Trying to show that $r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right)$ is an ellipse

Question

I am trying to prove that planets move in ellipses,

I watched this video: https://www.youtube.com/watch?v=DurLVHPc1Iw

and read this: https://arxiv.org/pdf/1009.1738.pdf .

But both sources end up with this as an equation for the paths that planets follow : $$ r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right), $$ where $C=G\frac{m_{1}+m_{2}}{h^{2}}$ which is just a constant, as is $\omega$, and $e$ is the eccentricity, so this is an ellipse with the focus at the origin as long as $e<1$ and when I put the equation in graphing software it gives me ellipse.

But I don't understand why it is an ellipse? Can anyone show me how it is an ellipse? Either from the normal polar coordinates of a ellipse centred at the origin and shifting it (I couldn't get the algebra to work out when I added ae to the x coordinate) or just by explaining it conceptionally?

Answer 1

Let's define $p=1/C$ (the semi-latus rectum) and set WLOG $\omega=0$. Let's construct then a line (the directrix) perpendicular to the $\theta=0$ axis and lying at a distance $FH=p/e$ from the the origin $F$.

An ellipse can be defined as the locus of points $P$ having a fixed ratio $e$ between their distance $PF$ from the focus and their distance $PN$ from a fixed lined (the directrix).

Let's take then a point $P$ whose distance from the focus is given by your equation:

$$ PF=r={p\over1+e\cos\theta}. $$

The distance of $P$ from the directrix is given by:

$$ PN=FH-FK={p\over e}-r\cos\theta= {p\over e}\left(1-{e\cos\theta\over1+e\cos\theta}\right)={r\over e}. $$

Hence $P$ lies on the ellipse having $F$ as focus and $HN$ as directrix.

Answer 2

We have $$ \begin{split} r&= \dfrac{1}{C} \left(\dfrac{1}{1+e \cos(\theta+\omega)}\right) \\ &= \dfrac{1}{C+Ce \cos(\theta+\omega)} \\ &= \dfrac{1}{C+Ce \cos(\theta+\omega)} \cdot \dfrac{1/C}{1/C} \\ &= \dfrac{1/C}{1+ e\cos(\theta+\omega)} \end{split} $$ Write $P= 1/C$. Then we have $$ \begin{split} r= \dfrac{P}{1+e \cos(\theta+\omega)} \\ r+er \cos(\theta+\omega)= P \\ r= P - er \cos(\theta+\omega) \end{split} $$ Now let $x= r \cos \theta$. Then noting that in our case, $r^2= x^2+y^2$, we have $$ \begin{split} r&= P - er \cos(\theta+\omega) \\ r&= P - ex \\ r^2&= P^2 - 2Pex + e^2 x^2 \\ x^2+y^2&= P^2 - 2Pex + e^2 x^2 \\ x^2 - e^2x^2 +2Pex + y^2&= P^2 \\ (1-e^2)x^2 +2Pex + y^2&= P^2 \\ x^2 + \dfrac{2Pex}{1-e^2} + \dfrac{y^2}{1-e^2}&= \dfrac{P^2}{1-e^2} \end{split} $$ Now it is just a routine complete the square in $x$ and we have $$ \left(x+ \dfrac{Pe}{1-e^2} \right)^2 + \dfrac{y^2}{1-e^2}= \dfrac{P^2}{(1-e^2)^2} $$ Of course, this is the same as $$ \dfrac{(x+Pe/(1-e^2))^2}{P^2/(1-e^2)^2} + \dfrac{y^2}{P^2/(1-e^2)} = 1 $$ So that we have an ellipse 'centered' at the point $(-Pe/(1-e^2), 0)$, semimajor axis length $P/(1-e^2)$, semiminor axis $P/\sqrt{1-e^2}$, and with foci at distance $$ \sqrt{A^2 - B^2}= \sqrt{\dfrac{P^2}{(1-e^2)^2} - \dfrac{P^2}{1-e^2}}= \dfrac{P|e|}{1-e^2} $$