The time it takes the Earth to go around the Sun.

  • Thread starter Jimmy B
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  • #26
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D H, I tried getting Earths orbital elements by using “telnet ssd.jpl.nasa.gov 6775”at the command line, but found it difficult to use.
Could you please give me some tips on how to do it?
 
  • #28
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Thanks for replying, but the web based version does not list all the orbital elements.
Could be why D H had to use the command line version to find Sun/Earth radius for a particular time and date.
The web based version gives the following.
JDCT Epoch Julian Date, Coordinate Time.
EC Eccentricity, e .
QR Periapsis distance, q (AU) .
IN Inclination w.r.t xy-plane, i (degrees) .
OM Longitude of Ascending Node, OMEGA, (degrees).
W Argument of Perifocus, w (degrees).
Tp Time of periapsis (Julian day number).
N Mean motion, n (degrees/day).
MA Mean anomaly, M (degrees).
TA True anomaly, nu (degrees).
A Semi-major axis, a (AU).
AD Apoapsis distance (AU)
PR Sidereal orbit period (day).

If I keep at it, I should figure out the command line version eventually.
 
  • #29
887
98
I've never tried using the telnet interface, so I can't help you there. If all you need is Sun/Earth radius it should be easy to calculate it from the coordinates. But I'm assuming you need more than that. Your grandson must be asking some very detailed questions. :)
 
  • #30
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I was wrong T M.
You can get most of the orbital elements by changing the preset values on the web based interface. It took me several attempts to do it though.

T M, what would say if someone asked you how long does it take the Earth to go round the Sun?
 
  • #31
887
98
T M, what would say if someone asked you how long does it take the Earth to go round the Sun?
365.25 days
 
  • #32
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While checking a generated ephemeris on the JPL horizons system, I notice the PR Sidereal orbital period seemed low at 365.2460079192060 days.
The sidereal year was equal to 365.256360417 days at noon on the 1 January 2000; this value is nearly 15 minutes less.
Why does NASA use this value for a Sidereal orbital period?
 
  • #33
jim mcnamara
Mentor
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As a small diversion - people have had issues with years and days in a year not working out the same. What you are wondering about on a smaller scale - why isn't the orbital time in seconds (tropical year or whatever) perfect? DH answered your question. Clearly. Please consider reading the answers you got.

The orbital period in days for calendrics is usually taken to be 365.24+ days. This why our Gregorian calendar has no intercalary day every 100 years, but does have one for years where mod(year, 400) == 0. This was not the case until 1572 where the Julian calendar had leap years every four years.

So:
Code:
if year is divisible by 400 then
   is_leap_year
else if year is divisible by 100 then
   not_leap_year
else if year is divisible by 4 then
   is_leap_year
else
   not_leap_year
Even more fun: Acceptance of the Gregorian Calendar as a civil calendar occurred at different times after it was adopted by Pope Gregory 1582, from then until 1923. This meant completely different dates in neighboring principalities in Europe for many years. And here in the US - Spanish provinces were using the Gregorian Calendar from the start , English colonies adopted in in 1752

Because England adopted the calendar in 1752. Have look at the output of the UNIX cal program for
Code:
cal 1752
check out September....

Code:
$ cal 1752
                               1752

       January               February                 March
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
          1  2  3  4                      1    1  2  3  4  5  6  7
 5  6  7  8  9 10 11    2  3  4  5  6  7  8    8  9 10 11 12 13 14
12 13 14 15 16 17 18    9 10 11 12 13 14 15   15 16 17 18 19 20 21
19 20 21 22 23 24 25   16 17 18 19 20 21 22   22 23 24 25 26 27 28
26 27 28 29 30 31      23 24 25 26 27 28 29   29 30 31

        April                   May                   June
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
          1  2  3  4                   1  2       1  2  3  4  5  6
 5  6  7  8  9 10 11    3  4  5  6  7  8  9    7  8  9 10 11 12 13
12 13 14 15 16 17 18   10 11 12 13 14 15 16   14 15 16 17 18 19 20
19 20 21 22 23 24 25   17 18 19 20 21 22 23   21 22 23 24 25 26 27
26 27 28 29 30         24 25 26 27 28 29 30   28 29 30
                       31
        July                  August                September
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
          1  2  3  4                      1          1  2 14 15 16
 5  6  7  8  9 10 11    2  3  4  5  6  7  8   17 18 19 20 21 22 23
12 13 14 15 16 17 18    9 10 11 12 13 14 15   24 25 26 27 28 29 30
19 20 21 22 23 24 25   16 17 18 19 20 21 22
26 27 28 29 30 31      23 24 25 26 27 28 29
                       30 31
       October               November               December
Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa   Su Mo Tu We Th Fr Sa
 1  2  3  4  5  6  7             1  2  3  4                   1  2
 8  9 10 11 12 13 14    5  6  7  8  9 10 11    3  4  5  6  7  8  9
15 16 17 18 19 20 21   12 13 14 15 16 17 18   10 11 12 13 14 15 16
22 23 24 25 26 27 28   19 20 21 22 23 24 25   17 18 19 20 21 22 23
29 30 31               26 27 28 29 30         24 25 26 27 28 29 30
                                              31


Message: do not expect consistency it sometimes is not there.
 
