Skip to main content
9 events
when toggle format what by license comment
Dec 24, 2021 at 23:32 vote accept WarpPrime
Dec 17, 2021 at 22:14 history edited notovny CC BY-SA 4.0
Completed truncated first sentence.
Dec 17, 2021 at 21:15 comment added notovny @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.
Dec 17, 2021 at 20:52 comment added AtmosphericPrisonEscape 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.
Dec 17, 2021 at 19:55 comment added notovny @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.
Dec 17, 2021 at 19:46 comment added AtmosphericPrisonEscape 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...
Dec 17, 2021 at 18:58 comment added planetmaker Very nice explanation. In essence: it's just jumping a frog off the comet to trigger it :)
Dec 17, 2021 at 17:54 history edited notovny CC BY-SA 4.0
added 804 characters in body
Dec 17, 2021 at 17:11 history answered notovny CC BY-SA 4.0