Orbital speed 0m.s ?

[Okhin]

Was wondering (in fact, was kept awaken by this question until late). Is there a way to be in a stable orbit with a speed of 0m.s in KSP ?

My reasoning is, since there's nothing but Kerbol SOI in KSP, and since the further you go from your orbiting body, the lower your orbital speed gets, is there a way to reach a point where you're "in orbit" but with an orbital speed of 0m.s ? I can imagine going at 10m.s or even 1m.s, but somehow, the 0m.s thing bugs me. I suppose that's reaching the end of Kerbol SOI, and being just at the limit of it, but is it doable ? Has it been done ? Does the Kraken allows for it ?

[Bej Kerman]

No, you'd have to be infinitely far away.

[Fraktal]

I think orbital speed is asymptotically converging on zero as altitude increases, but never quite reaches it unless you're so far out that it's a very small fraction of zero and KSP rounds it down.

[Okhin]

Mmm, so in KSP it might be doable (due to the necessary rounding, and the lack of SOI containing Kerbol SOI), but thanks for confirmng that it is a case where distance needs to be close to infinite. That's why my brain couldn't figures out what an orbit with a null orbital speed looked like.

Thanks

[Zhetaan]

In reality, an orbital speed of zero occurs at the 'endpoint' of a parabolic trajectory.  If there is any excess velocity, then the trajectory is hyperbolic (and in fact, the velocity beyond what is needed to escape is actually called the hyperbolic excess velocity), and if there is insufficient velocity, then there is an instantaneous point where the velocity is zero in the vertical direction (you're more familiar with that point as the apoapsis), so there is difficulty already in this because a parabolic trajectory is a pseudo-stable solution to the equation.

In other words, if you have any excess velocity beyond escape velocity, then your speed will never go to zero; it will go to whatever the hyperbolic excess is.

So to answer your question, no:  there is no stable orbit with zero velocity.  At best, it is a pseudo-stable co-orbital relationship where a nudge in any direction will either close the orbit or result in escape--and fun theatrics with Lagrange points don't count because those involve orbit about a third body.

In terms of the game, both infinity and zero are a bit more discrete.  On the one hand, there is an absolute limit to how far you can get from the sun before you have an addressing error in the coordinate system, and on the other hand, there is an absolute limit to the possible precision of a floating-point decimal such that the rounding errors will leave you with the wrong velocity.  On the gripping hand, you can probably get a good approximation out at the fringes of available space.

Edited July 28, 2020 by Zhetaan

[magnemoe]

20 hours ago, Zhetaan said:

In reality, an orbital speed of zero occurs at the 'endpoint' of a parabolic trajectory.  If there is any excess velocity, then the trajectory is hyperbolic (and in fact, the velocity beyond what is needed to escape is actually called the hyperbolic excess velocity), and if there is insufficient velocity, then there is an instantaneous point where the velocity is zero in the vertical direction (you're more familiar with that point as the apoapsis), so there is difficulty already in this because a parabolic trajectory is a pseudo-stable solution to the equation.

In other words, if you have any excess velocity beyond escape velocity, then your speed will never go to zero; it will go to whatever the hyperbolic excess is.

So to answer your question, no:  there is no stable orbit with zero velocity.  At best, it is a pseudo-stable co-orbital relationship where a nudge in any direction will either close the orbit or result in escape--and fun theatrics with Lagrange points don't count because those involve orbit about a third body.

In terms of the game, both infinity and zero are a bit more discrete.  On the one hand, there is an absolute limit to how far you can get from the sun before you have an addressing error in the coordinate system, and on the other hand, there is an absolute limit to the possible precision of a floating-point decimal such that the rounding errors will leave you with the wrong velocity.  On the gripping hand, you can probably get a good approximation out at the fringes of available space.

