An exercise that was left unsolved from last year's class gives me this equation :
$$
t-t_{p} = \sqrt{\frac{a^3}{\mu}}*(\arcsin(X) - e*X)
$$
where :
$$
X = \frac{\sqrt{1-e^2}*\sin(v)}{1+e*\cos(v)}.
$$
This is just Kepler's equation $M = E-e\sin E$, but written in terms of $X = \sin E$, where $E$ is the eccentric anomaly. We don't have a derivation of Kepler's equation on this site, so here goes. I'll start with a picture.
The above image portrays a body $P$ in an elliptical orbit about a body $F$ that occupies one of the foci of the ellipse. The ellipse has a semi-major axis $a$ along the horizontal axis and an eccentricity $e$. The center of the ellipse and its circumscribing circle are at $C$. The vertical projection of the current location onto the circumscribing circle is denoted $P'$.
Kepler's second law says that the area of the elliptical sector $ZFP$ is a linear function of time: $A(ZFP) = k(t-t_p)$, where $A(ZFP)$ is the area in question, $k$ is some constant, $t$ is the time at which the orbiting object reaches the position $P$, and $t_p$ is the time of periapsis passage. In a full orbit, the area swept by that elliptical sector is the area of the ellipse: $A(2\pi) = \pi a b$. Thus $\pi a b = kT$, where $T$ is the orbital period,or $k = \frac{\pi a b}T$. Kepler's third law combined with Newtonian gravity in turn tells us that $\frac{2\pi}T = \sqrt{\frac \mu {a^3}}$, where $\mu$ is the system's gravitational coefficient $G(M+m)$. Defining $n = \sqrt{\frac \mu {a^3}}$, we have $$A(ZFP) = \frac12ab\,n(t-t_p)\tag{1}$$
We need an expression for $A(f)$. To get there, it's best to introduce the concept of eccentric anomaly. This is portrayed in the image as the angle $E$. This is formed by projecting the point $P$ vertically to the intersection with the circumscribing circle, denoted $P'$. Given a point $x,y'$ on the circumscribing circle expressed relative to the center $C$, the corresponding point on the ellipse $x,y$ results by scaling the $y$ coordinate by $\frac b a$: $y=\frac b a y'$. This scaling means that the area of the elliptical sector $ZCP$ is the area of the circular sector $ZCP'$ scaled by the same scale factor. Since the area of the circular sector $ZCP$ is $\frac 1 2 a^2 E$ with $E$ expressed in radians, the area of the elliptical sector $ZCP$ is $\frac 1 2 ab E$.
The area in question, that of the elliptical sector $ZFP$, is the area of the elliptical sector $ZCP$ less the area of the triangle $FCP$. The latter is $\frac12 \, ae \, b\sin E$ (1/2 * base * height), or $\frac12 ab\,e\sin E$. Thus $A(ZFP) = \frac12 ab (E - e\sin E)$. Combining this with equation (1) yields
$$E-e\sin E = n(t-t_p) \equiv M \tag{2}$$
This is Kepler's equation. It provides a simple mechanism for computing time as a function of position. Computing position as a function of time requires inverting this transcendental function of the two variables $E$ and $e$. This inverse function cannot be expressed in terms of the elementary functions.
A very simple, guaranteed to work, method to find $E$ given $M$ and $e$ is to use the fixed point iteration scheme $E_{n+1} = M + e\sin E_n$. Any starting guess $E_0$ will do, but typically $M$ is used as the initial guess. This converges for all $M$ and all eccentricities between 0 (inclusive) and 1 (exclusive). The convergence is very slow, particularly for large eccentricities. A better approach is to use Newton's method, which exhibits quadratic convergence when it converges. A good initial guess is needed for large eccentricities to ensure convergence. Even better approaches, and even better initial guesses than $E_0=M$ have been found over the centuries. Kepler's equation is the subject of hundreds of scientific papers.
Inverting Kepler's equation gives us the eccentric anomaly as a function of time. But what about the true anomaly $f$? The relationship between $E$ and $f$ is easily found using the tangent half angle formula, $\tan^2 \frac x2 = \frac{1-\cos x}{1+\cos x}$. The coordinates of the point $P$ with respect to the focus $F$ are $x=a(\cos E-e)=r\cos f$, $y=a\sqrt{1-e^2}\sin E = r \sin f$. Thus
$$\tan^2 \frac f 2 = \frac{1+e}{1-e} \,\tan^2 \frac E 2$$
(Deriving this requires deriving $r=a(\cos E - e)$, not shown.) Since the true anomaly and eccentric anomaly are always on the same side of the $x$ axis, the tangents of their half angles will always have the same sign, yielding
$$\tan \frac f 2 = \sqrt{\frac{1+e}{1-e}} \,\tan \frac E 2 \tag{3}$$
What about using $X=\sin E$ instead of $E$, as is done in the question? That's a "don't do that, then" kind of situation (from the joke, "Doctor, it hurts when I hit myself like this:《 bonk 》.") This only gets less than half of the ellipse, and the convergence of $\arcsin X - e X = M$ is terrible (if it converges at all). Use Kepler's equation (equation (2)) to solve for $E$, then solve for the true anomaly $f$ (alternatively written as $\nu$ or $\theta$) via equation (3).
