mean value that changes from orbit to orbit. The orbit starting at midnight on Dec 31 2012 would be ~0.86 min longer than the mean value.

 

Did not know that. Thanks!

[J.C.MacSwell]

As I pointed out to this poster on another forum where he asked the same question, The sidereal year given is a mean value that changes from orbit to orbit. The orbit starting at midnight on Dec 31 2012 would be ~0.86 min longer than the mean value.

That is a fairly significant amount...almost a minute...what causes the fluctuation (Jupiter? just a guess...)

 

Happy...that point in the revolution....everyone

[Janus]

That is a fairly significant amount...almost a minute...what causes the fluctuation (Jupiter? just a guess...)

 

Happy...that point in the revolution....everyone

All the planets play some role in this. And actually, almost a minute is one of the smaller differences, 2010 was almost 16 mins longer than the mean, and in 2042 the difference will be over 24 min.

 

In addition, there is a smaller effect caused by the mean orbital period slowly changing. The Mean value given is for the epoch starting in 2000, by the year 6000, the mean tropical year will be 35 sec shorter and the sidereal year 12 sec shorter.