It is clear that particals in a repulsive central field given by a point mass at the origin are moving along hyperbolas, e.g. given by the expression ${x^2 \over a^2} - {y^2 \over b^2} = 1$ (after a change of coordinates). This also applies to three dimensions since the motion takes place in a plane due to conservation of angular momentum.

On the other hand, in three dimensions, the motion satisfies the initial value problem $$ \left\{\begin{array}{l} {d^2X \over dt^2} = C {X \over |X|^3}, \\ X(0) = x, \dot X(0) = v \end{array}\right. $$ for some given initial position and velocity $(x,v) \in \mathbb{R}^6$.

Unfortunately, while I can derive the trajectories, I have no idea, what can be said about the speed at which the particle moves at a certain time. In particular, I am interested in the quantitative behavior of $|V| = |\dot X|$ for given $x$ and $v$.

Are there any expansions or explicit expressions for the solution $X$ of the above initial value problem? Or is there any expansion or explicit expression for the absolute velocity $|V|$?

EDIT By conservation of energy we know that $$ \frac{1}{2}|V|^2 = \frac{1}{2}|v|^2 + C\left(\frac{1}{|x|} - \frac{1}{|X|}\right). $$ On the other hand, the behaviour of $\frac{1}{|X|}$ is well understood according to https://en.wikipedia.org/wiki/Kepler_problem#Solution_of_the_Kepler_problem $$ \frac{1}{r} = -\frac{km}{L^{2}} \left[ 1 + e \cos \left( \theta - \theta_{0}\right) \right]. $$ Actually, I don't quite understand how $\theta$ can be replaced by a time dependent expression and how the constants depend on the initial data $x$ and $v$...


1 Answer 1


You have rightly answered your own question in EDIT.

Putting $k m/ L^2 = 1/p $ where p is semi-latus rectum, the last equation becomes

$$ \frac{p}{r} = \left[ 1 + e \cos \left( \theta - \theta_{0}\right) \right] $$

in which $ \theta $ is the polar coordinate for classical polar conic equation.If $ \theta_0 $ is is non-zero, the conic is rotated so that major axis would not be along x- or y- axis.

You would notice that certain values of L,m k and total energy E decide eccentricity e. It will be found that high energy orbits force the orbit along non-returning hyperbolas.

Notice also that this same Newtonian formulation can result in hyperbolas, no need to have a separate repulsive differential equation. e =2 for hyperbolic orbits are sketched in:



In order to change to independent time variable from $\theta$ independent variable,

$$ u''+ u = -1/p, \theta^{'} = L u^2 $$

the latter part expresses conservation of angular momentum.

  • $\begingroup$ I don't know, how this answers my question. It's still mysterious, how the constants depend on $x$ and $v$ and we still only have a representation depending on $\theta$, while I'm interested in the evolution in time. $\endgroup$
    – thomas
    Mar 6, 2015 at 8:38
  • $\begingroup$ Actually the set of derivatives of $ x , v $ and the dynamic relations involve the constants. $\endgroup$
    – Narasimham
    Mar 6, 2015 at 9:41
  • $\begingroup$ It seems to me like the derivation of this solution in en.wikipedia.org/wiki/… is nonsense (it is asserted that $\omega$ is a conserved quantity). But the derivation on en.wikipedia.org/wiki/Kepler_orbit looks good. However, I still don't understand, how you get an expression in the time variable from this. $\endgroup$
    – thomas
    Mar 8, 2015 at 17:34
  • $\begingroup$ @thomas: Please see carefully it says that $ m \omega r^2 = L $, the angular momentum is conserved. There is no neat closed form expression for $ r = f(t) $ existing, may be possible in elliptic integrals,let me check.. The ellipse $ r- \theta $ relation is in closed form of course. $ r'' + \mu/r^2 - h^2 /r^3 = 0 $ is the radial ODE to be solved for $ r-time $ relation. $\endgroup$
    – Narasimham
    Mar 8, 2015 at 18:31
  • $\begingroup$ Oh, I see. You are right. I understood the $\equiv$-sign as "equals constantly", but they seem to use it in the sense "defined as" (which I usually denote by $:=$). So, possibly there is no expansion for $r=|X|$ in terms of $t$. But my question was actually targeted at some expansion of $|V|$ in terms of $t$, possibly only an estimate. I had some hope that at least that would be possible. $\endgroup$
    – thomas
    Mar 8, 2015 at 23:02

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