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In particle scattering there's a term called "impact parameter", which is the minimum distance a particle would pass a second particle at rest, if the attractive or repulsive force were ignored or "turned off".

In a flyby mission, is this also called impact parameter, or is there a different word for it?

below: From http://hyperphysics.phy-astr.gsu.edu/hbase/rutsca.html#c3

impact parameter

below: From http://www.astro.uwo.ca/~jlandstr/planets/webfigs/formation/slide5.html

enter image description here

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It's called $\boldsymbol B$ or $\vec B$, which is a vector representing not just the distance, but also the clock angle in the "B-plane", which is the plane through the center of the body and perpendicular to the approaching asymptote. That vector is also referred to as the "miss parameter". (Perhaps when you're talking about an expensive spacecraft, that term is a little less alarming than "impact parameter".)

I presume that they chose that letter because the magnitude of $\boldsymbol B$ is the semi-minor axis of the hyperbola, commonly referred to as $b$. Though I would never see $b$ written in the context of interplanetary trajectories, only $\boldsymbol B$, since that vector is the prime focus of the interplanetary navigator.

You can find this definition in the classic and indispensible Interplanetary Mission Design Handbook, on page 20. Here is the diagram copied from there:

B-plane image showing plane cut through the body perpendicular to the incoming trajectory direction, and the B vector ending at the point in the B-plane that that trajectory would go through if the body were not there or had no gravity

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    $\begingroup$ This is great, thanks! Yes, I like "miss" parameter better as well; spacecraft are less expendable than alpha particles. So in How to understand this complicated plot for Mariner 4's mid-course correction (to Mars)? I'm wondering then if that polar diagram might represent something analogous to the $\mathbf{B}$ vector, with the origin along Mariner 4's trajectory? $\endgroup$ – uhoh Sep 2 '18 at 16:32
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    $\begingroup$ Yes. That diagram is looking directly at the B-plane. The points marked 1, 2, 3, 4 are successive navigation solutions of the B vector after the mid-course maneuver, as they collected more data. $\endgroup$ – Mark Adler Sep 2 '18 at 16:46
  • $\begingroup$ Your link to the Interplanetary Mission Design Handbook is baroque, but these have similar names, would one be a suitable replacement? 1, 2, 3, 4 $\endgroup$ – uhoh Jun 19 at 2:15
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    $\begingroup$ Fixed. Thanks.. $\endgroup$ – Mark Adler Jun 19 at 4:13
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It's also called 'impact parameter' - or more commonly, semi-minor axis, in case of hyperbolic trajectories. It's not defined for elliptical orbits, and for bodies of negligible mass passing each other, it's 'separation at closest approach'.

enter image description here src

Formally, for hyperbolic orbits, 'impact parameter' is defined as 'distance between an asymptote and a focus'. The value, usualy denoted as $b$ is $$b = -a \sqrt{e^2-1}$$ where

  • $a$ is the semi-major axis (distance from center (intersection of asymptotes) and vertex (closest approach point on hyperbola))
  • $e$ is eccentricity, $e = {c \over a}$
  • $c$ is linear eccentricity (distance from center to focus). The focus being the location of the massive body by which the flyby is performed, or barycenter if both bodies have a considerable mass.
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@SF. and @MarkAdler gave excellent answers. There is another closely related parameter that appears in the diagram by Beiser in your question statement: $\theta$.

The astrodynamics equivalent is the bending angle, $\delta$. It is the angle from the direction of the approach asymptote (and thus the approach $V_∞$ vector) to the direction of the departure asymptote (and thus the departure $V_∞$ vector), making it the angle through which the trajectory "bends" as a result of the flyby.

The $\theta$ shown in the Beiser diagram is appropriate for repulsive forces, with the trajectory of the "projectile" bending away from the nucleus. Were the projectile negatively charged the trajectory would bend toward the nucleus, much as gravity bends the trajectory of an object flying by a massive body (the primary) toward the primary.

If you know the trajectory's eccentricity e, then finding $\delta$ is easy via the relation $$\sin\frac{\delta}{2} = \frac{1}{e}$$ yielding $$\delta = 2\sin^{-1}\frac{1}{e}$$

Various constraints on the trajectory can set the value of e, assuming a spherical primary. As an example, for a primary with gravitational parameter $\mu$ (= GM, where G is the universal gravitational constant and M is the mass of the primary), if you know the magnitude of the approach $V_∞$ and want a specific radius of closest approach $r_p$ (periapse radius), the eccentricity is then $$e = 1 + \frac{r_p}{\mu} {V_∞}^2$$ and b follows: $$b = r_p \sqrt{\frac{e+1}{e-1}}$$

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