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

[Jimmy B]

I used this equation to find the time in seconds for the Earth to complete 1 orbit around the Sun.

p=2\pi\sqrt\frac{\alpha^3}{\mu}

I put in 149598261000 metres for the semi-major axis, and 1.33E+20 for the gravitational parameter.
I got 3155391.8 seconds for the period.
This I believe is incorrect; I was expecting 315558432.4640 seconds, an anomalistic year.
Have I put the wrong values in?

 

[mathman]

Straightforward calculation 60x60x24x365.25 gives a number 10 times bigger than your calculation and 1/10 times your expectation.

 

[Jimmy B]

Sorry I made an error with both numbers.
The orbital period with that equation should have been 31553931.76540220 seconds, and the number in an anomalistic year should have been 31558432.4640 seconds.
That’s if number of days from perihelion to perihelion is correct at 365.259635 days (1994-2000).

I would also like to know the Suns true ecliptic longitude for the March Equinox on the 20/03/2013 at 11:02 UTC (11:02am).
Will it be 0 degrees or 360 degrees?

 

[D H]

Sorry I made an error with both numbers.
The orbital period with that equation should have been 31553931.76540220 seconds, and the number in an anomalistic year should have been 31558432.4640 seconds.
That’s if number of days from perihelion to perihelion is correct at 365.259635 days (1994-2000).

You used a value for the standard gravitational parameter that is good to just three decimal places. This means that the most you can quote as the answer is 365 days. That 31553931.76540220 seconds, or 365.207543581044 days, is ludicrous given your 1.33e20 m³/s² value for the gravitational parameter. If you want more precision you need more precise input values.

If you want more accuracy (precision and accuracy are not the same), you need to account for the fact that the Sun and Earth gravitate toward one another, you need to use values that are consistent with one another, and you need to use the right year (not the anomalistic year). If you want even more accuracy, you need to forego the simplistic formula that you used.

 

[marcus]

Jimmy, DH is right but just for fun I looked in wikipedia ("standard gravitational parameter")Body μ (km³s⁻²)
Sun 132,712,440,018(8)[1]
Mercury 22,032
Venus 324,859
Earth 398,600.4418(9)
Moon 4,902.7779
Mars 42,828

Neglecting the Moon, and the other planets, you could add the Sun and Earth figures together to get an estimate of μ
132,712,440,018 + 398,600 = 132,712,838,618

Remember the units here are km³ so if you want m³
you have a factor of 10⁹

So you might try again with a bit more precision in your μ.

http://en.wikipedia.org/wiki/Standard_gravitational_parameter

 

[Jimmy B]

Thanks for all the replies.
Marcus, when I times the mass of the Sun with the gravitational constant, the value I get is 1.98854692E+30*6.67E-11 =1.32712440E+20, the value wikipedia gives for the gravitational parameter for the Sun.
If I do the same for the Earth I get 3.986E+18.
Adding the Suns gravitational parameter to the Earths gives 1.36698444E+20.
Putting this value in for Mu into the equation in post 1, I get a period of 31094809.45 seconds for the Earth to go around the Sun once.
Converting it to days I get 359.8936279 days, that’s if I’ve worked it out right.

 

[Jimmy B]

DH, how many days does it take for the Earth to complete one orbit of the Sun?
Do you have a value Mu?
What other equations can I use to worked this out?

 

[D H]

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/?....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.

 

[Jimmy B]

Thanks D H, that's accurate enough for me.

 

[Jimmy B]

I’m stuck again.
If the Earth takes one sidereal year of 365.256404619 days to obit the Sun, why does it only take the Earth 365.2465278 days to get from this years Equinox which is on the 20/3/2013 at 11:02 am UTC, and next years Equinox which is on the 20/3/2014 at 16:57pm UTC?

I get it to about 14 minutes less than a sidereal year.
What have I done wrong this time?

 

[D H]

It's 20.4 minutes less, not 14. I don't know where you are getting your numbers from.

A mean sidereal year, 365.256360417 days, is the average (mean) amount of time it takes for the Earth-Moon system to complete one orbit about the Sun relative to the fixed stars.

The anomalistic year, 365.259636 days, is the time it takes for the Earth to go from one perihelion to the next. This is longer than the sidereal year by about 4.7 minutes because the perihelion date advances by about one calendar day every 60 years. This apsidal precession is caused mostly by gravitational influences of the other planets. (There's a small component due to general relativity.)

