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

1. Feb 23, 2013

### 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?

2. Feb 23, 2013

### mathman

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

3. Feb 23, 2013

### 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?

4. Feb 23, 2013

### D H

Staff Emeritus
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.

5. Feb 23, 2013

### 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

6. Feb 24, 2013

### 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.

7. Feb 24, 2013

### 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?

8. Feb 24, 2013

### D H

Staff Emeritus
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.

9. Feb 24, 2013

### Jimmy B

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

10. Feb 25, 2013

### 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?

11. Feb 25, 2013

### D H

Staff Emeritus
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.

12. Feb 25, 2013

### 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.

13. Feb 26, 2013

### 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?

14. Feb 26, 2013

### D H

Staff Emeritus
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.

15. Feb 27, 2013

### 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.

16. Feb 27, 2013

### 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.

17. Feb 28, 2013

### 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 realise it is a very difficult thing to achieve.

18. Mar 2, 2013

### 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

19. Mar 3, 2013

### D H

Staff Emeritus
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.

20. Mar 4, 2013

### 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?

21. Mar 6, 2013

### 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?

22. Mar 7, 2013

### 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?

23. Mar 8, 2013

### Jimmy B

After a lot of searching on the internet I eventually found a government web site that had what I required to workout the Earth’s orbit around the Sun.
All I had to do was put the date in to 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 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?

24. Mar 8, 2013

### D H

Staff Emeritus
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.)

25. Mar 8, 2013

### 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.