How does one calculate the time to go from one point in an orbit to another? I am playing Kerbal Space Program, and I am trying to find out how long it takes to go from one point in an orbit to another. Given the periapsis, apoasis, the gravitation parameter, and the two points distances from the center of the body (and there true anomalies), how do I calculate the time to go from one point in an orbit to another?
 A: As mentioned in a comment, there's not enough information in the question (version 1.0) to solve the problem. To provide enough information, one needs to know uniquely specify where the two points in question lie on the orbit.
The standard way to do this is to use the true anomaly, the angular displacement between the point in question and the periapsis point, with the angle measured through the focus at which the central mass lies, and with angle measured positively in the direction of the orbital travel.
Before I get to the standard way, I'll look at the short and easy way. Suppose you know the mean anomaly for those two points, $M_1$ (initial) and $M_2$ (final). Then the time of flight from point 1 to point 2 is $\frac{M_2-M_1}{n}$, where $n$ is the mean motion: $n = \sqrt{\frac{\mu}{a^3}}$. You know $\mu$ because that's a given. The semi-major axis $a$: That's a simple calculation given the periapsis and apoapsis distances $r_p$ and $r_a$: $a = \frac {r_p + r_a}{2}$.
All you need to do then is to know how to transform those true anomaly values to mean anomaly values. This is done in two stages for each true anomaly value. First compute the eccentric anomaly via $\tan \frac E2 = \sqrt{\frac{1-e}{1+e}} \tan\frac\theta2$, where $e$ is the eccentricity $e=\frac{r_a-r_p}{r_a+r_p}$. Then compute the mean anomaly via Kepler's equation $M=E-e\sin E$, where the mean anomaly $M$ and eccentric anomaly $E$ are expressed in radians. (Not degrees!)
There are some gotchas in the above, which is that the expression $\tan \frac E2 = \sqrt{\frac{1-e}{1+e}} \tan\frac\theta2$ produces a value between $-\pi$ and $\pi$ (exclusive) for angles $\theta$ that are not of the form $(2k+1)\pi$. One problem: What if you want to start or end at apoapsis? The answer is to use an eccentric anomaly of $\pm\pi$ for these cases. A second problem: What if you want to go from 179° to 183°? I'll address this next.
Suppose, for sake of simplicity, that $\mu$ is one length unit3 per time unit2, and the semi-major axis is one length unit, and the eccentricity is 1/3. The mean motion is thus $\sqrt{\frac{\mu}{a^3}}$ = 1/time unit. We want to go from $\theta_1=179^{\circ}=\frac{179}{180}\pi$ radians to $\theta_2=183^{\circ}=\frac{183}{180}\pi$ radians. Subtracting 360° ($2\pi$) from $\theta_2$ yields a value of $-\frac{177}{180}\pi$ radians. These correspond to eccentric anomalies of 3.116910597126455 and -3.067561512594607 radians, respectively. Applying Kepler's equation yields mean anomalies of 3.108684080302416 and -3.0429070002978842483, respectively. The difference between the two is -6.151591080600301 radians. We need to add $2\pi$ to address that $2\pi$ was subtracted from $\theta_2$, resulting in a mean motion change of 0.131594226579285 radians. Dividing by the mean motion results in 0.131594226579285 time units, or 1/50th of an orbit. Note that this is almost twice as much time as the it would take to move by 4 degrees in a circular orbit.
A: As David Hammen stated in his comment, for the given information: semi-major axis $a$, eccentricity $e$, gravitation parameter $\mu$, initial radius $r_1$ and final radius $r_2$, there is not a conclusive answer. Assuming the orbit has an eccentricity greater than zero (apoapsis is not equal to periapsis), then $r_1$ corresponds to two point on the orbit (or one if it is exactly equal to the radius of the periapsis or apoapsis). Namely one after periapsis and before apoapsis passage and the second one after apoapsis and before periapsis passage. The same can be said about $r_2$.
Assuming for now that they are both after periapsis and before apoapsis passage and $r_1<r_2$, then there are multiple ways of calculating the time between the two.
One would be to convert the radii into eccentric anomalies,
$$
E(r) = \cos^{-1}\left(\frac{a - r}{e\,a}\right).
$$
And then use Kepler's equation,
$$
M(r) = E(r) - e\, \sin E(r),
$$
where $M$ is the mean anomaly, from which the time since periapsis passage can be calculated with,
$$
t_p(r) = \sqrt{\frac{a^3}{\mu}} M(r).
$$
The time between radii can then be found by taking the difference, since they both have the periapsis as reference,
$$
\Delta t = t_p(r_2) - t_p(t_1).
$$
Another one would be to convert the radii into true anomalies,
$$
\nu(r) = \cos^{-1}\left(\frac{a (1 - e^2) - r}{e\,r}\right).
$$
And then use a similar equation to get the time since periapsis passage,
$$
t_p(r) = \sqrt{\frac{a^3}{\mu}} \left(2\tan^{-1}\left(\sqrt{\frac{1-e}{1+e}}\tan{\frac{\nu(r)}{2}}\right)-\frac{e\sqrt{1-e^2}\sin{\nu(r)}}{1+e\cos{\nu(r)}}\right).
$$
Or another one by combining either of the pair of equations directly,
$$
t_p(r) = \sqrt{\frac{a^3}{\mu}} \left(\cos^{-1}\left(\frac{a-r}{e\,a}\right)-\sqrt{e^2-\left(1-\frac{r}{a}\right)^2}\right).
$$
If either of the radii are positioned on the other half of the orbit then you just have to negate its $t_p(r)$ and take $\Delta t$ modulo the orbital period to get the actual time between the two points.
A: You can use Kepler second law, which states: A line joining a planet and the Sun sweeps out equal areas during equal intervals of time.
So you can calculate the area of the sector $S$ created by your two points, which is not funny at all : (Ellipse sector.) EDIT: It must be used the sector through the focus not the center, focus sector, it looks like they do it here, I didn't check the math.
Than you have $t=\frac S{A}P$
Where $A=ab\pi$ is the ellipse area and $P=2\pi\sqrt{\frac {a^3}{G(M_1+M_2)}}$ is the period of the total orbit.
