MHB Formula for Ellipse 'Change of Distance by Change of Theta'?

[SquishyWeRRa]

I am working on planetary orbits in an ellipse where the sun is at the foci, not the centre of the ellipse.

I need a formula that describes the 'Change of Distance by Change of Theta angle' where Distance represents the distance between a planet to the sun (Planet is on parameter, Sun is at foci) and Theta represents the angle swept out when a planet moves from one position to the next.

E.g. Let Theta change by 1 degree, what is the change of Distance from the planet to the sun?

Since the orbit is ellipse, the change of distance by change of radius is not constant.

Furthermore, the distance is not calculated from the centre of the ellipse, but at the foci of the ellipse’

Kepler's 2nd law states that the area swept over a specific amount of time is constant.

Assuming that each sector's segment is negligible, the sector becomes a triangle

Area of triangle= 1/2 ab sin⁡C
dA=1/2 (Old Radius)(New Radius)sin⁡(dθ)
Using small angle approximation, sin(dθ)=dθ
dA/dt=(Old Radius)(New Radius)(dθ/dt)
Summation of dA is Area of ellipse πab
Summation of dt is Orbital period
πab/P=(Old Radius)(New Radius)/2 * (dθ/dt)

I know a, b and P. I can let dθ be increments of 1 degree. I need to know how New Radius changes to Old Radius with consideration of dθ. I need to find dt, which is the time taken to sweep that specific 1 degree.
Thank you

 

[I like Serena]

Hi SquishyWeRRa! Welcome to MHB! (Smile)

Here's how an ellipse with its foci looks like, including the relevant formulae.
\begin{tikzpicture}
\draw[gray, very thin,->] (-6,0) -- (6,-0); % x-axis
\draw[gray, very thin,->] (0,-4) -- (0,4); % y-axis
\draw[gray, very thin] (-5,-3) rectangle (5,3);
\draw[red,line width=2pt] (-4,0) -- (-4,1.8) node[right=1pt] {$\ell=\dfrac{b^2}{a}$}; % semi latus rectum
\draw[domain=-180:180,smooth,variable=\t,line width=2pt] plot ({5*cos(\t)},{3*sin(\t)});
\node at (1.2,-2) {$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}=1$ };
\node at (-1.5,-2) {$r=\dfrac{b^2}{a - c \cos\theta}$ };
\node at (1.5,-3.5) {$(a \cos u, b \sin u)$ };
\draw (-4,0) -- (0,3) -- (4,0);
\node at (2,3pt) {c};
\node at (-4pt,1.5) {b};
\node at (2.2,1.65) {a};
\draw[<->, green] (-5,-0.3) -- (0,-0.3);
\node[green] at (-2.5,-5pt) {a};
\fill (-4,0) circle (0.1); % focus
\fill (4,0) circle (0.1); % focus
\end{tikzpicture}
It looks like you're looking for the formula:
$$r=\dfrac{b^2}{a - c \cos\theta}$$
The change in distance $dr$ if we change $\theta$ by $d\theta$ is given by the derivative.
And the relationship with $dt$ follows from Kepler's law.