
#1
Feb2313, 01:29 PM

P: 34

I used this equation to find the time in seconds for the Earth to complete 1 orbit around the Sun.
[tex] p=2\pi\sqrt\frac{\alpha^3}{\mu} [/tex] I put in 149598261000 metres for the semimajor 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
Feb2313, 03:29 PM

Sci Advisor
P: 5,935

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




#3
Feb2313, 06:47 PM

P: 34

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 (19942000). 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
Feb2313, 09:27 PM

Mentor
P: 14,432

The time it takes the Earth to go around the Sun.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
Feb2313, 09:42 PM

Astronomy
Sci Advisor
PF Gold
P: 22,792

Jimmy, DH is right but just for fun I looked in wikipedia ("standard gravitational parameter")
Body μ (km^{3}s^{−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 km^{3} so if you want m^{3} you have a factor of 10^{9} So you might try again with a bit more precision in your μ. http://en.wikipedia.org/wiki/Standar...onal_parameter 



#6
Feb2413, 03:04 AM

P: 34

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*6.67E11 =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
Feb2413, 03:22 AM

P: 34

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
Feb2413, 06:56 AM

Mentor
P: 14,432

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 semimajor axis. That value is wrong. If you use 149597887.5 km as the semimajor axis length and a combined Sun+Earth+Moon gravitational parameter of 132712440018+398600.4418+4902.7779 km^{3}/s^{2}, 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...+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
Feb2413, 01:10 PM

P: 34

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




#10
Feb2513, 07:37 AM

P: 34

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
Feb2513, 09:36 AM

Mentor
P: 14,432

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 EarthMoon 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
Feb2513, 02:04 PM

P: 34

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
Feb2613, 02:24 PM

P: 34

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
Feb2613, 08:58 PM

Mentor
P: 14,432

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 EarthMoon 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
Feb2713, 06:12 AM

P: 34

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
Feb2713, 06:00 PM

P: 734

A few months ago I modified my nbody 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
Feb2813, 09:33 AM

P: 34

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
Mar213, 02:32 PM

P: 34

There seems to be a lot of these on the internet.
Length of Tropical year ” =365.24218966980.00000615359*((JD245145)/36525)0.000000000729*((JD245145)/36525)^2+0.00000000264*((JD245145)/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 


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