Last edited:
  • #34
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Thanks Jim for replying.
It is very confusing how we keep time. The Gregorian calendar is a prime example.

The time it takes the earth to orbit once around the sun with respect to the fixed stars is called a Sidereal year.
Now as far as I know the precise length of this year is unknown, would you agree?
 
  • #35
D H
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What is the precise diameter of a lumpy Russet potato?

You are asking for the precise value of a quantity that is known to vary. Does that make any sense?
 
  • #36
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Ask Gordon Ramsey, he’ll know that one.

Do you know how much the Sidereal year varies D H? According to NASA it now is 365.246007919206 days. See post 32.
What fixed stars are used as a reference for the Earth to complete one orbit? They did use the star Sirius until they found it had proper motion.
 
  • #37
D H
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I don't know where you got that value of 365.246. I suspect that your value of 365.246 day resulted from using something other than 365.25 days per (Julian) year as a conversion factor.

I see values of 1.0000174 years (this page: http://ssd.jpl.nasa.gov/?planet_phys_par) or, when I use the site to generate an ephemeris for the Earth, I see values of 1.0000174 years or 365.25636 days.

1.0000174 years and 365.25636 days are the same number. Multiply 1.0000174 by 365.25 and you'll get 365.25635535, or 365.25636. Astronomers use the Julian year (exactly 365.25 days of exactly 86,400 seconds each) to measure time.


JPL does not use those values to compute their ephemerides. Those values for the sidereal period are presented for informational purposes only. Here's what they use instead:

JPL starts with initial estimates of the masses of the bodies that comprise the solar system combined with initial estimates of the positions and velocities of those bodies at some key point in time, the epoch time. They next use numerical integration techniques to propagate these initial estimates forwards and backwards in time according to the laws of physics. They then develop a set of Chebyshev polynomial-based lookup tables from this forward and backward propagation. These Chebyshev polynomials let them quickly compute the states (position and velocity) of any modeled celestial body at some given time.

Next they use this lookup table to compute what an observation of some body at a given point in time would yield. They have lots (lots and lots) of real observations on hand. Comparing the observed values with the predicted ones gives them an idea of how good / how bad these lookup tables are, and also gives them an idea of how to improve them. So they improve them by tweaking those initial estimates of mass, epoch position, and epoch velocity. Then they repeat the process, and keep repeating until the observed vs prediction errors become sufficiently small.

When you use the Horizons system to generate an ephemeris, you are using the end result of this huge numerical grinding process.
 
  • #38
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I use this ephemeris.
Ephemeris Type ELEMENTS.
Target Body Earth Moon Barycenter [EMB] [3].
Center Sun [Bodycenter].
Time Span Start=2013-03-04 Stop=1 d.
Table settings default.
Display Output default.
Formatted HTML.
http://ssd.jpl.nasa.gov/horizons.cgi#results

This ephemeris gives the date of this year’s perihelion as, Julian date 2456295.7720, 03/01/2013 06:31:45, was this correct? How much does the Sidereal year vary by, not this much surly? What star or point in space did they start all these calculation from this year?
 
  • #39
D H
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Those are osculating orbital elements. They are computed from position and velocity by assuming that the orbit is Keplerian. This Keplerian assumption makes for a (somewhat) easy set of transformations from position and velocity to orbital elements.

The problem is that the orbit is not Keplerian. The Earth-Moon barycenter doesn't follow an Keplerian ellipse. It's close to elliptical, but that "close to" means that some funky results will arise.
 
  • #40
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Now you’ve mentioned ellipses. How far does the Earth travel around the Sun each year?

365.25636004 days * Earths average velocity 29784.813 m/s = 939.9536 million kilometres.

Or?

(360 degrees / 365.25636004 days) * 939.9536 million kilometres = 939.89623 million kilometres
 
  • #41
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1
You made another error somewhere in your computation of the Earth's gravitational parameter. That number is just wrong.

You also made a mistake in multiplying the mass of the Sun by the gravitational constant. The standard gravitational parameters are observables. The mass of the Sun is computed by dividing the solar gravitational constant by the universal gravitational constant. Use the wrong value of G (e.g., the value provided by google) and you'll get a wrong value. Use inconsistent values of the solar mass and G (e.g., the values provided by wikipedia) and you'll get a wrong answer.