A mean tropical year, 365.242190419 days, is the average amount of time it takes to pass from one solstice to the next (or from one equinox to the next). This is shorter than the sidereal year by about 20.4 minutes because the Earth's rotational axis isn't fixed. It instead precesses (a large but slow motion) and nutates (a bunch of small but fast motions) with respect to inertial space. This axial precession is caused mostly by gravitational torques on the Earth by the Moon and the Sun.

 

[Jimmy B]

Yikes!
This is more complicated than I thought it was a couple of weeks ago.
It’s going to take a while for the penny to drop on that lot.
Thanks D H.

 

[Jimmy B]

I get the total number of days from Equinox 20/3/2013 11:02 UTC to Equinox 20/3/2014 16:57 UTC as 365.2465277777810 days.
About 14 minutes short of a Sidereal year of 365.2564360417 days.
This should be 20.4 minutes; I’ve lost 6 minutes somewhere.
Have I added it up right?

 

[D H]

That's why I was very careful the use terms such as "mean tropical year" as opposed to just "tropical year" in my previous post.

Years, as measured from equinox to equinox, or from solstice to solstice, vary in length. The same applies to years as measured from perihelion to perihelion (anomalistic year) and years as measured by some star appearing to have moved by 360 degrees (sidereal years). The Earth orbits the Sun, but it also orbits about the Earth-Moon center of mass. The phase of the Moon near the equinox, or wherever you are marking the start / end of a year, changes the length of a "year" a bit.

 

[Jimmy B]

Is there any way of working out these astronomical events, like the precise time it takes the Earth to go from Perihelion to Perihelion?
The equation that I gave in post one now looks a bit silly.
There must be a method that astronomers use to calculate these events.

 

[TurtleMeister]

A few months ago I modified my n-body program to calculate the perihelion advance of Mercury. At first I thought there was a problem with my code, because the variation I was getting from one orbit to the next was greater than I had anticipated. After doing some research I discovered that this variation is normal and is caused by the varying influences of all the other bodies in the solar system. I can only assume that this variation occurs with the Earth as well. So knowing the precise time it takes Earth to go from one perihelion to the next depends on how precise you know the details of the other bodies that influence it.

 

[Jimmy B]

If you don’t mind me asking, what would you suggest that I do to work out the date and time that the Earth takes to complete one orbit of the Sun?
It seems such a simple question, but now I realize it is a very difficult thing to achieve.

 

[Jimmy B]

There seems to be a lot of these on the internet.

Length of Tropical year ” =365.2421896698-0.00000615359*((JD-245145)/36525)-0.000000000729*((JD-245145)/36525)^2+0.00000000264*((JD-245145)/36525)^3”

It returned the length of the Tropical year as 365.24218885959300 days, for the 02/03/2013 11:00:00 UTC.
Is this correct?
I then change the year to 02/03/4013 11:00:00 UTC, two thousand years in the future.
The length of the Tropical year was 365.24206764874700 days, 10.47261709227 seconds less.
Is this correct?
JD = 2456353.95833333

 

[D H]

What that polynomial represents is an approximation of the (short-term) mean tropical year. It does not give the length of this, or any other, tropical year.

The reason that the duration of the mean tropical year varies is the 26,000 year long axial precession period. Right now, we're in a phase of that cycle where the mean length of the tropical year, averaged over several years, is decreasing. It will be the other way around 13,000 years from now.

What that third order polynomial represents is the variation of duration of the duration of a tropical year over a small span of that 26,000 year cycle, maybe 4,000 years at most based data spanning from ancient Chinese, Egyptian, and Mesopotamian times to the present. That polynomial is just an ad hoc curve fit. Something that approximates the length of (say) the year 2025 is even hairier (and is even more ad hoc) than are those third order polynomials that astronomers are keen to use.

 

[Jimmy B]

What is the length of this year’s Tropical year? I can’t find it, all I can find is the average Tropical year.
How many years have astronomers kept a record of it, is it a good average?
What was the exact length of time last year for the Sun to go from zero degrees to zero degrees on the ecliptic?

