4
$\begingroup$

The title refers to the promotional slogan "What happens in Vegas, stays in Vegas."

The question How much CO2 would city-to-city rocket flight produce compared to airliners? seems to be focused on CO2 contribution to the Earth's atmosphere.

As an exercise to better understand the effects of rocket exhaust on the atmosphere, is there some point above which most rocket exhaust would not become part of Earth's atmosphere?

For example, at sea-level, all exhaust remains, and probably by the time one is passing the moon, nearly none of it has a chance to enter the atmosphere due mostly to geometry.

But roughly where would be the cross-over point for a typical LOX/RP-1 launch vehicle's exhaust, where roughy half escapes and half remains?

$\endgroup$
  • 1
    $\begingroup$ Related: thelastmanonearth.blogspot.com/2008/06/… $\endgroup$ – Organic Marble Jan 4 '18 at 1:35
  • 3
    $\begingroup$ Depends on which way you're thrusting. Rocket exhaust goes backward relative to the rocket at ~3000-4500 m/s; for typical LEO/GEO insertions, that leaves the exhaust in an Earth-suborbital trajectory, so you should expect the bulk of it to reenter. Even Apollo's translunar injection adds ~3200 m/s to LEO velocity using an exhaust velocity of ~4100 m/s! Obviously unless you have an infinitely long nozzle (= absolute zero exhaust temperature) there will still be some loss from plume dispersion, though. $\endgroup$ – Russell Borogove Jan 4 '18 at 2:10
  • 2
    $\begingroup$ Rocket exhaust has a high temperature, so some molecules will be at escape velocity. $\endgroup$ – Hobbes Jan 4 '18 at 9:38
  • 1
    $\begingroup$ @Hobbes however expansion $\rightarrow$ cooling, until the point where the density is so low that there are no further collisions, at which point the endpoint temperature will be "remembered" by the distribution of final velocities of the individual molecules with respect to the average CM exhaust velocity. There will be a ~4k m/s component behind the nozzle with a ~1k m/s transverse component, but the characteristic $k_B T$ of the exponential tail will reflect the temperature of the "cooled-off" gas after expansion. $\endgroup$ – uhoh Jan 4 '18 at 10:07
  • 2
    $\begingroup$ @Hobbes The expansion cools the exhaust gas mightily; SSME has a chamber temperature of ~3400K and an exhaust temperature of only ~685K (~412C) (according to some random dude on the internet who sounded plausible to me). $\endgroup$ – Russell Borogove Jan 4 '18 at 16:05
2
$\begingroup$

It’s meaningless to talk about orbital dynamics for gases. The mean free path up through and including cis-lunar space is much shorter than any useful orbital path. For example, figure 14-6 in Champion et al shows the mean free path as a function of altitude:

enter image description here

It grows to only a few thousand km by lunar orbit. So any ejected gas is going to be brought into equilibrium to the ambient gas before it can realistically escape, or even orbit for any significant time.

So then, the answer to this question

is there some point above which most rocket exhaust would not become part of Earth's atmosphere?

depends on what one means by atmosphere.

  • If you mean “all the gas gravitationally bound to Earth, even the outer-space-like parts”, the answer is: “If it’s emitted in that atmosphere, it stays in that atmosphere”

  • If you mean the common definition of atmosphere, the stratosphere and below (I.e. where “air” is), then one has to consider the motion of the emitted-in-space gas through the Karman line on the way to and perhaps across the stratosphere.

The answer to that second bullet depends on timescale somewhat.

There's a lot of research on vertical mixing times in the high stratosphere. Unfortunately, it's mostly paywalled, but this 1997 paper gives you an idea of the complexity and issues: There are multiple mechanisms involved, different parts of the globe have different conditions, and the distribution of drivers is (still) not clear.

The bottom line is that mixing up to 30-50km definitely happens, and the timescale is years to a few decades.

Above that, at low orbital altitudes, the atmosphere's mean free path gets long. But it's not long compared to the size of the atmosphere, and the inter-collision time isn't long compared to years. It's clear diffusive mixing takes place, but that's slow, slower than observation. It's not clear which bulk mixing mechanisms are dominant (in particular, the role of collisions with ions is getting a lot of attention and more than a few academic arguments), but over decades bulk mixing is clearly present.

What's not clear in the literature, at least to my knowledge, is the answer to exactly the question asked here: What is the timescale for mixing across the bottom of the thermosphere and across the top of the stratosphere?

If it's solely driven by vertical-balance diffusion, then the time scale would likely be centuries. But there might be mechanisms driven by i.e. solar and ionic interactions that speed the mixing up significantly. It's a tough region to probe, and it's not clear when this will get sorted out.

