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Flavio Manna

Is there a value or something else that indicates the orientation of an orbit?

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I start by saying i'm not good at science. i'm tying to do a solar system simulation, i alredy did the gravity simulator, the first problem now comes with the planets' orbit.  

Let's presume that we're looking at the solar system from above and that there is a cartesian plane where a the center of it there is our Sun: 

1379070012_2020-05-11(2).png.c51bf8b651d772ce9d0ef846e306ff25.png

 

And now let's add Mercury's orbit: 

738577194_2020-05-11(3).png.18479e6adc24fbaddc29c0adc9955985.png

 

So, thanks to the first law of Keplero we know that the sun is actually on one of the two Mercury orbits's foci, but the orbit could even like this: 125293180_2020-05-11(4).png.0ca5d02bd0bc29f253e05e62413ee4da.png

 

So my question is, is there a value, a  data or something else that indicates the orentation (or direction) of an orbit? Problably i'm searching for the rotation around the y-axis.

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4 minutes ago, Flavio Manna said:

Actually i did, honestly i didn't understand it because of my very poor english... i'll try to read again these wikipedia's pages, thank you.  

Look on the left side of the wiki pages and you will see a 'Languages' section. Choose yours.

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2 minutes ago, StringJunky said:

Look on the left side of the wiki pages and you will see a 'Languages' section. Choose yours.

Unfortunately they are not one-to-one translations and English version is usually the richest in knowledge..

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5 minutes ago, Sensei said:

Unfortunately they are not one-to-one translations and English version is usually the richest in knowledge..

OK, Sensei. Cheers for that.

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2 hours ago, Flavio Manna said:

So my question is, is there a value, a  data or something else that indicates the orentation (or direction) of an orbit? Problably i'm searching for the rotation around the y-axis

Here is an example of a elliptical orbit where we can see the other focus more clearly than your drawings:

ellipse.png

(From: https://www.schoolsobservatory.org/learn/astro/esm/orbits/orb_ell)

If we take the line between the Sun and the other focus, I think you are asking: "what is the angle of that line?" (Here it is aligned with the X-axis, but it could be rotated at any angle) Is that correct?

 

If there is only a single planet (as in that diagram) then it doesn't really matter what orientation you choose because there is nothing to compare it to. If you have several planets, then their ellipses will all be aligned differently. The only way you can know the actual alignments is by making measurements of the planetary orbits.

Here is a more realistic diagram of several orbits in the solar system:

OSC_Astro_03_04_Solar.jpg

From: https://courses.lumenlearning.com/astronomy/chapter/orbits-in-the-solar-system/  - that looks like a good page that might answer a lot of your questions.

As you can see, the planetary orbits are pretty close to circular.

Also, the relative alignment of the ellipses will change over time as the interaction of the bodies will cause the orbits to deviate from perfect ellipses. Like this:

260px-Perihelion_precession.svg.png

From: https://en.wikipedia.org/wiki/Apsidal_precession

Note that if you are simulating the orbits by calculating the positions after shot time intervals, you need to be sure your simulation converges. For example, if your time steps are too large, then the simulation will be inaccurate and the orbit will become unstable and you will fling the Earth off into space.

 

48 minutes ago, Flavio Manna said:

Here how can i find the other focus of Mercury's orbit?

Not easy, because the eccentricity the orbit is so small.

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There are a number of elements that are used to fully describe a planet's orbit.:

The semi-major axis of the orbit (average orbital radius)

The eccentricity ( how elliptical the orbit is)

The orbital inclination

The longitude of perihelion

The longitude of the ascending node.

 

The last three are measured with respect to the Earth's orbit around the Sun. The imaginary plane which passes through the  Earth's orbit  is the ecliptic. 

The orbital inclination is how much the planet's orbit is tilted relative to this plane.*

The longitude of perihelion is measured relative to the Vernal equinox. (draw a line from the Sun that passes through the Earth's orbit where the Earth would be on the date of the spring equinox.)  If you draw a line between planet and Sun when the planet is at perihelion ( the closest point of its orbit around the Sun, the angle between this line and the line above  will give you the longitude of perhelion. (which also gives you how the major axis of the orbit is aligned).

The longitude of the ascending node gives you where the plane of the planet's orbit and the ecliptic cross each other.

 

*Because the Earth is subject to various gravitational disturbance from other planets, this plane can change. To keep things simple, the ecliptic is defined as the plane of the Earth orbit on the starting date of the epoch we are presently in.  That way, we aren't always adjusting the inclinations of the other orbits due to the Earth own orbital changes.)

 

All of these elements are subject to change over time due to being perturbed by other bodies in the solar system.

The Horizons web site can give the these values and the position of a planet for any date:

https://ssd.jpl.nasa.gov/horizons.cgi#top

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Strange and Janus already have given you excellent illustration and glossary.

Just in case what you are looking for is something more mechanical, there are several ways to make ellipses with mechanical elements.

One of them is the "nothing grinder" or "do nothing machine," also called "elliptic trammel":

Elliptic%20Trammel.jpg?itok=FySGeU8S

But I'm not sure it tells you anything about the foci. I don't think it does. In Italian I think it is "Trammel di Archimede."

And the other one is a couple of tacks and a piece of string.

Drawing%20an%20ellipse.jpg?itok=V6N9Zx6J

The tacks' positions are the foci.