 

[Jimmy B]

If I take the rate of precession as 5,028.796 arc seconds per Julian century or 50.28796195 arc seconds per Julian year.
72 years is only 1 degree of precession 3600/50.28796195 =71.58770927 years.
Multiply by 360 degrees to get 25,771.5 years, 360 x 71.58770927 = 25,771.57534 years, so to find out an accurate difference between a sidereal year and a tropical year, 50.28796195/1,296,000 x 365.25 x 24 x 60 = 20.40853122 minutes.
A Sidereal day and a Solar day have an angular duration of 1296, 000”.
This gives a Tropical year as (Sidereal year 365.256363004 – 20.40853122 minutes) 365.24219041287500 days.
Is this a good average for a Tropical year?

 

[Jimmy B]

Has anyone just for the fun of it ever used this equation to find the velocity of the Earth at any instance?

V={\mu}\big(\frac2r-\frac1a\big)

The problem I had was trying to work out what the value of “r” the radius at any instance.
I checked on the internet and the only values for the radius that I could find where when the Earth is at perihelion and aphelion.
Is there an official body that collects this type of data?

 

[Jimmy B]

After a lot of searching on the internet I eventually found a government website that had what I required to workout the Earth’s orbit around the Sun.
All I had to do was put the date into find the Earth’s orbital statistics.

It gave the following for today.

Julian Day 2456359.5
Geom/Mean/Longitude Sun 345.8698 degrees.
Geom/Mean/Anom Sun 5102.706 degrees
Eccentricity Earth Orbit 0.0167
Sun Equation of centre 1.717115
Sun True Longitude 347.5869 degrees – 180 gives Earths True Longitude as 167.5869 degrees.
Sun True Anomaly 5104.423 degrees.
Sun Radius Vector 0.9935661 AU’s.
Sun Apparent Longitude 347.4359 degrees.
Mean Obliquity Ecliptic 23.43758 degrees.
Obliquity Corrected 23.435593 degeees.
Sun Right Ascension 101.419 degrees.
Sun Declination -4.90522 degrees.
Var Y 0.043022.

Plugging 14848579.764114 km into the equation on post above gives a velocity of 30.006981 km for the Earth.
Are the values accurate?

 

[D H]

I've had a very busy and very long work week. Sorry for the delay in response.

Regarding your equation v=\mu\left(\frac 2 r - \frac 1 a\right), you have an error in that expression. The left hand side should be v², not just v. Correcting that yields the vis viva equation, v^2 = \mu\left(\frac 2 r - \frac 1 a\right).

As far as web sites go, I suggest you look at JPL's Horizons system, http://ssd.jpl.nasa.gov/?horizons.

Finally, 30 km/s is about right. That's just a bit higher than the Earth's mean orbital velocity, and since we're just a couple months past perihelion, our velocity should be just a bit higher than average. Using the Horizons system, the Earth's velocity vector in the ICRF frame at JD 2456359.5 was (-6.986, -29.188, 0.000) km/s, which has a magnitude of 30.013 km/s. (There's no reason to go below meters/second.)

 

[Jimmy B]

Thanks DH.
The only reason why on here is because my grandson has been asking me all sorts of questions ever since he got an astronomy book for Christmas.
He’s been writing them down in notepad and will coming to see me tomorrow.
Thankfully with your help I might be able to answer a couple of them.

 

[Jimmy B]

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?

 

[TurtleMeister]

You can also use the web-interface:
http://ssd.jpl.nasa.gov/horizons.cgi

 

[Jimmy B]

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.

 

[TurtleMeister]

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. :)

 

[Jimmy B]

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?

 

[TurtleMeister]

T M, what would say if someone asked you how long does it take the Earth to go round the Sun?

365.25 days

 

[Jimmy B]

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?

 

[jim mcnamara]

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:

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

cal 1752

check out September...

$ 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.

 

[Jimmy B]

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?

 

[D H]

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?

 

[Jimmy B]

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.

 

[D H]

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.

 

[Jimmy B]

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?

 

[D H]

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.

 

[Jimmy B]

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

 

[sshai45]

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/?....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?

 

[BobG]

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.

 

[Jimmy B]

What is the Earths average velocity around the Sun? I just plucked that value of the internet. Each website that I went on had a slightly different average velocity. I was spoilt for choice.
They also had their own method of calculating it.

 

[Jimmy B]

This is one of many.

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



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?

 

[D H]

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?

 

[Jimmy B]

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?

 

[Jimmy B]

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.

 

[Jimmy B]

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?

 

[Jimmy B]

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?

 

[Jimmy B]

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.