It seems it is very difficult to have e=1 perfectly in nature. The final state (being captured or running away) of a celestial body with a parabolic trajectory, is determined by minor perturbation?
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asked Dec 23, 2020 at 8:40
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1$\begingroup$ Would not a projectile launched from an airless body at less than escape velocity follow a parabolic trajectory? Neglecting perturbations from other bodies, and inhomogenieties in the bodies themselves, of course. $\endgroup$– jamesqfDec 23, 2020 at 20:47
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1$\begingroup$ No, elliptical, but the ellipse would intersect with the surface, and if the velocity is much less then escape, then that very thin ellipse is well approximated by a parabola. $\endgroup$– James KDec 24, 2020 at 0:32
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2$\begingroup$ @jamesqf You're probably thinking of the result from introductory physics courses that objects under constant acceleration follow a parabolic trajectory. When dealing with, say, a baseball thrown across a baseball field, gravity is nearly constant. However, in astronomical mechanics, gravity can't be considered constant. And if the Earth were a point source of gravity, so that the baseball's trajectory weren't terminated by colliding with the ground, the parabolic trajectory would be a small scale approximation to the overall orbit, which would be elliptical. $\endgroup$– AcccumulationDec 24, 2020 at 2:09
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1$\begingroup$ OP: "Perfect" with what margin of error? If something were following what should be a parabolic path but there were a single excess massive particle in some direction in the visible universe then the path would no longer be parabolic... even if in practice measuring the deviation was impractical or even theoretically impossible. $\endgroup$– Mark Morgan LloydDec 24, 2020 at 13:45
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$\begingroup$ I don't even understand this question? $\endgroup$– FattieDec 24, 2020 at 18:31
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4 Answers
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No, but nor do elliptical or hyperbolic trajectories. These are features of a model we have for gravity, not part of nature.
The modelling process involves constructing a mathematical framework that describes a natural system and allows for predictions. It is often said that "no models are correct, but some models are useful". For gravity, one common model is the Newtonian model. In the Newtonian model, you can set up two particles in such a way that one follows a parabolic curve (in some inertial frame) Parabolic trajectories exist in the Newtonian model.
In nature, there are more than two particles in the universe and so there are no parabolic, elliptical nor hyperbolic trajectories. Moreover, the Newtonian model is known to only approximate gravity (even General Relativity is probably an approximation, gravity is probably some kind of quantum interaction. But the details are not yet known)
Nevertheless. The Newtonian model is a very useful model for solar system dynamics. Long-period comets, falling from the distant Oort cloud, will have orbits that are very long and are well approximated by parabolas. (but only approximated, as comets, in particular, have significant non-gravitational accelerations due to reaction forces from outgassing, in addition to gravitational perturbations of the planets, and effects from the non-spherical shape of the sun)
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5$\begingroup$ The analogy is a bit off because whilst Newtonian 2-body applies to many systems with asymptotic accuracy (in a limit of extreme mass and/or scale relations), parabolic orbits are a boundary case. A randomly picked 3-body configuration has a decent likelyhood of being well approximated by a superposition of elliptical and hyperbolic trajectories, whereas actual 2-body systems have vanishingly small likelihood of parabolic behaviour. $\endgroup$ Dec 23, 2020 at 23:45
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$\begingroup$ I think I'll leave my answer as is. Parabolic trajectories do exist in a newtonian model, they don't exist in reality. I think that is the main difference, but feel free to upvote uhoh's answer that makes exactly this point. $\endgroup$– James KDec 24, 2020 at 0:34
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1$\begingroup$ I did! — The point is, though both hyperbolic/elliptic strictly speaking “don't exist in reality” any more than parabolic, good approximations of these are actually extremely common in reality, whereas parabolic trajectories can only be approximated by unlikely coincidence (or intelligent planning). $\endgroup$ Dec 24, 2020 at 0:49
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$\begingroup$ "Long-period comets, falling from the distant Oort cloud, will have orbits that are very long and are well approximated by parabolas." Are parabolas a better approximation than hyperboles? $\endgroup$ Dec 24, 2020 at 2:17
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$\begingroup$ Are hyperbolas a better approximation than parabolas (for an orbit with e = 1.00+-0.01) $\endgroup$– James KDec 24, 2020 at 2:42
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Does a parabolic trajectory really exist in nature?
This is a great question! But let's ask two simpler questions first, then move back to yours.
- Do Keplerian orbits (circular, elliptical, parabolic, hyperbolic) exist in nature?
- If they did, what fraction would be parabolic?
The answer to #1 is no. If there were only two spherical objects in the universe, each would be affected only by the other. In that case their trajectory would be Keplerian.
The answer to #2 is zero, and that's for the reason you suspected; eccentricity would have to be exactly unity, and for any finite number of eccentricities that can occur, there's no finite probability that any exact value would come up. There's be zero chance of an eccentricity of 0, 0.5 1, 2 or any specific number.
But to your question about a population of real-world orbits, Those that have eccentricities of say 0.01 to 0.99 we can safely call them elliptical, because in the short run at least they will not be affected by the gravity of other solar system objects enough to push them into exactly zero, or to and beyond 1.
