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Voyager 2 will leave the dominating gravitational influence of the Sun for a region where other stars have comparable influence on the craft. By the time it gets that far away, how much will it have slowed down due to the Solar system pulling backward on it?

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  • $\begingroup$ The escape velocity falls with the root of the distance from a gravitating body, so at the current distance of approx. 109/133 AU from the sun the Voyager probes are still subject to a solar escape velocity of approx. 1/10-th of the solar escape velocity at Earth's orbit, i.e. something on the order of 4km/s. $\endgroup$ – CuriousOne Sep 8 '15 at 20:30
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You could argue about exactly where that location is. But if we ignore the influence of neighboring systems, you can calculate the speed it has at any distance from the sun.

But for simplification, we can just calculate the speed it would have at infinity. It will retain that excess until it encounters some other body, and it won't be going much faster than that.

In January 2015, Voyager 2 had a speed (relative to the sun) of $15.497 km/s$, at a distance of $1.6\times10^{10} km$. (From Voyager weekly reports)

The escape speed from the sun at that distance is: $\sqrt{\frac{2GM}{r}}$ or $4.07 km/s$. And for a hyperbolic trajectory, $v^2 = v_{esc}^2 + v_{inf}^2$, or rearranging:

$$ v_{inf} = \sqrt{v^2 - v_{esc}^2}$$ $$ v_{inf} = \sqrt{(2.4\times10^{8} m^2/s^2) - (1.66\times10^7m^2/s^2)}$$ $$ v_{inf} = 14.95 km/s$$

The craft has a speed so much in excess of escape that it will lose very little to the sun the rest of the way to anywhere.

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