2.1.3 Kepler's Equation:

This equation will give you the position of any orbiting object, the equation is

${M}={E}-{e}{\sin{{\left({E}\right)}}}$

Where

M = Mean anomaly

E = Eccentric Anomaly

e = eccentricity

This equation is Transcendental because it mixes a free term E with a term inside of a function sin(E) this means that it can only be solved numerically or approximated with methods like Newton-Raphson.

Derivation:

Consider an eclipse with semi-major axis a and semi minor-axis b and eccentricity e. The focus of the eclipse is a distance ae from the center for this ellipse you circumscribe a circle.

For a point P on the eclipse you project a line outwards from the center of the ellipse to the circle that you circumscribed into the object this is point Q. The angle from the periapsis (far right of the ellipse) to the point Q is E, or the eccentric anomaly

Because point P is directly below point Q the x coordinates are the same and because the x coordinate at point Q is equal to r*cos(E) point is equal to

$x = \ a\ Cos(E) - ae$

you subtract ae to make the focus be the point of reference

The y coordinate on the other hand works out to be

$y = b\ Sin(E)$

This is because an ellipse is just a compressed circle so every y value on an eclipse is just compressed by a value of $\frac{b}{a}$ so we have $\frac{b}{a}*aSin(E)$ which works out to our equations

The only other thing that we need to include in the set up is Kepler's 2nd law Kepler's second law states that an object sweeps equal areas in equal times. The total area of your ellipse is $\pi ab$ so that means that we can use the rule of 3

$\frac{Area\ Swept}{\pi ab} = \frac{t}{T}$

Where T is total time needed to complete an orbit

M (Mean anomaly is defined as)

$M = \frac{2\pi t}{T}$

So were left with

$Area\ Swept = \frac{1}{2}ab*M$

The Area of elliptical sector from center is given by

$\frac{b}{a}*\frac{1}{2}*a^{2}E = \frac{ab}{2}E$

Now the area of the triangle created by the triangle formed by the focus and the center of the eclipse is given by

$Triangle\ area = \frac{1}{2}bh$

${b}={a}{e}$

h = the height/y value

$Triangle\ area = \frac{1}{2}aey$

And we know that

$y = b\ Sin(E)$

So

$Triangle\ area = \frac{1}{2}abeSin(E)$

Subtracting that area from the total area to get the area of the focus

$Total\ area\ of\ arc\ from\ focus = \frac{ab}{2}E - \frac{1}{2}abeSin(E)$

factorizing

$Total\ area\ of\ arc\ from\ focus = \frac{ab}{2}(E - eSin(E))$

Now equalizing this equation to Kepler's second law

$\frac{ab}{2}(E - eSin(E)) = \frac{1}{2}ab*M$

We get

${E}-{e}{\Sin{{\left({E}\right)}}}={M}$

Or Kepler's equation