So the answer is "No, not if you wait long enough, but we don't know how long that is yet"

$\endgroup$
3
$\begingroup$

Virtually all exhaust will stay inside the atmosphere, most of it in the very high atmosphere. In order for it to leave, it would have to reach escape velocity. It's actually going the wrong direction, and thus won't have the energy to escape. Some of it that is light enough, like leftover hydrogen, might escape, but it won't be that much. The only exhaust that might escape Earth is for interplanetary missions, which depending on the trajectory the final bit of exhaust might escape the atmosphere, or retrograde exhaust from an orbital mission.

I suspect that almost all launches from Earth will leave the rocket exhaust in Earth orbit, soon to return to Earth, particularly because most rocket firings happen nearly in the prograde direction, which means the exhaust will have less energy then the rest of what's around.

Of all of the environmental affects that are on the watch list for rocket launches, Ozone depletion is the largest concern, followed by soot/ particles, and then on to CO2 emissions.

$\endgroup$
  • $\begingroup$ Can you address "... is there some point above which most rocket exhaust would not become part of Earth's atmosphere?" specifically? I think this answer is only considering the sub-orbital hopping discussed in the linked question. A deep space mission would deposit thrust at much higher altitudes. $\endgroup$ – uhoh Jan 5 '18 at 15:12
  • $\begingroup$ Added a sentence. Sub-orbital will remain in the atmosphere. Only interplanetary, or retrograde burning for orbital missions, has a good chance of leaving the atmosphere. $\endgroup$ – PearsonArtPhoto Jan 5 '18 at 15:26
  • 1
    $\begingroup$ It's not the altitude that matters, it's the velocity. Rocket exhaust is just like any orbiting particle, if it's going slow enough, it will fall back down to Earth. The higher you are will affect that somewhat, but the fundamental rule stays the same.And really it would have to escape Earth's gravity entirely to not return to Earth, or else be really really high (HEO orbit at least) $\endgroup$ – PearsonArtPhoto Jan 5 '18 at 18:27
  • 2
    $\begingroup$ Of course altitude makes a difference in the orbital velocity, but there is no magic altitude where the gas will stay away just because... $\endgroup$ – PearsonArtPhoto Jan 6 '18 at 1:04
  • 2
    $\begingroup$ The number depends on what direction the thrust is going, what is the velocity of the thrust, and the altitude, not to mention a few other things like where the Moon is at. The real question was will the suborbital exhaust stay, and that answer is unequivocally yes. $\endgroup$ – PearsonArtPhoto Jan 6 '18 at 3:40
0
$\begingroup$

Let's look at a transfer orbit, it's orbital velocity at apoapsis, escape velocity and exhaust velocity.

For an ellipse with periapsis and apoapsis $r_{peri}, r_{apo}$

$$a = \frac{r_{peri} + r_{apo}}{2}$$

$$ v_{apo} = \sqrt{G M_E \left( \frac{2}{r_{apo}} - \frac{1}{a} \right)} $$

$$ v_{circ} = \sqrt{ \frac{G M_E}{a}} $$

$$ v_{esc} = \sqrt{2}v_{circ} = \sqrt{ \frac{2 G M_E}{a}} $$

and the critical exchaust velocity is that which shooting out the back of a spacecraft at apoapsis, moving at $v_{apo}$ would still have escape velocity:

$$v_{ex, crit} = v_{apo} + v_{esc}$$

Plotting those, you can see that for the second impulse at apoapsis, for exhaust velocities of 2000, 3000 and 4000 m/s, (roughly Isp's of 200, 300 and 400), the exhaust would have escape velocity for apoapsese of roughly 80,000 130,00 and 260,000 kilometers.

There are certainly spacecraft put in Earth orbit out that far, but they are rare.

For most satellites put in orbit closer to Earth, the exhaust does not reach escape velocity. Instead, the exhaust orbits the Earth and will slowly mix its momentum with other atoms of the atmosphere and begin to thermalize and mix physically with the atmosphere.

It's a whole 'nuther ball of wax for electric propulsion! Back at the "turn of the century" (i.e. 2001) satellites to GEO were all (or nearly all) sent with conventional chemical propulsion. These days all-electric GEO telecommunications satellites are all the rage because it saves so much weight. In addition to using electric propulsion for station keeping which was developed earlier using arcjets, now ion propulsion is common both for station-keeping and going from LEO to GEO by spiraling outwards.