For orbits so close to the hairy edge like an eccentricity of 0.99999999, or 1.00000001, we might call them "parabolic" sometimes, but in those cases we would have an understanding that there is no such thing as an exactly parabolic orbit and that the obit may easily change between bound and unbound states due to a weak perturbing force by another body.
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1$\begingroup$ I understand there are no perfect Keplerian orbits, but e>0.99&& e<1.01 should happen. A change between bound and unbound states is widespread? Do you have examples? $\endgroup$ Dec 23, 2020 at 13:46
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3$\begingroup$ consider the several discarded boosters that temporarily change from a solar orbot to captured in earth orbit, then change back to solar later. Perturbing bodies can switch binding easily. $\endgroup$ Dec 23, 2020 at 18:47
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$\begingroup$ @questionhang No and no need for any. The answer cites "orbits so close to the hairy edge like an eccentricity of 0.99999999, or 1.00000001" as a case when someone might apply the term parabolic with the understanding that it might be elliptical or might be hyperbolic, and being so close, that could change. I don't think any astronomer would ever officially label an orbit as parabolic, but if that has ever happened, it would probably be a case like this. $\endgroup$– uhohDec 23, 2020 at 19:08
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1$\begingroup$ A hypothetical example might be a rogue asteroid entering the solar system slightly unbound and through multi-body interactions with one or more larger planets become slightly bound to the Sun and enter a highly eccentric elliptical orbit. As I discuss in the answer, assigning Keplerian labels to orbits when there's more than two bodies in the universe is always approximate, but it could spend some time slightly bound to the Sun before getting kicked again by Jupiter or somebody else, either to an unbound or a more strongly bound orbit. Stuff happens even when we're not looking. $\endgroup$– uhohDec 23, 2020 at 19:12
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Short answer: Yes, orbits with a value $e=1$ certainly exist. An object perturbed out of a bound state from $e<1$ to $e>1$ must at some time have a value of $e=1$.
Long answer: The intermediate value theorem states that for a continuous function $f$, if $f(t_1)=x_1$ and $f(t_2)=x_2$, then for any value $x$ in the interval $(x_1,x_2)$, there exists a value $t$ in the interval $(t_1,t_2)$ such that $f(t)=x$.
For real bodies, we can assume that eccentricity is a continuous function of time since position and velocity are continuous functions of time (we assume no teleportation or infinite acceleration engines).
If we apply the intermediate value theorem to the eccentricity of an orbit as it passes through a bound state at $t_1$ where eccentricity $e_1 = 0.999$ to an unbound state at $t_2$ where eccentricity $e_2=1.001$, then there must be at least one value $t$, $t_1<t<t_2$ such that the eccentricity is $1$ at time $t$.
Notes:
- The above proof works for bodies perturbed out of or into a gravitationally bound state (with obligatory XKCD comic):
The resulting orbital path won't appear exactly parabolic in most coordinate systems since it will have a value of $e=1$ for an infinitely short period of time.
We can also define a coordinate system in which an orbit is perfectly parabolic using the same methods I've used here.
answered Dec 23, 2020 at 17:16
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1$\begingroup$ Oh this is interesting! Yes, an osculating orbit can indeed pass through $\varepsilon=1$ but the time it spends there is: $$0$$ $\endgroup$– uhohDec 23, 2020 at 22:27
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$\begingroup$ Can we say that an instantaneous condition that is over as soon as it happens has an existence? I'd say no, but I suppose that's a debatable topic. $\endgroup$– uhohDec 23, 2020 at 22:35
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$\begingroup$ @uhoh I agree that it's a debatable topic, but I am on the other side of the debate. The trajectory of any body can be thought of as an collection of positions and velocities in time, perhaps as the center of mass. If none of them exist, aren't we left with no trajectory at all, and further, no existence in the universe? $\endgroup$– Connor Garcia ♦Dec 23, 2020 at 23:27
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1$\begingroup$ Just because it happens in the math doesn't mean it exists in real life, as space isn't infinitely dividable. Continuous functions are models, not reality -- just ask Zeno! $\endgroup$– epsDec 24, 2020 at 2:00
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Sure. Go to the moon, and throw a rock.
When you're on a planet with no atmosphere, like the Moon, gravity can be considered as a field oriented in a constant direction, straight down, and its force is a effectively constant value in a constant direction. When this occurs, and there is no atmosphere to cause wind resistance, then the trajectory of an object thrown into space and experiencing no other forces will be parabolic in shape as it goes up and then comes back down, as long as its initial speed doesn't exceed the escape velocity of the planet, and it doesn't travel enough distance (vertically or horizontally) for the force of gravity to notably change.
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3$\begingroup$ Actually that's more of an ellipse (the part of which which you can see until it's intersected by the moon surface can be approximated by a parabola). $\endgroup$ Dec 25, 2020 at 16:15
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1$\begingroup$ Trajectories of thrown objects are only parabolic on an infinite featureless plain. On a spherical body, it's an ellipse, though most bodies are large enough and throws are slow enough that the easier-to-calculate parabolic approximation is usable. $\endgroup$– MarkDec 25, 2020 at 22:53
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$\begingroup$ @Mark Trajectories are parabolic if you disregard changes to the direction and magnitude of the gravity vector, because the thrown body covers a small enough distance in angle and distance from the center of the body. $\endgroup$ Dec 26, 2020 at 11:03