In this example we have a celestial body following Keplar's laws of planetary motion, this is done as a function of time, an example of this can be found here on Wikipedia. (Calculations are done in 2 dimensions so we can ignore the z axis.) With this we also have all the data needed to calculate the orbit, (epi/perihelion distance, mass, etc.) in addition we also have an initial position at time t0.
The rocket is traveling at a constant velocity, we know the velocity, but not the vector in which it travels. The purpose of finding the time of intersection is to find this out. Like with the celestial body we also know its initial position at time t0. We also know t0
Is it possible to find either the time at which the rocket and the celestial body would intersect (the ideal scenario), or the velocity vector itself for the rocket to travel the shortest possible distance, and if so, how could this be calculated?
A Kepler orbit in the plane with greater axis has the orbit form
$$r(\phi) = \frac{a(1-\varepsilon^2)}{1+ \varepsilon \cos( \phi-\alpha)}$$
$\alpha$ is the angle of orientation of all orbits of the same excenticity $\varepsilon$ and axis parameter $a$.
The orbit is an ellipse for $\varepsilon<1$.
Since a rocket will follow an elliptic orbit too, that typically starts nearly tangential to the earth orbit, because fast rockets on hyperbolic orbits need velocities high enough to leave the solar system, are to heavy and too expensive.
The ballistic rockets start below 8km/s, the earth bound satelites below 11 km/s and the cosmic velocity for a hyperbolic orbit to leave the solar system is about 17 km/s
With three parameters $\alpha, a, \varepsilon$ its easy find the set of second ellipses between an arbitrary point on the first ellipse and an arbitrary starting point with an variing arbitrary tangent vector. The numerical or algebraic integration of time against radius is simple.
As always in mechanics, high energy motion is free and connects two points by a straigth line in no seconds as the limiting orbit of hyperbolas.
By energy and angular momentum conservation the time integral derives from energy by
$$\frac{r'^2}{2}+ \frac{L^2}{2 r^2} - \frac{GM}{r} =- E_{tot}$$
$$ dt = \frac{dr}{\sqrt{\frac{2 GM}{r}-\frac{L^2}{r^2} - 2 E_{tot} }}$$
The indefinite integal has a closed algebraic useless form, that could be used for mononotone invertible parts, but its complex form makes its use as a definite integral difficult.
$$t=\frac{-8 P+\sqrt{2} \text{gM} \left(i \log \left(8 \text{En} \text{gM} r-8 \text{En} \left(4 \text{En} r^2+L^2-2 \sqrt{2} P r\right)+\text{gM}^2\right)+2 \tan ^{-1}\left(\frac{2 \sqrt{2} P-4 i \text{En} r}{\text{gM}}\right)\right)}{8 \text{En}^{3/2}}$$ with $P = \sqrt{ En ( gM r -L^2 - 2 En r^2)}$