User @antlersoft wrote a nice answer to my question on the difference between barycentric and heliocentric models of the solar system when applied to comets (edge cases of the systems). In a comment, @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?


1 Answer 1


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:


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:


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. 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.

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    $\begingroup$ Very nice explanation. In essence: it's just jumping a frog off the comet to trigger it :) $\endgroup$ Dec 17, 2021 at 18:58
  • $\begingroup$ 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... $\endgroup$ Dec 17, 2021 at 19:46
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    $\begingroup$ @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. $\endgroup$
    – notovny
    Dec 17, 2021 at 19:55
  • $\begingroup$ 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. $\endgroup$ Dec 17, 2021 at 20:52
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    $\begingroup$ @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. $\endgroup$
    – notovny
    Dec 17, 2021 at 21:15

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