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I'm trying to derive the universal formulation of the time-of-flight equation that appears in the Kepler problem (following Bate Fundamentals of astrodynamics and Vallado Fundamentals of astrodynamics and applications), and I don't see a point about the hyperbolic case.

Starting from the energy equation:

$$ \dot{r}^2 = -\frac{\mu p}{r^2} + \frac{2\mu}{r} -\frac{\mu}{a} $$

after a Sundman transformation ($\dot{\chi} = \frac{\sqrt{\mu}}{r}$ ) and separating we have

$$ d\chi = \frac{dr}{\sqrt{-p + 2 r - \frac{r^2}{a}}}$$

Then both Bate and Vallado proceed and integrate assuming the elliptic case ($a > 0$), yielding

$$ \chi + c_0 = \sqrt{a} \arcsin{\frac{\frac{r}{a} - 1}{\sqrt{1 - \frac{p}{a}}}}$$

Later it is justified that this formulation is valid for elliptical, parabolic and hyperbolic orbits using the Stumpff functions, even though the integration was not performed for the case $a < 0$ ($a = 0$ is trivial). Furthermore, Vallado states "this case results in a hyperbolic sine solution, which we won't use" but after several trials I got a hyperbolic cosine solution instead:

$$ \chi + c_0 = \sqrt{-a} \cosh^{-1}{\frac{\frac{r}{-a} + 1}{\sqrt{1 - \frac{p}{a}}}}$$

I expected a sinh though, to keep simmetry with the other case.

My main question is: how come can you start from a certain case and then prove the formulation is valid for all of them? And as an aside: how would the hyperbolic derivation be?

Note: I know Battin An Introduction to the Mathematics and Methods of Astrodynamics has probably more detailed math but there's no way I can get one until the end of the holidays.

astrojuanlu
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  • This is maybe a related question. And to answer your main question, I would suggest testing it. For example comparing it to a numerical simulation, such as ode45. – fibonatic Dec 29 '13 at 22:26
  • Are you familiar with the complex relation $\sinh ix=i\sin x$? This implies $i\sinh^{-1}x = \sin^{-1}ix$. – Pulsar Dec 29 '13 at 23:55
  • @fibonatic The point is that at the end of the derivation both introduce the Stumpff functions (which work for both positive and negative values of a) and then go backwards and show the formulation works for all cases, so in principle it's already tested. Thanks for linking your question but I don't think it's related to mine. – astrojuanlu Dec 30 '13 at 12:49
  • @Pulsar yes I am, and it's what motivates the definition of the Stumpff functions, but it's of little help at this stage. As I said above, I'm trying to go the inverse path (which, for me, seems the direct path). – astrojuanlu Dec 30 '13 at 12:50
  • All right, I matematically proved it. I'm answering my own question this year or the next :) – astrojuanlu Dec 30 '13 at 17:02

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The hyperbolic cosine solution is actually correct, contrary to what Vallado seems to state. The key here is in the definition of the Stumpff functions (which Bate calls universal functions with a change of variable).

If we solve for $r$ from the elliptic case result, we have $r = a \left(1 + e \sin{\frac{\chi + c_0}{\sqrt{a}}} \right)$, whereas for the hyperbolic case it is $r = a \left(1 - e \cosh^{-1}{\frac{\chi + c_0}{\sqrt{-a}}} \right)$. Both Bate and Vallado derivate an expression for $r$ in terms of $\chi$ and $\psi = \chi^2 / a$, which is:

$$r = \chi^2 C + \frac{\mathbf{r}_0 \cdot \mathbf{v}_0}{\sqrt{\mu}} \chi (1 - \psi S) + r_0 (1 - \psi C)$$

where $S$ and $C$ are the second and third Stumpff functions, in this case taking positive arguments. I then followed exactly the same derivation starting from the hyperbolic solution and arrived to the very same equation, with the Stumpff functions taking negative arguments this time. Plus it's easy to see that going from one form to the other only takes using the properties of the complex numbers, and everything works. Problem solved.

astrojuanlu
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