<p>It's a consequence of the reversibility of orbits that anything that free-falls from a near-interstellar distance arrives at nearly escape velocity, or faster. Take the Vis-Viva equation, the formula for the velocity of objects in Keplerian orbits/trajectories:</p>
<p><span class="math-container">$$v=\sqrt{\mu\left(\frac{2}{r}-\frac{1}{a}\right)}$$</span></p>
<p>Where <span class="math-container">$r$</span> is the radial distance, <span class="math-container">$a$</span> is the semi-major axis, and <span class="math-container">$\mu$</span> is the standard gravitational parameter.</p>
<p>When <span class="math-container">$a$</span> is very, very large compared to <span class="math-container">$r$</span>, things start to look like this:</p>
<p><span class="math-container">$$v=\sqrt{\mu\left(\frac{2}{r}\right)}$$</span></p>
<p>Which is the formula for escape velocity.</p>
<p>So basically, almost <em>anything</em> falling from the Oort cloud to the inner solar system is going to be scraping the underside of escape velocity in the inner system, at minimum.</p>
<p>For an example, the inner edge of the Oort cloud is believed to start at about <span class="math-container">$2\,000\, \mathrm{AU}$</span> from the Sun, according to <a href="https://solarsystem.nasa.gov/solar-system/oort-cloud/overview/" rel="noreferrer">NASA's Solar System Overview</a>. As such an object in a highly-elliptical orbit that takes it out that far would have a semimajor axis of about <span class="math-container">$1\,000\,\mathrm{AU}$</span>.  Jupiter's Semimajor axis is about <span class="math-container">$5 \,\mathrm{AU}$</span>.</p>
<p>The Vis-viva calculation puts an object orbiting the Sun with a semi-major axis of <span class="math-container">$1000\, \mathrm{AU}$</span> moving at <span class="math-container">$18\,814\, \mathrm{m/s}$</span> when it reaches <span class="math-container">$5\, \mathrm{AU}$</span>. Solar Escape velocity at <span class="math-container">$5\, \mathrm{AU}$</span> is <span class="math-container">$18\,837\,\mathrm{m/s}$</span>, a difference of <span class="math-container">$23 \,\mathrm{m/s}$</span> that would be swallowed whole if I'd used proper significant figures in the calculation.</p>

CC BY-SA 4.0
2021-12-17T18:58:01.393Z
Very nice explanation. In essence: it's just jumping a frog off the comet to trigger it :)
planetmaker
CC BY-SA 4.0
2021-12-17T19:46:08.470Z
Any object obeying the vis-viva equation is bound to the star, anything bound has a factor of $1/\sqrt(2)$ between its perihelion velocity and the escape velocity at perihelion...
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-12-17T19:55:21.393Z
@AtmosphericPrisonEscape  Both of your statements are incorrect.  The Vis-viva equation works for _all_ Keplerian orbits and trajectories, as along as you're using negative semi-major axis for hyperbolic trajectories.   Escape velocity /$\sqrt{2}$  =  the _circular orbit velocity_ for the chosen distance, only.
notovny
CC BY-SA 4.0
2021-12-17T20:52:57.430Z
Just that velocity at perihelion is not a radial one, but a tangential one, as the object has to orbit. Furthermore, the approximation for large a is only valid in the approximation of infinite eccentricity, which is not given for all small solar system bodies.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-12-17T21:15:43.820Z
@AtmosphericPrisonEscape If an object at perihelion $r_p$ is moving at $v_p=\sqrt{\frac{\mu}{r_p}}$, then that orbit _must_ be circular.  For an elliptical orbit, it will always be moving _faster_ than that speed at perihelion, which invalidates your $1/\sqrt{2}$ ratio claim.  Under the Keplerian-Newtonian two-body simplifications, _any_ velocity of magnitude  less than the local escape velocity, in any direction results, in an elliptical orbit.
notovny
CC BY-SA 4.0
2021-12-17T17:11:36.587Z
2021-12-17T22:14:11.557Z
2021-12-17T22:14:11.557Z
notovny
<p>User @antlersoft wrote a nice answer to my question on <a href="https://astronomy.stackexchange.com/questions/47823/31410">the difference between barycentric and heliocentric models of the solar system</a> when applied to comets (edge cases of the systems). <a href="https://astronomy.stackexchange.com/posts/comments/104183">In a comment</a>, @DavidHammen mentions that</p>
<blockquote>
<p>Planetary perturbations near perihelion will change the aphelion for most highly-eccentric objects, regardless of the coordinates used.</p>
</blockquote>
<p>Because most comets have inclinations different than that of the ecliptic plane and do not intersect the orbit of a planet, with the famous exception of Comet Bowell in a close swingby of Jupiter, with an inclination of only 1.66 degrees. As a result, why are such small perturbations of planets several AU away enough to take a comet from a closed to an open orbit? How close to escape velocity are most Oort-cloud comets?</p>

CC BY-SA 4.0
2021-12-18T07:20:10.850Z
See section 3 of this [paper](https://academic.oup.com/mnras/article/304/3/579/1022485). Also see: http://www2.ess.ucla.edu/~jewitt/oort2.html
Nilay Ghosh
<orbit><gravity><comets>
How close to escape velocity are most Oort-cloud comets?
WarpPrime
# How close to escape velocity are most Oort-cloud comets?

