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Take the staring point as the origin for vectors. Then the trajectory is given by $$\vec r(t)=\vec{v}_0t+\vec{g}t^2/2$$ Suppose now that your target is at position $\vec{r}_1$, so we have to ensure that there exist a solution for $t$ of $$\vec{r}_1=\vec{v}_0t+\vec{g}t^2/2$$ Lets look at this equation solved for $\vec{v}_0$: $$... 4 You're assuming that the Kruskal–Szekeres (U,V) coordinates have to be defined in terms of the Schwarzschild (r,t) coordinates, but there is nothing special or fundamental about the Schwarzschild coordinates. General covariance says that we can use any coordinates we like. If the K-S coordinates had been the ones originally chosen by Schwarzschild, then ... 1 The Schwarzschild metric is, in -+++ sign convention and units of c = 1 is$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \frac{\mathrm{d}r^2}{1-\frac{2M}{r}} + r^2\left(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2\right)\text{.}$$We can index the coordinates arbitrarily, but let's take them in the typical order: (U^0,U^1,U^2,U^3) ... 2 1) Spline curves are designed for just this sort of thing. A spline is basically a set of cubic (typically) polynomials at each section of your data that go exactly through your points. But since they're just cubic polynomials, they're pretty smooth. Assuming the path is a reasonably small portion of the earth, and the points are close together on a ... 1 You have to be careful when you derive your equations of motion in cylindrical versus cartesian coordinates. For example, if your Hamiltonian (I'm ignoring the z direction because it is the same in cylindrical to cartesian) is  H = \frac{p_x^2}{2} + \frac{p_y^2}{2} + V(x, y) then in cylindrical coordinates you get  H = \frac{p_r^2}{2} + ... 3 Consider an equation that you have written in Cartesian coordinates x,y,z which gives you these coordinates as a function of time or something like that. Then you want to write it in, say, cylidrical coordinates. Then you write something like:$$ x=r\cos\phi\\ y=r\sin\phi\\ z=z  and plug it into your equation. Now just forget the word coordinates. It is ...