As far as I can tell, a trajectory with zero excess velocity would place it (essentially) in nearly the same orbit around the sun as earth (assuming it even departs far away enough for the sun’s gravity well to dominate). I can’t think of any possible mission objectives that would make such a trajectory desirable, but perhaps my imagination is limited. Why would one want to choose zero excess velocity upon escape?

Has anyone ever attempted/achieved such a trajectory? If so, what were the mission objectives that made such an escape desirable over a traditional hyperbolic escape?


Why zero excess velocity? Well, with almost zero excess velocity you can stay near Earth, but not too near. For example, the Spitzer space telescope did this to communicate with Earth while avoiding radiant heat from Earth. It's been drifting away, but slowly enough that other factors first reduced its effectiveness.


Why would one want to choose zero excess velocity upon escape?

If you aren't in a great hurry, and you have a small delta-V budget, and you want to visit the Trojan points L4 or L5, you can do so by getting just outside of Earth's sphere of influence, then lowering or raising your solar orbit very slightly to get ahead of or fall behind Earth. You wouldn't want exactly zero excess velocity for this, but maybe pretty close to it.

  • $\begingroup$ Not exactly: even terrestrially, one can't be on time for an appointment, just too early or too late. $\endgroup$ – Camille Goudeseune Sep 30 '19 at 2:37
  • $\begingroup$ Your move to L4 or L5 is low delta-v only in the orbital plane. If you want to get to something that librates out of plane around the Lagrange point, this can happen. $\endgroup$ – Oscar Lanzi Sep 30 '19 at 21:18

Parabolic escape trajectory is only theoretical, it only "works" in a two body system, and in a two body system "escape" is a meaningless practice anyways.

In a multi body system the forces from other bodies, especially around the edge of the gravity well, make parabolic escape impossible: before you'd have zero velocity the other body would've already dominated the gravitation acceleration/force (different locations(!)).

You can however escape with "minimal velocity": such an escape would put you in an orbit around the sun, an orbit very much like earth but slightly smaller/larger (depending on the direction of your escape).

This is for example useful for solar observation satellites, which need to be in a stable orbit around the sun/not blocked by earth, yet staying near earth is important for signal strength.

But it's impossible anyways (not only due to mechanical inaccuracies but also above impossible to solve problem), and for calculations it's better to just assume hyperbolic velocity. I'd strongly prefer not doing too much at the edge of the gravity well, and just speeding past it and later correcting for the differences.

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    $\begingroup$ Once you get far from the solar system, you can consider it a mass point and a parabolic velocity is (reasonably) well defined. The impacts of the structure of the solar system fall off as $1/r^3$ or faster. $\endgroup$ – Ross Millikan Sep 30 '19 at 15:50
  • $\begingroup$ @RossMillikan of course, however you then cannot ignore the gravitational effect of the other stars/milkyway. - Hence I'm not really sure the point you're trying to make. If the goal is escape from sun (not earth) indeed it is correct to combine the solar system mass to a point, however at the solar system edge of the gravitational well the same problem occurs, with all the other star's bodies. $\endgroup$ – paul23 Oct 1 '19 at 13:06
  • $\begingroup$ I think there is enough room between the stars that you could reasonably talk about being on a parabolic trajectory from the solar system without worrying about the impact of the other stars. Yes, when or if you pass by them the velocity will change and if you reference back to the solar system it may not be parabolic any more. My point is to disagree that you have to worry about the planets when you define parabolic velocity leaving the solar system. $\endgroup$ – Ross Millikan Oct 1 '19 at 14:41
  • $\begingroup$ @RossMillikan that is obvious: but gravity is actually the force that doesn't reduce with distance. So long before you actually achieve "zero" velocity relative to the sun, the sum of all other stars in the milkyway have already taken over and are accelerating you differently, just like the sun itself experiences gravity. If you do not care about reaching "zero" you do not care about parabolic: hyperbolic or elliptoid trajectories are equally good there. $\endgroup$ – paul23 Oct 1 '19 at 20:10
  • $\begingroup$ There's nothing to agree or disagree, this is cold hard math, unless you have proof it happens differently? In which case I'm really eager to hear it. $\endgroup$ – paul23 Oct 1 '19 at 20:45

I can't think of one. For mission planning there is nothing special about parabolic velocity. There is something special from the perspective of teaching orbital mechanics as it is the boundary between closed and open orbits, but from a practical mission point of view it looks just like a very long ellipse or a barely open hyperbola for many, many years. It could be that a desirable slingshot maneuver results in a velocity that is close to parabolic, but it would be an accident. It could be that you are trying to escape the solar system and barely have a big enough rocket to do so. You could even send the probe out very slightly short of parabolic velocity, then pick up a gravitational assist from a nearby star that keeps you from falling back.


As everyone else has mentioned, there doesn't seem to be a mission for a true parabolic escape, especially since an exact parabolic trajectory is a target of zero size and therefore there is a zero chance of hitting it exactly.

Also, a true parabolic orbit only makes sense in a two-body model. Once you consider the gravity of anything else, the orbit elements, including eccentricity, are no longer constant. A trip to Earth/Sun L1 might start out hyperbolic, but change to elliptical on the way out, due to the Sun's gravity. At some instant in between, it will have an eccentricity of exactly 1, but no one cares, because that is a two-body concept in a three-body problem.

Some missions have a trajectory very close to parabolic, but not exact:

  • All the press reports on Apollo talk about the mission needing to achieve escape velocity in order to leave Earth and head to the moon. In fact, the speed achieved after translunar injection is just slightly short of escape velocity. The orbit is very stretched out, with its periapse at around 200km altitude, while its periapse is well past the moon, maybe 500,000km out (and sensitively dependent on the exact accuracy of the burn). For instance, the Apollo 11 flight report says that the TLI burn achieved a speed of 10,841m/s at an altitude of 320.2km. At that altitude, escape speed is 10,914m/s, so Apollo 11 was 73m/s short of escape, which would have taken something like a 3-second longer TLI burn. Of course this apoapse is only academic, since the burn targeted the moon, and the moon will of course change the orbit before the spacecraft reaches the first apoapse.
  • Trips to Earth/Sun L1 or L2 might be the closest to exactly parabolic -- they (in a sense) go up to balance on the edge of the Earth's sphere of influence. There are several solar observing spacecraft at L1, and several astronomical observers at L2.
  • Often the parabolic case (C3=0) is used in launch vehicle payload planner's guides. They will list a table or graph of how much mass they can put onto a parabolic escape trajectory. If you are designing an interplanetary mission, your spacecraft must weigh less than that. However, even in this case, most payload guides will have curves of achievable C3 (which is the square of the hyperbolic excess speed, in km^2/s^2) given a certain mass. If you are planning a mission to Mars in a particular year and that year's transfer requires a C3 of 10km^2/s^2, then you use that value to check what your maximum mass is on a given launch vehicle.
  • MESSENGER to Mercury was initially launched into an orbit around the sun which had almost exactly the same period as that of the Earth. It did in fact return to Earth space exactly one year later, and performed a flyby which changed the orbit so that it would intersect that of Venus. From there it used flybys of Venus and Mercury to climb down the gravity well and eventually orbit Mercury. I don't think its departure C3 was zero, but it might have been close. What I wonder about is if they actually got any advantage from the first Earth flyby. The flyby arrival and departure C3 must have been close to the same as the launch C3, so why didn't they just wait a year to launch, onto the post-flyby trajectory? Is it just a case of getting the spacecraft into space, because it is much harder to cancel a mission after it is launched?

Satellites put into geostationary orbit tend to use zero excess velocity trajectories to conserve more fuel for later positioning/adjustment


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