<p>No, but nor do elliptical or hyperbolic trajectories.  These are features of a model we have for gravity, not part of nature.</p>
<p>The modelling process involves constructing a mathematical framework that describes a natural system and allows for predictions. It is often said that &quot;no models are correct, but some models are useful&quot;.  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.</p>
<p>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)</p>
<p>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)</p>

CC BY-SA 4.0
2020-12-23T23:45:55.480Z
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.
leftaroundabout
CC BY-SA 4.0
2020-12-24T00:34:39.463Z
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.
James K
CC BY-SA 4.0
2020-12-24T00:49:04.690Z
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).
leftaroundabout
CC BY-SA 4.0
2020-12-24T02:17:36.773Z
"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?
Acccumulation
CC BY-SA 4.0
2020-12-24T02:42:03.487Z
Are hyperbolas a better approximation than parabolas (for an orbit with e = 1.00+-0.01)
James K
CC BY-SA 4.0
2020-12-24T16:17:25.933Z
But the Newtonian model fails for Mercury, it took General Relativity to account for the error
Barmar
CC BY-SA 4.0
2020-12-23T09:21:28.313Z
2020-12-23T15:31:48.193Z
2020-12-23T15:31:48.193Z
James K
<blockquote>
<p>Does a parabolic trajectory really exist in nature?</p>
</blockquote>
<p>This is a great question! But let's ask two simpler questions first, then move back to yours.</p>
<ol>
<li>Do Keplerian orbits (circular, elliptical, parabolic, hyperbolic) exist in nature?</li>
<li>If they did, what fraction would be parabolic?</li>
</ol>
<p>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.</p>
<p>The answer to #2 is zero, and that's for the reason you suspected; eccentricity would have to be <em>exactly</em> 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.</p>
<p>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.</p>
<p>For orbits so close to the hairy edge like an eccentricity of 0.99999999, or 1.00000001, we might call them &quot;parabolic&quot; sometimes, but in those cases we would have an understanding that there is no such thing as an <em>exactly parabolic orbit</em> and that the obit may easily change between bound and unbound states due to a weak perturbing force by another body.</p>

CC BY-SA 4.0
2020-12-23T13:46:41.500Z
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?
questionhang
CC BY-SA 4.0
2020-12-23T18:47:08.047Z
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.
Ross Presser
CC BY-SA 4.0
2020-12-23T19:08:57.827Z
@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.
uhoh
CC BY-SA 4.0
2020-12-23T19:12:52.587Z
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.
uhoh
CC BY-SA 4.0
2020-12-23T13:34:53.230Z
2020-12-23T13:34:53.230Z
uhoh
<p><strong>Short answer:</strong> Yes, <a href="https://www.researchgate.net/figure/Principal-sketch-of-different-orbit-types-depending-on-eccentricity-e-An-orbit-with_fig1_260163290" rel="nofollow noreferrer">orbits</a> with a value <span class="math-container">$e=1$</span> certainly exist.  An object perturbed out of a bound state from <span class="math-container">$e&lt;1$</span> to <span class="math-container">$e&gt;1$</span> must at some time have a value of <span class="math-container">$e=1$</span>.</p>
<p><a href="https://i.sstatic.net/HTl6p.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HTl6p.png" alt="enter image description here" /></a></p>
<p><strong>Long answer:</strong> The <a href="https://en.wikipedia.org/wiki/Intermediate_value_theorem" rel="nofollow noreferrer">intermediate value theorem</a> states that for a continuous function <span class="math-container">$f$</span>, if <span class="math-container">$f(t_1)=x_1$</span> and <span class="math-container">$f(t_2)=x_2$</span>, then for any value <span class="math-container">$x$</span> in the interval <span class="math-container">$(x_1,x_2)$</span>, there exists a value <span class="math-container">$t$</span> in the interval <span class="math-container">$(t_1,t_2)$</span> such that <span class="math-container">$f(t)=x$</span>.</p>
<p>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).</p>
<p>If we apply the intermediate value theorem to the eccentricity of an orbit as it passes through a bound state at <span class="math-container">$t_1$</span> where eccentricity <span class="math-container">$e_1 = 0.999$</span> to an unbound state at <span class="math-container">$t_2$</span> where eccentricity <span class="math-container">$e_2=1.001$</span>, then there <em>must</em> be at least one value <span class="math-container">$t$</span>, <span class="math-container">$t_1&lt;t&lt;t_2$</span> such that the eccentricity is <span class="math-container">$1$</span> at time <span class="math-container">$t$</span>.</p>
<p><strong>Notes:</strong></p>
<ol>
<li>The above proof works for bodies perturbed <em>out of</em> or <em>into</em> a gravitationally bound state (with obligatory <a href="https://xkcd.com/1189/" rel="nofollow noreferrer">XKCD</a> comic):</li>
</ol>
<p><a href="https://i.sstatic.net/M9hoH.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/M9hoH.png" alt="enter image description here" /></a></p>
<ol start="2">
<li><p>The resulting orbital path won't appear exactly parabolic in most coordinate systems since it will have a value of <span class="math-container">$e=1$</span> for an infinitely short period of time.</p>
</li>
<li><p>We can also define a coordinate system in which an orbit is perfectly parabolic using the same methods I've used <a href="https://astronomy.stackexchange.com/questions/39652/what-are-non-keplerian-orbits-what-are-some-familiar-examples-in-our-solar-sy/40221#40221">here</a>.</p>
</li>
</ol>

