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Basically, what the title asks. To my understanding, New Horizons is currently the fastest spacecraft that's moving away from the Sun, with arguably only Helios-A and Helios-B possibly faster still, but in an elliptical orbit around the Sun, so they won't make it out of the heliosphere.

So I'm curious, when will New Horizons traverse larger distance from the Sun than the Voyager 1 will in that time, and at approximately what distance?

I'm not entirely sure this can yet be calculated with any precision, since we don't know which Kuiper belt object will be selected for New Horizons visit after its Pluto flyby (or do we?), so feel free to make broad assumptions in your calculations.

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2 Answers 2

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New Horizons will never overtake Voyager 1.

Although New Horizons is currently faster than any other man-made object, it won't be by the time it reaches the outer corners of the solar system.

From the John Hopkins University/Applied Physics Laboratory New Horizons page:

Though New Horizons will also reach 100 AU, it will never pass Voyager 1, because Voyager was boosted by multiple gravity assists that make its speed faster than New Horizons will travel. Voyager 1 is escaping the solar system at 17 kilometers per second. When New Horizons reaches that same distance 32 years from now, propelled by a single planetary swingby, it will be moving about 13 kilometers per second.

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    $\begingroup$ So New Horizons doesn't get any speed boost from slingshooting past Pluto? Or is the plan to even reduce relative velocity to assume orbit in the Kuiper belt? $\endgroup$
    – TildalWave
    Commented Jan 28, 2014 at 11:54
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    $\begingroup$ Pluto is gravitationally tiny and won't have any significant influence on New Horizons' orbit. Perhaps a tiny correction but I wouldn't think so. New Horizons won't go into orbit anywhere. I don't know if they plan to use any fuel to slow down at all (but I could look it up). I wouldn't think so. $\endgroup$
    – gerrit
    Commented Jan 28, 2014 at 12:09
  • $\begingroup$ @gerrit: Presumably its path through the Pluto system will be selected so it can reach its next target, a Kuiper Belt object that, as far as I know, has not yet been selected. $\endgroup$ Commented Dec 12, 2014 at 23:48
  • $\begingroup$ never is a very long time. Both the voyager 1 probe and the new horizons probe are likely to orbit the center our galaxy for as long as they exist. Surely at some point in one of these many orbits over the next few billion years, new horizons will be further away from the earth than voyager 1. $\endgroup$
    – Katie
    Commented Apr 29, 2021 at 1:10
  • $\begingroup$ @Katie Feel free to add an answer calculating how many millions or billions years in the future that will happen :-) $\endgroup$
    – gerrit
    Commented Apr 29, 2021 at 8:16
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Here's a nice graph of Voyager 2's speed, and the difference made by the gravitational assists:
Voyager speed graph
You can see that the probe slowed down between assists. New Horizons would follow a similar graph, but with fewer assists its speed will end up below Voyager 2's, as @gerrit said.

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    $\begingroup$ It's interesting that the Neptune fly-by resulted in gravity braking - not an assist. $\endgroup$ Commented Jan 28, 2014 at 22:48
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    $\begingroup$ To get a Triton flyby, a high-priority science objective, Voyager 2 had to go over the North pole of Neptune. That resulted in a net reduction in heliocentric energy. $\endgroup$
    – Mark Adler
    Commented Mar 25, 2014 at 1:22
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    $\begingroup$ The graph looks like it left Earth with slightly more speed than the escape velocity needed to leave the solar system. But then it dropped below escape velocity again. I would assume a probe at escape velocity would stay at escape velocity. What am I missing? $\endgroup$
    – kasperd
    Commented Jul 15, 2015 at 14:30
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    $\begingroup$ @Hobbes That explains why the speed can be drawn as a function of distance. Each such graph would either reach zero velocity at some finite distance or converge towards some constant at infinite distance. But the voyager 2 graph doesn't follow such a line, since it starts about one pixel above the escape velocity and then drops below escape velocity. $\endgroup$
    – kasperd
    Commented Jul 15, 2015 at 14:53
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    $\begingroup$ @kasperd, I asked your question here as I'm also curious what the reason is $\endgroup$
    – neelsg
    Commented Jul 29, 2015 at 7:54

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