As per @kasperd's comment on another question.

Looking at the graph below, it appears that Voyager 2 started at just over escape velocity of the solar system. Just before the Jupiter flyby, it was travelling much slower than escape velocity. Presumably, travelling to Jupiter the only thing that would have changed the probe's velocity is the Sun's gravity. As the probe went further away from the Sun, you would expect it to slow down, but why would it drop below escape velocity if it started above it?

Graph on Voyager 2 velocity


You are correct that Voyager did not change from above escape velocity to below escape velocity shortly after launch. The plot is misleading in that it is just not very accurate right there at 1 AU. The plot lines are kind of thick and a smidge off.

Now that I look at it more closely, the escape velocity line in that plot is wrong in other places as well.

Here is a better plot:

Voyager 2 and solar escape velocities

  • $\begingroup$ Is this graph saying that voyager 2 was launched at a speed of 36km per second? Because I have other sources saying that New Horizons holds the record for the fastest launch velocity which is about 58,536 km per hour (16,260 km/s) not even half the speed of which is shown in the graph. $\endgroup$ – Matthew Aug 27 '17 at 2:57
  • 8
    $\begingroup$ Whenever you talk about a velocity, you have to ask relative to what? The graph is showing velocities relative to the Sun. The New Horizons velocity you're quoting is relative to Earth. (By the way, you put a comma in your comment that should be a decimal point.) $\endgroup$ – Mark Adler Aug 27 '17 at 8:15
  • 1
    $\begingroup$ So how does this value convert into velocity relative to the Earth? I am struggling to find out what is the launch velocity of the voyager 2. My bad with the coma. $\endgroup$ – Matthew Aug 27 '17 at 9:19
  • 7
    $\begingroup$ Earth is going about 30 km/s around the Sun, so that is most of the difference. However the velocity relative to Earth drops rapidly after injection as it departs Earth so it depends on when and where (Earth altitude) this so-called "launch velocity" is measured, and it depends on the departure direction to subtract the Earth orbital velocity vector, so there isn't a simple additive conversion. You should ask a new question. $\endgroup$ – Mark Adler Aug 27 '17 at 16:05

The image (original at Wikimedia Commons) is only an approximation, as evidenced by the noticeable change in shape of the solar system escape velocity line at 14AU. The line is only defined with three points, and my guess would be that the creator of the graph tried to shape the curve manually.

According to Wikipedia the solar system escape velocity at the earth's orbit (1 AU) is 42.1km/s, considerably more than the ~35.5km/s of this graph. The graph's author notes that it is based on this image, which despite being from NASA, seems to share the same inaccuracies.

According to another Wikipedia page:

The only objects to date to be launched directly into a solar escape trajectory were the New Horizons spacecraft, its third stage and its two small de-spin masses.

  • 2
    $\begingroup$ Someone should graph an improved version! $\endgroup$ – gerrit Jul 29 '15 at 9:28
  • 1
    $\begingroup$ Someone's working on it ;-) it looks like the two curves should run parallel. I'll know more tonight. $\endgroup$ – Hobbes Jul 29 '15 at 9:31
  • 1
    $\begingroup$ @Hobbes Who's doing that? It would also be great to have one for Voyager 1 (and the other probes too). $\endgroup$ – curiousdannii Jul 29 '15 at 9:34
  • $\begingroup$ I'm working on it. The Voyagers will be more complicated because they weren't launched on hyperbolic trajectories. $\endgroup$ – Hobbes Jul 29 '15 at 9:49
  • 2
    $\begingroup$ I'm not even getting a 404 error at that original image (NASA) link. It seems NASA isn't maintaining their history. Sigh. (no replays of original the Apollo 11 tapes either...) $\endgroup$ – FKEinternet May 6 '18 at 6:06

At first glance, both orbital velocity and escape velocity are proportional to $\sqrt {\frac1{distance to the sun}}$, so I expected the curves to run parallel, with an offset for the launch velocity of the probe, plus further offsets for each gravity assist. Mark's answer shows it's more complicated than this.

Escape velocity

For a spherically symmetric massive body such as a (non-rotating) star or planet, the escape velocity at a given distance is calculated by the formula:

$$V_{Escape} = \sqrt {\frac{2GM}{r}}$$

where G is the universal gravitational constant ($G = 6.67×10^{−11} m^3 kg^{−1} s^{−2}$), M the mass of the body, and r the distance from the point in space to its center of mass.

Plugging this into Excel gets me this graph for escape velocity vs. distance:

Escape velocity

I'm still working on orbital velocity data, to be continued.

  • $\begingroup$ Looks as if an escape mission would do well to refuel at Saturn, at the knee of that L-shaped escape velocity curve. $\endgroup$ – LocalFluff Jul 30 '15 at 9:00
  • $\begingroup$ if you do that, you'd lose the Jupiter and Saturn gravity assists. $\endgroup$ – Hobbes Jul 30 '15 at 10:00
  • $\begingroup$ One might still swing by Saturn. Voyagers' gain from Jupiter to Saturn is however of the same size as the decrease in escape velocity (6-7 km/s). Anyway, curiousdannii here has just explained that it is simply a poor plot. Maybe Solar system escape is something you, Hobbes, would feel inspired to illustrate somehow? $\endgroup$ – LocalFluff Jul 30 '15 at 10:07

voyager's speed dropped below escape velocity b/c its escape path was convoluted. It was a planetary flyby mission, necessitating extra delta V (and gravity assists) so that voyager could 'weave' its way out, rather than a straight escape path.

  • 3
    $\begingroup$ This is incorrect. $\endgroup$ – kim holder May 12 '18 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.