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When doing some back of a napkin calculations for a recent question Where would the Pascal B manhole cover be now?, I realised a quick way of solving these class of orbital mechanics question is to first work out periapsis distance and velocity from the initial state vector, which is simple (much simpler than a full set of orbital elements, for instance).

But then, is there some simple way to calculate the angle between the current velocity vector and the asymptotic velocity vector (the "Kentucky windage"), expressed by just $\mu$, $r_P$, $v_P$ and $r$? Any equivalent set of values, like deflection angle, impact parameter, $v_{\infty}$ or eccentricity would of course also be perfectly valid.

The answer is of course yes, but I can't see I have any immediately applicable formulas lying around, so I would have to derive it myself if there aren't anyone out there who already has it readily available.

kentucky windage

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    $\begingroup$ I always love seeing questions arise from attempts to answer other questions. I considered answering that question, until I considered how I don't actually know the relevant maths. $\endgroup$
    – Ingolifs
    Aug 27 at 4:40
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Using Perifocal polar coordinates, where the x-axis points from the central body to the periapsis, and the polar equation for conic sections: $$r=\frac{a(1-e^2)}{1+e \cos(f)}$$

Perifocal-coordinate image for calculation of "Kentucky windage" angle

Provided parameters
$\mu$ Standard Gravitational Parameter of the central body
$r_p$ Periapsis Distance (Point P in the diagram)
$v_p$ Speed at Periapsis
$r$ Radial distance at an arbitrary point post-periapsis. (Point P' in the diagram)

$r_p$ and $v_p$ define the specific orbital energy ($\epsilon$): $$\epsilon=\frac{v_p^2}{2}-\frac{\mu}{r_p}$$

And since they're perpendicular at periapsis, their product is the specific relative angular momentum ($h$): $$h =r_p v_p \sin\frac{\pi}{2} = r_p v_p$$

Semi-major axis ($a$) can be found from specific orbital energy : $$a=-\frac{\mu}{2\epsilon}$$

Orbital eccentricity ($e$) from semimajor axis and periapsis distance : $$ e=1-\frac{r_p}{a}$$ Given the radial distance at $P'$, you can solve for true anomaly ($f$) at $P'$:

$$f=±\arccos\left(\frac{a(1−e^2)−r}{er}\right)$$

And use the vis-viva equation to get $v$, the speed at $P'$ : $$v=\sqrt{\mu\left(\frac{2}{r}-\frac{1}{a}\right)}$$

Direction of the velocity vector ($\arg(\vec{v})$) at $P'$ can be found by conservation of angular momentum and the true anomaly: $$\arg(\vec{v})=f+\arcsin\frac{h}{rv}$$

Direction of the asymptote ($\theta_{\infty}$) is: $$\theta_{\infty}=\arccos\left(-\frac{1}{e}\right)$$

And your "Kentucky windage" angle ($\kappa$) is: $$\kappa =\theta_{\infty}-\arg(\vec{v})$$

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    $\begingroup$ Whoops, drawing the diagram in GeoGebra made me realize I'd flipped a sign on Kentucky Windage. Should be good, now. $\endgroup$
    – notovny
    Aug 27 at 11:23
  • $\begingroup$ You math people are an enduring source of wonder. $\endgroup$ Aug 31 at 18:04

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