(Asked by: username_0 on 2021-12-17)

User @antlersoft wrote a nice answer to my question on [the difference between barycentric and heliocentric models of the solar system](https://astronomy.stackexchange.com/questions/47823/31410) when applied to comets (edge cases of the systems). [In a comment](https://astronomy.stackexchange.com/posts/comments/104183), @DavidHammen mentions that

> Planetary perturbations near perihelion will change the aphelion for most highly-eccentric objects, regardless of the coordinates used.

Because most comets have inclinations different than that of the ecliptic plane and do not intersect the orbit of a planet, with the famous exception of Comet Bowell in a close swingby of Jupiter, with an inclination of only 1.66 degrees. As a result, why are such small perturbations of planets several AU away enough to take a comet from a closed to an open orbit? How close to escape velocity are most Oort-cloud comets?

**Tags:** orbit, gravity, comets

> **Comments:**
> **username_1**: See section 3 of this [paper](https://academic.oup.com/mnras/article/304/3/579/1022485). Also see: http://www2.ess.ucla.edu/~jewitt/oort2.html  

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## 1 answers

### Answer by username_2 on 2021-12-17. Score: 17 | Accepted answer

It's a consequence of the reversibility of orbits that anything that free-falls from a near-interstellar distance arrives at nearly escape velocity, or faster. Take the Vis-Viva equation, the formula for the velocity of objects in Keplerian orbits/trajectories:

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

Where $r$ is the radial distance, $a$ is the semi-major axis, and $\mu$ is the standard gravitational parameter.

When $a$ is very, very large compared to $r$, things start to look like this:

$$v=\sqrt{\mu\left(\frac{2}{r}\right)}$$

Which is the formula for escape velocity.

So basically, almost *anything* falling from the Oort cloud to the inner solar system is going to be scraping the underside of escape velocity in the inner system, at minimum.

For an example, the inner edge of the Oort cloud is believed to start at about $2\,000\, \mathrm{AU}$ from the Sun, according to [NASA's Solar System Overview](https://solarsystem.nasa.gov/solar-system/oort-cloud/overview/). As such an object in a highly-elliptical orbit that takes it out that far would have a semimajor axis of about $1\,000\,\mathrm{AU}$. Jupiter's Semimajor axis is about $5 \,\mathrm{AU}$.

The Vis-viva calculation puts an object orbiting the Sun with a semi-major axis of $1000\, \mathrm{AU}$ moving at $18\,814\, \mathrm{m/s}$ when it reaches $5\, \mathrm{AU}$. Solar Escape velocity at $5\, \mathrm{AU}$ is $18\,837\,\mathrm{m/s}$, a difference of $23 \,\mathrm{m/s}$ that would be swallowed whole if I'd used proper significant figures in the calculation.

> **Comments:**
> **username_3**: Very nice explanation. In essence: it's just jumping a frog off the comet to trigger it :)  
> **username_4**: Any object obeying the vis-viva equation is bound to the star, anything bound has a factor of $1/\sqrt(2)$ between its perihelion velocity and the escape velocity at perihelion...  
> **username_2**: @username_4  Both of your statements are incorrect.  The Vis-viva equation works for _all_ Keplerian orbits and trajectories, as along as you're using negative semi-major axis for hyperbolic trajectories.   Escape velocity /$\sqrt{2}$  =  the _circular orbit velocity_ for the chosen distance, only.  
> **username_4**: Just that velocity at perihelion is not a radial one, but a tangential one, as the object has to orbit. Furthermore, the approximation for large a is only valid in the approximation of infinite eccentricity, which is not given for all small solar system bodies.  
> **username_2**: @username_4 If an object at perihelion $r_p$ is moving at $v_p=\sqrt{\frac{\mu}{r_p}}$, then that orbit _must_ be circular.  For an elliptical orbit, it will always be moving _faster_ than that speed at perihelion, which invalidates your $1/\sqrt{2}$ ratio claim.  Under the Keplerian-Newtonian two-body simplifications, _any_ velocity of magnitude  less than the local escape velocity, in any direction results, in an elliptical orbit.  

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