At the top of your parabolic arch you will have zero vertical speed but you will have an horizontal one. Reference frame also matter here, if using surface speed an parabolic trajectory with Ap at GEO will have zero speed. In KSP an zero orbital speed might be possible because of rounding but this will probably be so far out it break the game. 
Think you can not have an orbit elliptical orbit of more than 80 something year in KSP with an Pe at Kerbin orbit before escape trajectory. 
Seems like the 80 year orbit was an very old version, newer ones has thousands of year to Ap, think old versions had serious issues if running for 200 year or more. 

Edited July 29, 2020 by magnemoe

[Zhetaan]

2 hours ago, magnemoe said:

At the top of your parabolic arch you will have zero vertical speed but you will have an horizontal one. Reference frame also matter here, if using surface speed an parabolic trajectory with Ap at GEO will have zero speed. In KSP an zero orbital speed might be possible because of rounding but this will probably be so far out it break the game. 
Think you can not have an orbit elliptical orbit of more than 80 something year in KSP with an Pe at Kerbin orbit before escape trajectory. 
Seems like the 80 year orbit was an very old version, newer ones has thousands of year to Ap, think old versions had serious issues if running for 200 year or more. 

I suppose you're right that I shouldn't assume others will take a non-rotating reference frame as a given.

Further, you are correct that older versions could not handle dates after a century or two.

Escape velocity from the sun at Kerbin's altitude is given by:

v_esc = √ (2μ / r)
v_esc = √ (2 [1.1723328 x 10¹⁸] / [13,599,840,256])
v_esc = √ (2.3446656 x 10¹⁸ / 13,599,840,256)
v_esc = 13,130 m/s

An eighty-year solar orbit with a periapsis at Kerbin has a semimajor axis of:

a = ³√ (μT² / 4π²)
a = ³√ [([1.1723328 x 10¹⁸] * [9203545 * 80]²) / 4π²]
a = ³√ [([1.1723328 x 10¹⁸] * [5.4211353962896 x 10¹⁷]) / 4π²]
a = ³√ (6.355374838311 x 10³⁵ / 4π²)
a = ³√ (1.609835252771 x 10³⁴)
a = 252,499,474,052 metres

And an apoapsis of:

a = 252,499,474,052 metres
2a = 504,998,948,103
2a - Pe = Ap = 504,998,948,103 - 13,599,840,256
Ap = 491,399,107,847 metres, which is a bit over four times beyond the apoapsis of Eeloo (note that this is true apoapsis, not surface altitude).

The periapsis velocity of this orbit is:

v² = μ * [(2 / r) - (1 / a)]
v² = 1.1723328 x 10¹⁸ * [(2 / 13,599,840,256) - (1 / 252,499,474,052)]
v² = 1.1723328 x 10¹⁸ * (1.4310014659636 x 10^-10)
v² = 167,760,995.5397
v = 12,952 m/s

Which is 177 m/s less than escape velocity at that altitude.  It would take very little to push this orbit to escape, so your recollection is probably accurate.

[steuben]

The usual unstated constraint is "unpowered" orbit. If we allow "powered" orbits it might be do able. Yes I know it will be _very_ dv expensive.

[VoidSquid]

For such calculations I have the Simple Orbit Calculator mod, which still does work, although it's unfortunately abandoned for a while now.

https://github.com/stevehead/ksp-SimpleOrbitCalculator/releases

[Hannu2]

You can not achieve it by playing, because it would take insanely long time, but maybe you can use some mod to move your craft so far from sun that orbital velocity is rounded to zero. I think you have to test does such trick work or just call Kraken. All other objects have finite SOI and such trick is impossible.

[swjr-swis]

Okhin: I wonder if a 0.0 m/s orbit is possible? I can't sleep, help!

Everyone in this forum: <tries to answer through amazing feats of reason, logic, math and historic references>

Me, an intellectual: <spends 15 seconds to cheat a plane to orbit at increasing digits of 9 until the orbital velocity is 0.0 m/s>

Spoiler