<p>The (<em>x</em>, <em>y</em>, <em>z</em>) position of an object on its orbit with reference to a reference plane are given by:</p>
<p><em>u</em> = ω + ν</p>
<p><em>x</em> = <em>r</em> (cos Ω cos <em>u</em> − sin Ω sin <em>u</em> cos <em>i</em>)</p>
<p><em>y</em> = <em>r</em> (sin Ω cos <em>u</em> + cos Ω sin <em>u</em> cos <em>i</em>)</p>
<p><em>z</em> = <em>r</em> sin <em>i</em> sin <em>u</em></p>
<p>(MEEUS, Jean. <em>Astronomical Algorithms</em>, Second Edition, p. 233)</p>
<p>This should help you find what you need.</p>

CC BY-SA 4.0
2020-11-22T19:08:53.817Z
2020-11-22T19:08:53.817Z
Pierre Paquette
<p><a href="https://i.sstatic.net/UaxFd.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/UaxFd.png" alt="Question" /></a></p>
<p>An ellipse E is arbitrarily oriented in 3-D space with the origin at one of its foci. Use the standard elements provided in the figure. Assume the reference direction <span class="math-container">$\gamma$</span> or <span class="math-container">$\gamma'$</span> is the positive Y-axis direction. Find the projected position vector of the ellipse onto the plane of reference as a function of the angle <span class="math-container">$\nu$</span>.</p>
<p>After some calculation, I obtained the position vector of the celestial body to be
<span class="math-container">$$r=\frac{a(1−e^2)}{1+e\cos(\nu)}$$</span>
Where <span class="math-container">$a$</span> is the semi-major axis and <span class="math-container">$e$</span> is the eccentricity. Now, I begin with the <span class="math-container">$(r,ν)$</span> coordinates in the plane of the ellipse and project it onto the reference plane using the <span class="math-container">$\cos(i)$</span> factor. How, then, can I find the orthogonal components <span class="math-container">$(x,y)$</span> of this projected position vector on the reference plane? A parameterized solution would be appreciated as well.</p>

CC BY-SA 4.0
2020-06-23T21:40:54.747Z
The exam problem is not very clear to me.  I'll forgive the use of \gamma' for the Aries symbol.  What is don't understand is the expression "position vector of the ellipse"  Having thought about it, I still don't understand what that is supposed to mean.  Points have postion vectors. Ellipses don't have position vectors.
James K
CC BY-SA 4.0
2020-06-24T03:56:43.603Z
@JamesK When they say position vector of the ellipse, they mean the a general vector (meaning all the points can be represented by a particular parameter)
Arnav Das
CC BY-SA 4.0
2020-11-26T11:17:39.150Z
@JamesK any elliptical orbit will have what's called an "eccentricity vector" but I don't know if that's what's meant here. It's not sufficient to completely define an orbit, but it does give the line of apses and which apse is which.
uhoh
<orbit><gravity><exoplanet><orbital-mechanics><orbital-elements>
How to project an arbitrary elliptical orbit onto the night sky plane?
Arnav Das
# How to project an arbitrary elliptical orbit onto the night sky plane?

(Asked by: username_0 on 2020-06-23)

[![Question](https://i.sstatic.net/UaxFd.png)](https://i.sstatic.net/UaxFd.png)

An ellipse E is arbitrarily oriented in 3-D space with the origin at one of its foci. Use the standard elements provided in the figure. Assume the reference direction $\gamma$ or $\gamma'$ is the positive Y-axis direction. Find the projected position vector of the ellipse onto the plane of reference as a function of the angle $\nu$.

After some calculation, I obtained the position vector of the celestial body to be
$$r=\frac{a(1−e^2)}{1+e\cos(\nu)}$$
Where $a$ is the semi-major axis and $e$ is the eccentricity. Now, I begin with the $(r,ν)$ coordinates in the plane of the ellipse and project it onto the reference plane using the $\cos(i)$ factor. How, then, can I find the orthogonal components $(x,y)$ of this projected position vector on the reference plane? A parameterized solution would be appreciated as well.

**Tags:** orbit, gravity, exoplanet, orbital-mechanics, orbital-elements
> **Comments:**
> **username_1**: The exam problem is not very clear to me.  I'll forgive the use of \gamma' for the Aries symbol.  What is don't understand is the expression "position vector of the ellipse"  Having thought about it, I still don't understand what that is supposed to mean.  Points have postion vectors. Ellipses don't have position vectors.  
> **username_0**: @JamesK When they say position vector of the ellipse, they mean the a general vector (meaning all the points can be represented by a particular parameter)  
> **username_2**: @JamesK any elliptical orbit will have what's called an "eccentricity vector" but I don't know if that's what's meant here. It's not sufficient to completely define an orbit, but it does give the line of apses and which apse is which.  

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## 1 answers

### Answer by username_3 on 2020-11-22. Score: 1 | Accepted answer

The (*x*, *y*, *z*) position of an object on its orbit with reference to a reference plane are given by:

*u* = ω + ν

*x* = *r* (cos Ω cos *u* − sin Ω sin *u* cos *i*)

*y* = *r* (sin Ω cos *u* + cos Ω sin *u* cos *i*)

*z* = *r* sin *i* sin *u*

(MEEUS, Jean. *Astronomical Algorithms*, Second Edition, p. 233)

This should help you find what you need.


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