It's all in this website: https://americanhistory.si.edu/blog/ellipsographs

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On 5/11/2020 at 3:17 PM, Flavio Manna said:

i'm tying to do a solar system simulation, i alredy did the gravity simulator, the first problem now comes with the planets' orbit.  

 

So my question is, is there a value, a  data or something else that indicates the orentation (or direction) of an orbit? Problably i'm searching for the rotation around the y-axis.

If you're simulating Newtonian gravity, then all you need to get an elliptical orbit is to give the smaller mass a velocity. Then calculate updated positions and velocities, using gravity for acceleration. The data would be mass of the sun, location of the planet relative to center of sun, and velocity at that location. Or if accurate relative positions aren't important, just distance from the center of the sun at apogee or perigee, and speed at that point (oriented 90 degrees from the sun) will do. Simply updating position, velocity, and acceleration iteratively ("Euler method") can give a good approximation for simple visualization.

I'd be curious to see if you can find eg. the max deviation in distance between Mercury and the sun, based only on those data. You would know you have a small enough step size, when making it a lot smaller (eg. half) doesn't change the results much.

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On 5/11/2020 at 4:47 PM, Flavio Manna said:

Here how can i find the other focus of Mercury's orbit? 

 

On 5/11/2020 at 5:36 PM, Strange said:

Not easy, because the eccentricity the orbit is so small.

Couldn't you just find the distance from the first focus to Mercury at perihelion, and the distance from the first focus to Mercury at aphelion, then take a vector from the first focus toward the point of aphelion, with a length that's the difference of the two distances? Even if those values are determined by an accurate-enough simulation, that should be easy.

According to wikipedia, the difference is 23.8 million km, so that's approximately how far apart the foci should be (neglecting the sun's wobble). That's more than half the distance between Mercury and the sun at perihelion.

Mercury's orbit eccentricity is huge (0.2056) compared to Earth's (0.0167).

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9 hours ago, md65536 said:

 

Couldn't you just find the distance from the first focus to Mercury at perihelion, and the distance from the first focus to Mercury at aphelion, then take a vector from the first focus toward the point of aphelion, with a length that's the difference of the two distances? Even if those values are determined by an accurate-enough simulation, that should be easy.

According to wikipedia, the difference is 23.8 million km, so that's approximately how far apart the foci should be (neglecting the sun's wobble). That's more than half the distance between Mercury and the sun at perihelion.

Mercury's orbit eccentricity is huge (0.2056) compared to Earth's (0.0167).

I may have misread the question as to how you could do it from a diagram like the one in the post. 

But you are absolutely right, it can be calculated from the orbital parameters.

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Just to expand a bit on my last post:

If you look at this diagram, you can see how to orient Mercury's orbit.

https://en.wikipedia.org/wiki/Longitude_of_the_periapsis#/media/File:Orbit1.svg

For Mercury, the reference direction is the vernal equinox.  It is a point on the celestial equator, just a bit South of Pegasus.

For this line, you would measure the longitude of the ascending node. For the epoch starting on J2000 ( astronomy uses Julian years), for Mercury, this is 48.33076593 degrees. Using this as a reference, you tilt the the orbit by the inclination ( 7.000497902 degrees).  Then  you measure out the longitude of periapsis ( longitude of perihelion for Mercury: 77.45779628 degrees)

A line passing through this point and the Sun follows the line of apsides ( which passes through both foci).

Now, as already mentioned, the values given are not constant. So for any given date, you will have to take into account that the inclination changes by -0.00594749 deg/century, the longitude of the ascending node by -0.12534081 deg/cen, and the longitude of perihelion by 0.16047689 deg/cen.

And even these values are only good for the interval of 1800 AD-2050 AD

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On 5/17/2020 at 4:08 AM, Strange said:

I may have misread the question as to how you could do it from a diagram like the one in the post.

To do it from a diagram like that, you can rotate the image 180 degrees, then overlap the ellipses. Here the two suns are the foci. Obviously the accuracy will be limited by image quality but for example you can see that the foci of Mercury's ellipse are farther apart than Earth's, and that Mercury is on the order of 3/2 as far from the sun at aphelion as it is at perihelion.

Mercury: image.png.4aeab43bb2eaeb91504e5692821645f0.png

Earth: image.png.bae83108677487c1b94d111343cb6a3e.png

 

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5 minutes ago, md65536 said:

To do it from a diagram like that, you can rotate the image 180 degrees, then overlap the ellipses.

Neat.

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Posted (edited)
On 5/12/2020 at 4:34 PM, studiot said:

One other thing (english word parameter) is which way round the Sun the planets travel Venus goes the other way form Earth.

Venus orbits the same direction as all the other planets.  But it spins on its own axis backwards, so it has one more day per year than its sidereal spin rate, as opposed to one less like all the other planets.  Earth for instance experiences 365.25 days per year, but spins at a rate of about 366.25 times.

Venus spins so slow that it goes around about once per Venus year, giving it just two days in its year.

Edited by Halc

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On 5/21/2020 at 5:44 PM, Halc said:

Venus orbits the same direction as all the other planets.  But it spins on its own axis backwards, so it has one more day per year than its sidereal spin rate, as opposed to one less like all the other planets.  Earth for instance experiences 365.25 days per year, but spins at a rate of about 366.25 times.

Venus spins so slow that it goes around about once per Venus year, giving it just two days in its year.

You are quite right. Thank you for that correction.  +1

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