I saw these two links, and they gave me a question:

If the solar probes Helios I and II can be 4.1x faster than Voyager, and if it would take a Voyager-like probe 70,000 years to transit the 40-trillion kilometers to Alpha Centauri, what is the shortest plausible time to send a probe to Alpha Centauri with a solar-slingshot?

Is it possible that we could have a probe half-way to Alpha Centauri in 25 years using only today's technology?


The claim about very high Helios probes speed is made for their perihelion i.e. in the point of their orbit closest to the Sun. It originates from the Sun gravity as the probe "falls" closer and is eaten as it goes farther. Here's the depiction of their orbit at Wikipedia:

Helios probe trajectory

As the probe can't go far from the Sun this high speed is completely useless for interstellar trip. Instead you should look what speed the probe will have when it goes (if it does) to the boundary of the Solar system when its decrease due to the Solar gravity becomes very small.

Also seems that you heard about gravitational slingshot maneuvers however don't understand it correctly. The gravitational slingshot uses planet relative motion to the sun with some speed $u$. When probe approaches the planet relatively to the latter it goes in the hyperbolic trajectory approaching at speed $\vec{v}_0$ and leaving its sphere of influence at speed $\vec{v}_f$ similar in module $|\vec{v}_0|\simeq |\vec{v}_f|$ but with different direction. However relatively to the Sun you have initial speed $\vec{v}_0+\vec{u}$ and the final speed $\vec{v}_f+\vec{u}$. If the direction changes significantly in the planet reference frame it will lead to significant change in the speed relatively to the Sun. You can't however make any "slingshot" if simply orbit around the sun — all speed boost in the perihelion is eaten as you rise to the aphelion.

If our Solar system had massive planet closer to the Sun we could use this boost to perform nice gravity slingshot at that planet. However both Venus and Mercury are quite light and are not well-suited for such a maneuver.

What you can get from this boost may come from using Oberth effect — the changes of speed when it's high lead to higher changes of energy.

Anyway all these gains are of order of km/s, 10 km/s at max. The distance to Proxima Centauri is 4.26 light-years. To get there in 50 years you need to have about 8% of lightspeed. It's about 25000 km/s not 25 km/s!

| improve this answer | |
  • $\begingroup$ 25 Mm/s then. Nice answer. Thank you. $\endgroup$ – EngrStudent Feb 8 '17 at 18:05
  • 1
    $\begingroup$ With a(n imaginary) Solar sail a slingshot trip by the Sun would make better sense. Using it not for gravity assist but as a source of propulsion. $\endgroup$ – LocalFluff Feb 9 '17 at 3:48

This is a supplemental answer to help show how to calculate the speeds and see that the high speed of spacecraft passing close to the sun is only available close to the sun - and not to suggest that the 2011 Centauri Dreams article reprinted in Gizmodo is just a little confusing if not misleading about this by citing the speed of a spacecraft near the sun and comparing it to a spacecraft that has left the solar system. (There are of course slingshot maneuvers, but then that's beyond the spirit of the original article)

Total energy:

$$E_{tot}=\frac{1}{2}mv^2 - \frac{GMm}{r}$$

Divide through by the mass of the satellite, it becomes specific orbital energy;

$$\mathscr{E}_{tot} = \frac{1}{2}v^2 - \frac{GM}{r}$$

Make hand-waving gestures, or read here about basic orbital mechanics:

$$\mathscr{E}_{tot} = - \frac{GM}{2a}$$

where $a$ is the semi-major axis. The total energy is constant, and can be calculated from the semi-major axis alone (don't need eccentricity). Put them together;

$$- \frac{GM}{2a} = \frac{1}{2}v^2 - \frac{GM}{r}$$

...and you get the very handy vis-viva equation.

$$v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$$

From Wikipedia, Helios-B perihelion and aphelion are 0.29 and 0.98 AU, or about 43 and 147 million km, and the semi-major axis $a$ is half of the sum (half of the major axis), or about 95 million km.

The Standard Gravitational Parameter of the sun is about 1.33E+11 km^3/s^2 (1.33E+20 km^3/s^2 in standard mks units), so you can calculate the velocities are about 69 km/s at periapsis, but drop to about 20km at apoapsis.

| improve this answer | |
  • 1
    $\begingroup$ If it is hyperbolic vs elliptic, they fail to describe sign convention for "a". I don't see how a negative length exists as a physical scalar like length. In the hyperbolic trajectory, do the $2/r$ and $1/a$ add? $\endgroup$ – EngrStudent Feb 9 '17 at 16:17
  • $\begingroup$ @EngrStudent take a look again at my answer, where I've said "Make hand-waving gestures, or read here about basic orbital mechanics." $\endgroup$ – uhoh Feb 9 '17 at 16:51
  • 1
    $\begingroup$ For reference, the New Horizons probe took approximately 9.5 years to reach Pluto which was about 4.4 light hours away at flyby. By comparison, Alpha Centauri is about 4.4 light years away. At the NH rate it would take over 83,000 years to reach Alpha Centauri! $\endgroup$ – Vince 49 Feb 11 '17 at 17:44

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.