# How to project an arbitrary elliptical orbit onto the night sky plane? 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.

• 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. Jun 23 '20 at 21:40
• @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) Jun 24 '20 at 3:56
• @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
Nov 26 '20 at 11:17

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)