You'll also get a wrong answer if you use 1.00000261 au (value provided by wikipedia) as the length of the Earth's semi-major axis. That value is wrong.

If you use 149597887.5 km as the semi-major axis length and a combined Sun+Earth+Moon gravitational parameter of 132712440018+398600.4418+4902.7779 km3/s2, you'll get a value for the period that is within 3.8 seconds of the sidereal year (*not* anomalistic year): http://www.wolframalpha.com/input/?i=2*pi*sqrt((149597887.5+km)^3/((132712440018+398600.4418+4902.7779)+km^3/s^2))+-+1+sidereal+year.

That's about as good as you're going to get with this simplistic formula that ignores the effects of the other planets and that ignores relativistic effects.
I'm curious then -- what's the "next-more-complicated but as simple as possible" formula or method that would allow one to get to just within sub-second accuracy? Also, how much accuracy does one need to demand before one has to start getting into General Relativity?
 
  • #42
BobG
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Now you’ve mentioned ellipses. How far does the Earth travel around the Sun each year?

365.25636004 days * Earths average velocity 29784.813 m/s = 939.9536 million kilometres.

Or?

(360 degrees / 365.25636004 days) * 939.9536 million kilometres = 939.89623 million kilometres
Is that really your "average" velocity? Or is it what the Earth's speed would be if the Earth were in a circular orbit?

Most people that ask about the Earth's average velocity (or the average velocity of any other object) are really asking for a ballpark figure and it's common to just give the circular velocity for objects that have low eccentricities. Calculating the actual "average" velocity is hard to do, so most people don't bother. (The difficulty arises in calculating the circumference of an ellipse - it becomes even more difficult if you're calculating the circumference of a perturbed ellipse.)

In other words, if you really do care (why when it's not used for anything?), make sure your source actually calculated the average velocity.
 
  • #43
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What is the Earths average velocity around the Sun? I just plucked that value of the internet. Each web site that I went on had a slightly different average velocity. I was spoilt for choice.
They also had their own method of calculating it.
 
  • #44
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This is one of many.

[tex]vp=\sqrt((1+e)\mu/(1-e)a)[/tex] [tex]va=\sqrt((1-e)\mu/(1+e)a)[/tex]



vp = velocity at perihelion and va = velocity at aphelion.

vp = 30.25375649 km and va =29.32291236 km

mean velocity = vp+va/2

I got a mean velocity of 29.7883344272490 km.
have I done it right?
Is it correct?
 
  • #45
D H
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No. You have *way* too many digits, for one thing.

For another, how are you defining "mean"? Suppose you take a two hour drive. You drive a steady 30 MPH for the first hour, then 60 MPH for the second. What's your average speed?
 
  • #46
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Thanks for the reply, I’ll discount that one.
Here’s another.
Average speed of the Earth around the Sun = (2*pi)*(a/p) = 29784.8132 m/s.
a = 1.49598E+11 Earths semi-major axis, and p = 31558149.76 seconds in a sidereal year.
Earths orbital perimeter = 365.256363007 days in a Sidereal year * 29784.8132 average speed = 9.39954E+11 meters (939.9536 million kilometers).
Is this any better?

Will not be long now until the Vernal Equinox, according to US Naval Observatory it will be on the 20/03/2013 at 11:02 UTC.
Is this correct?
 
  • #47
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I'm curious then -- what's the "next-more-complicated but as simple as possible" formula or method that would allow one to get to just within sub-second accuracy? Also, how much accuracy does one need to demand before one has to start getting into General Relativity?
Same here, if you ever find out please let me know.
 
  • #48
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Mean Anomaly

Hi, all. I’m back with another question.
If you want to find the Earths mean anomaly using this equation “M = n t”.
What value should “n” be, the daily motion, if “t = 0”?
Should it be 360/Sidereal year, 360/Tropical year, 360/Anomalistic year, or is it none of these?
 
  • #49
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Wikipedia has the mean anomaly of 357.51716 degrees, is this correct?
If it is, the only way I can satisfy this equation “M = Mo + nt” that is on Wikipedia is, Mo = 357.5176 degrees, n = 0.98560025850 degrees, and “t” = 365.259644606150 days.
“t” has a very near value of a anomalistic year. Is the anomalistic year used for this equation?
 
  • #50
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I also got M = 6.2832173611570 radians (360.001836557620 degrees) using this equation from wiki.

M = n t = sqrt ( G M / a^3) * t

My values were, G M = 1.33E+11, a^3 = 149598261.0 km^3, t = 31558432.541760 seconds in a anomalistic year.
There does seem to be a relationship. Getting somewhere at last; who knows I might be able to answer one of my grandson question after-all this.
 

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