CC BY-SA 4.0
2020-12-23T22:27:44.690Z
Oh this is interesting! Yes, an *osculating orbit* can indeed pass through $\varepsilon=1$ but the time it spends there is: $$0$$
uhoh
CC BY-SA 4.0
2020-12-23T22:35:12.043Z
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.
uhoh
CC BY-SA 4.0
2020-12-23T23:27:26.507Z
@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?
Connor Garcia
CC BY-SA 4.0
2020-12-24T02:00:44.093Z
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!
eps
CC BY-SA 4.0
2020-12-24T02:07:53.377Z
Positions don't have eccentricities.
uhoh
CC BY-SA 4.0
2020-12-24T03:00:59.780Z
@eps If "space isn't infinitely dividable", just how dividable is it? In regards to Zeno's paradox, my answer above is more akin to saying that if an arrow hits a target, it had to pass through a plane between the archer and target, with center of mass as a point on that plane at some time.
Connor Garcia
CC BY-SA 4.0
2020-12-23T17:16:39.207Z
2020-12-23T19:13:01.310Z
2020-12-23T19:13:01.310Z
Connor Garcia
<h3>Sure. Go to the moon, and throw a rock.</h3>
<p>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.</p>

CC BY-SA 4.0
2020-12-25T16:15:35.983Z
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).
Paŭlo Ebermann
CC BY-SA 4.0
2020-12-25T22:53:47.650Z
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.
Mark
CC BY-SA 4.0
2020-12-26T11:03:22.400Z
@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.
nick012000
CC BY-SA 4.0
2020-12-25T11:02:16.720Z
2020-12-25T11:02:16.720Z
nick012000
<p>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 <a href="https://en.wikipedia.org/wiki/Parabolic_trajectory" rel="noreferrer">parabolic trajectory</a>,
is determined by minor perturbation?</p>

CC BY-SA 4.0
2020-12-23T20:47:31.700Z
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.
jamesqf
CC BY-SA 4.0
2020-12-24T00:32:22.653Z
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.
James K
CC BY-SA 4.0
2020-12-24T02:09:32.710Z
@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.
Acccumulation
CC BY-SA 4.0
2020-12-24T13:45:21.817Z
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.
Mark Morgan Lloyd
CC BY-SA 4.0
2020-12-24T18:31:17.050Z
I don't even understand this question?
Fattie
CC BY-SA 4.0
2020-12-25T11:50:39.977Z
Nice question but you could also ask about the simpler case of circular orbits with $e = 0$.
badjohn
<orbit>
Does a parabolic trajectory really exist in nature?
questionhang
# Does a parabolic trajectory really exist in nature?

(Asked by: username_0 on 2020-12-23)

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](https://en.wikipedia.org/wiki/Parabolic_trajectory),
is determined by minor perturbation?

**Tags:** orbit
> **Comments:**
> **username_1**: 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.  
> **username_2**: 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.  
> **username_3**: @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.  
> **username_4**: 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.  
> **username_5**: I don't even understand this question?  
> **username_6**: Nice question but you could also ask about the simpler case of circular orbits with $e = 0$.  

---

## 4 answers

### Answer by username_2 on 2020-12-23. Score: 28

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)

> **Comments:**
> **username_7**: 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.  
> **username_2**: 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.  
> **username_7**: 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).  
> **username_3**: "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?  
> **username_2**: Are hyperbolas a better approximation than parabolas (for an orbit with e = 1.00+-0.01)  
> **username_8**: But the Newtonian model fails for Mercury, it took General Relativity to account for the error  

---

### Answer by username_9 on 2020-12-23. Score: 10

> 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.

1. Do Keplerian orbits (circular, elliptical, parabolic, hyperbolic) exist in nature?
2. 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.

> **Comments:**
> **username_0**: 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?  
> **username_10**: 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.  
> **username_9**: @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.  
> **username_9**: 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.  

---

### Answer by username_11 on 2020-12-23. Score: 5

**Short answer:** Yes, [orbits](https://www.researchgate.net/figure/Principal-sketch-of-different-orbit-types-depending-on-eccentricity-e-An-orbit-with_fig1_260163290) 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 $e=1$.

[![enter image description here](https://i.sstatic.net/HTl6p.png)](https://i.sstatic.net/HTl6p.png)

**Long answer:** The [intermediate value theorem](https://en.wikipedia.org/wiki/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 such that the eccentricity is $1$ at time $t$.

**Notes:**

1. The above proof works for bodies perturbed *out of* or *into* a gravitationally bound state (with obligatory [XKCD](https://xkcd.com/1189/) comic):

[![enter image description here](https://i.sstatic.net/M9hoH.png)](https://i.sstatic.net/M9hoH.png)

2. 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.
3. We can also define a coordinate system in which an orbit is perfectly parabolic using the same methods I've used [here](https://astronomy.stackexchange.com/questions/39652/what-are-non-keplerian-orbits-what-are-some-familiar-examples-in-our-solar-sy/40221#40221).

> **Comments:**
> **username_9**: Oh this is interesting! Yes, an *osculating orbit* can indeed pass through $\varepsilon=1$ but the time it spends there is: $$0$$  
> **username_9**: 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.  
> **username_11**: @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?  
> **username_12**: 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!  
> **username_9**: Positions don't have eccentricities.  
> **username_11**: @eps If "space isn't infinitely dividable", just how dividable is it? In regards to Zeno's paradox, my answer above is more akin to saying that if an arrow hits a target, it had to pass through a plane between the archer and target, with center of mass as a point on that plane at some time.  

---

### Answer by username_13 on 2020-12-25. Score: -1

### 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.

> **Comments:**
> **username_14**: 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).  
> **username_15**: 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.  
> **username_13**: @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.  

---

