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It's going to take a lot of money, time and resources to design & launch interstellar probes. It would certainly be helpful if we could develop a mission profile that would lead to an orbital capture rather than a flyby. There may be many exotic trajectories to insert a probe into an orbit, but it seems logical to me to start with the simplest: thread the needle right between the two stars with a speed less than escape velocity.

Note: if my somewhat simple grasp of orbital mechanics is in error & this is not the simplest trajectory to achieve orbital insertion, I would appreciate an explanation.

Also, for the purposes of this question, let's ignore the various schemes proposed for deceleration.

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    $\begingroup$ Yes, one has to assume that any practical orbital insertion maneuver would include deceleration on the inbound leg, if nothing else than a "drag" solar sail deployed behind the craft once it enters the stellar system's sphere of influence. My specific question is an attempt to place an upper bound on what final velocity needs to be after deceleration. $\endgroup$ – Jerard Puckett Feb 11 '14 at 13:53
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This is a good question, but I'm afraid that the answer isn't anything particular, it's only constrained by the direction vector. In case of a vector exactly perpendicular to both bodies of the binary system, escape velocity is combined escape velocity of each of the two bodies at the distance to them. So:

$$v_{e}={\sqrt {{\frac {2GM_\alpha}{r_\alpha}}}} + {\sqrt {{\frac {2GM_\beta}{r_\beta}}}}$$

Where $G$ is the universal gravitational constant $6.67384(80)\cdot 10^{{-11}}\ {\mbox{m}}^{3}\ {\mbox{kg}}^{{-1}}\ {\mbox{s}}^{{-2}}$. This function of ours is describing combined gravitational potential of two bodies at their common barycenter that we need to escape from, and is smallest with a direction vector exactly perpendicular to the axis between them, keeping their gravitational potential equipotent. Changing vector even slightly in direction of either one will only result in greater required escape velocity, since you start approaching one of them and decreasing radius $r$, while losing gravitational potential in the opposite vector by the other one, but it would be somewhat more forgiving in direction towards the less massive body.

In our case, the common barycenter is the saddle point, or $L_1$ Lagrange point at which the gravitational potential of the two bodies is equipotent. But to get that distance relative to both bodies, since Alpha Centauri binary stars α Cen A and α Cen B (our binary system Alpha Centauri AB) aren't exactly of the same mass, we'll have to first calculate it. Wikipedia on Alpha Centauri tells us of their distance between each other this:

During the pair's 79.91-year orbit about a common center, the distance between them varies from about that between Pluto and the Sun to that between Saturn and the Sun.

This is not too encouraging with respect to precision, but since we're lowering the bar already and calculating averages, it allows us to neglect gravitational effects of Proxima Centauri (Alpha Centauri C). So the α Cen AB are distant to each other from 778,547,200 km (5.204267 AU) up to 5,874,000,000 km (39.264 AU), which are semi-major axes of Jupiter and Pluto, respectively. Let's split this in geometric half, at 3,326,273,600 km (22.2341335 AU). Now, α Cen A has a mass of $M_\alpha = 1.100\ \mbox{M}_{\odot }$ and α Cen B $M_\beta = 0.907\ \mbox{M}_{\odot }$, where the unit is in solar masses $\mbox{M}_{\odot }=(1.98855\ \pm \ 0.00025)\ \cdot 10^{{30}}{\hbox{ kg}}$.

To find the distance to common barycenter, we'll have to solve a two-body problem. For the distance to the primary star, α Cen A, our solver looks like this:

$$r_{\alpha}=a\cdot {M_{\alpha} \over M_{\alpha}+M_{\beta}}={a \over 1+M_{\alpha}/M_{\beta}}$$

Where $a$ is the distance between the centers of the two bodies (3,326,273,600 km). Solving this, we get $r_{\alpha} = 1,823,069,735.92\ \mbox{km}$ and since $r_{\beta} = a - r_{\alpha}$, we get $r_{\beta} = 1,503,203,864.08\ \mbox{km}$. OK, we now have all the information we require to plug it in our first solver, and we get:

$$v_{e\alpha}={\sqrt {{\frac {2 \cdot 6.67384\cdot 10^{{-11}}\ {\mbox{m}}^{3}\ {\mbox{kg}}^{{-1}}\ {\mbox{s}}^{{-2}} \cdot 1.100 \cdot 1.98855 \cdot 10^{{30}}{\hbox{ kg}}}{1,823,069,735,920\ \mbox{m}}}}}$$

$$v_{e\beta}={\sqrt {{\frac {2 \times 6.67384\times 10^{{-11}}\ {\mbox{m}}^{3}\ {\mbox{kg}}^{{-1}}\ {\mbox{s}}^{{-2}} \cdot 0.907 \cdot 1.98855 \cdot 10^{{30}}{\hbox{ kg}}}{1,503,203,864,080\ \mbox{m}}}}}$$

or $v_{e\alpha} = 17,064.14\ {\mbox{m s}}^{{-1}}$ and $v_{e\beta} = 12,655.11\ {\mbox{m s}}^{{-1}}$. Assuming I plugged all the right numbers at all the right places, and there isn't any major flaw in my approach, we get:

$$v_{e} = v_{e\alpha} + v_{e\beta} = 29,719.25 \pm 469.90\ {\mbox{m s}}^{{-1}}$$

These are of course all just ballpark averages, and the escape velocity would lower when α Cen A and B are more distant to each other, and increase as they approach. It also doesn't include position of Proxima Centauri with respect to the Alpha Centauri binary and approach vector, so don't send anything there just yet. :)

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    $\begingroup$ At that speed, it looks like an approach vector that soaks up some of Proxima's radial velocity toward the Sun of 21.7 km/s is well worth a look. $\endgroup$ – Jerard Puckett Feb 10 '14 at 23:24

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