You can estimate the exhaust velocity of an ion engine using

$$\frac{v}{c} = \sqrt{\frac{2E}{m_0 c^2}}. $$

Choose $E=$ 100 keV and $m_0 c^2=$ 931 MeV times 50 to 200 AMU and you get between 0.2 and 0.1% of the speed of light, which is way beyond escape velocity. You can safely assume that any angular momentum gained by electric propulsion in Earth orbit at or beyond reasonable LEO is compensated by an equal and opposite angular momentum of ions flying out of the back of the spacecraft. Depending on the scenario they could become trapped by the Earth's magnetic field and spiral into Earth's poles, thereby assuring that the angular momentum is transferred to the Earth's solid body, or if far enough away, leak out into the solar system after bending in the Earths' magnetic field, making it more complicated.

So I've just asked Where do ion propulsion's ions go? Do they remain in the solar system or shoot out into interstellar space?

Python script for math and plots: https://pastebin.com/47wBu6sJ

enter image description here

enter image description here

$\endgroup$
  • $\begingroup$ I think this is probably the best answer, but nobody else does ;-) $\endgroup$ – uhoh May 18 at 0:25
-2
$\begingroup$

Orbit (EDIT: more like escape velocity)

Well, kind of. Gas is similar to anything in space: if it's in orbit, it won't come down until the atmosphere or solar radiation decays it's trajectory enough. But no matter what, it will settle onto the top of the atmosphere if it stays in orbit. If it escapes the Earth's gravitational pull then it will never come down.

It's not necessarily a matter of altitude, but more of the orbital trajectory of the particles.

As for the crossover value, it depends on the engine, namely its exhaust velocity and plume dispersion, as well as the ascent profile of the rocket. So the answer is that there isn't really any set threshold that will define, for all rockets on LP-1/LOX or any other fuel combination, when 50% of the exhaust will fall back to Earth and 50% will not. It is entirely dependent on the particular rocket launch.

EDIT:

Let me rethink my answer...

Let's assume a Merlin 1D from SpaceX as an example, running on kerosene and oxygen.

In a vacuum, Merlin 1D has a specific impulse of 311 seconds, so our exhaust velocity $v_e=g_0\times I_{sp}$, so about 3,050 m/s.

If the rocket is at the end of its orbital insertion burn into LEO, then it will be traveling at orbital velocity, 7,800 m/s. If the exhaust leaves the engine the other way at 3,050 m/s, then the final velocity of the exhaust would be 4750 m/s, well under orbital velocity. It will be recaptured by the atmosphere.

If the exhaust is to leave the Earth forever, then it must be traveling away from Earth at escape velocity, 11,200 m/s, and then some to account for other forces. That means that the rocket itself must be traveling at escape velocity plus the exhaust velocity: 14,250 m/s. The rocket must be well on its way out of the Earth's gravitational influence if it is to stay in space.

Again, this leads to the same answer that there is no practical 50%. It is more of a binary returns-to-earth or doesn't-return-to-earth outcome. Altitude is not important, it is velocity and direction of the exhaust that matters.

$\endgroup$
  • $\begingroup$ I didn't down vote your answer, but it's really not a helpful answer. I agree that there is not a specific, single altitude, but right now the subject is totally "up in the air", and I think working through the physics may bring out a rough answer. It's probably much higher than 100 km, but lower than 1,000,000 km, and I mentioned a particular class of engine as a guideline specifically to avoid the "it depends on the engine" excuse for leaving a non-answer as an answer. Let's see if we can work the problem as an exercise and try to narrow it down to less than four orders of magnitude. $\endgroup$ – uhoh Jan 5 '18 at 14:13
  • 1
    $\begingroup$ No, a gas molecule is not just like a satellite. Due to the incredibly small mass of a molecule, it's trajectory can be and is dramatically affected by solar radiation. As well as its integrity, often being broken up in to atoms. This is how planets lose their atmospheres. $\endgroup$ – Mark Adler Jan 5 '18 at 14:37
  • $\begingroup$ @uhoh I have updated my answer. $\endgroup$ – Elijah Seed Arita Jan 5 '18 at 21:46
  • $\begingroup$ Yep! But remember that escape velocity depends on distance. You can't use a value that applies to the surface (or LEO) at other locations. Further, once you get farther, the Sun's gravity comes into play as well. So there will definitely be a place where the cone of expanding exhaust plume will no longer predominantly become part of Earth's atmosphere. It's unfortunate that this answer has collected three down votes, which sting a lot for a new user. You might consider in this particular case to delete it and post a new, separate answer (but don't do this repeatedly). $\endgroup$ – uhoh Jan 6 '18 at 1:08
  • 1
    $\begingroup$ @uhoh I’ll take the blow. I wrote an incomplete answer and the votes reflect that. As long as the edited question gains votes back then I will be okay. $\endgroup$ – Elijah Seed Arita Jan 6 '18 at 11:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.