Can an object from a natural process escape earth gravitation?

Question

I'm no expert, but I once studied basic from advanced physics and understand

gravity
action/reaction
escape velocity of 11.2 km/s from the earth surface
escape velocity changing as the object go far away
Karman line for space (100 km)

What I want to know, can a natural object, from a natural process, reach the escape velocity 11.2 km/s and get free of earth gravitation?

I have in mind some volcano explosions. Tornados seem weak. Or also a Comet impact. Question is not limited in time, maybe it was possible millions of years before, when activities and gravity might have been different.

By omitting the constraint on the escape velocity, hydrogen and helium are two likely candidates for "natural" objects escaping the Earth's gravity. — polfosol ఠ_ఠ, Aug 09 '21 at 12:54
Well [putting my tongue in my cheek], since oxidation and catalytic hydrazine decomposition are both natural processes, Voyager2 qualifies... — Peter - Reinstate Monica, Aug 09 '21 at 13:54
There are many rocks from Mars on Earth. If they could escape Mars gravitation, why couldn't comparable processes on Earth go the other way? — Barmar, Aug 09 '21 at 15:14
"I once studied basic from advanced physics". That sentence puzzles me. (It does not make grammatical sense, and don't want to change the meaning via an incorrect edit.) — RonJohn, Aug 09 '21 at 19:30
@Barmar it's easier to escape small, barren Mars than big, atmosphere-encircled Earth. Being closer to the Sun might also make it harder. — RonJohn, Aug 09 '21 at 19:41
@RonJohn, I'm from a French background, so my english might surely not be on top. There is physics at school level, and university level. So I had some courses that covered basic Physics that you'd find at university level (that I called advanced physics). — KeitelDOG, Aug 11 '21 at 15:55
@KeitelDOG "my english might surely not be on top." I figured as much, which I why I questioned, not criticized/mocked you. A better sentence would be "I once studied basic to advanced physics". — RonJohn, Aug 11 '21 at 16:11
Once you get into high orbit, you can get anywhere in the solar system, and even outside of it, for free. You just have to try many, many, many different initial trajectories and allow a lot of time to happen. If you KSP, take a peek at this: https://forum.kerbalspaceprogram.com/index.php?/topic/63369-a-preliminary-study-into-the-migratory-habits-of-the-common-octo-2 — PcMan, Aug 11 '21 at 20:10

score 38 · Accepted Answer · edited Aug 10 '21 at 20:34

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Yes, it is not only possible, but has almost certainly happened on Earth. The asteroid that killed the dinosaurs is thought to have produced these high velocity fragments on the order of one-thousandth of the mass of the impactor according to Poveda and Cordero [2008]. In fact, they estimate the volume of escape velocity ejecta from the Chicxulub event as quite substantial in that:

the number of fragments with sizes larger than 10 cm and 2 cm is about 4x10^10 and 2x10^12, respectively

The authors of the above paper call these "Chicxulubites" and speculate on the possibility of finding them on the Moon and Mars.

The larger the impactor radius and velocity, the higher a percentage of its overall mass will reach escape velocity. In an asteroid impact, the fastest moving ejecta comes from near the center of the crater. The ejecta closer to the edge of the crater has much lower velocity:

The equation governing the velocity of the ejecta $V_{ej}$ is therefore partly a function of $r$, the distance from the center of the crater.

$$V_{ej} = \frac{2\sqrt{Rg}}{1+\epsilon }\left(\frac{r}{R}\right)^{-\epsilon}$$

Where $\epsilon$ is a material coefficient of 1.8 for hardpack soil, $g$ is the surface gravitational acceleration, $R$ is the total radius of the equator, and $r$ is the radius at which the ejecta was ejected.

Notes:

I go into more detail in this answer: What percentage of a lunar meteor strike is blown back into space?.
Practically, an object near sea level must achieve higher than escape velocity in order to overcome atmospheric drag as it escapes.

edited Aug 10 '21 at 20:34

Mark Omo

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answered Aug 09 '21 at 00:24

Connor Garcia

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related: Could rocks from Earth have reached the Kuiper belt, or Neptune at least? If so, how? – uhoh Aug 09 '21 at 01:02
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@uhoh I suspect that so much small material in such a wide variety of escape velocity trajectories is similar to a monte carlo simulation, in that various pieces end up all over the solar system. – Connor Garcia Aug 09 '21 at 02:56
Many thanks for all the explanations, didn't know ejecta was the word for it. I suppose then that the ejecta will be a combination of both earth and asteroid particles, and transformed particles from impact, fusion and chemical reactions. – KeitelDOG Aug 09 '21 at 17:02
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@uhoh very useful answer, thanks – KeitelDOG Aug 09 '21 at 17:03
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Good answer. It's tough to find a natural store of energy that can hold enough energy to make a reasonably sized object move at escape velocity - a much larger object that's already moving at escape velocity is a great solution! – Nuclear Hoagie Aug 10 '21 at 14:38
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@KeitelDOG Ejecta is indeed composed of both Earth and asteroid material. The high concentration of iridium in the KT-Boundary sediment layer on Earth is what led Louis and Walter Alvarez to hypothesize the Chicxulub event. – Connor Garcia Aug 10 '21 at 15:59

uhoh · Answer 2 · 2021-08-09T23:12:58.257

If the object can be as small as an atom, then it's happening constantly. Atoms in a hot gas mixture will have an average velocity related to the average kinetic energy which is in turn related to temperature.

tl;dr: The Earth loses 95,000 tonnes of hydrogen and 1,600 tonnes a year of helium in the form of atoms or molecules.

For example, in a standard atmosphere near the surface, the average speed of a molecule is also roughly the speed of sound, or about 330 m/s for an average molecular mass of 28. For helium atoms in air with a mass of 4 the speed will be $\sqrt{7}$ larger, or about 870 m/s.

That still sound very far from escape velocity, but that's the average. There's always a distribution in velocities and for a given velocity component the Maxwell-Boltzman velocity distribution is

$$f_v(v_i) = \sqrt{\frac{m}{2 \pi k_B T}} \exp\left( \frac{-mv_i^2}{2 k_B T} \right)$$

But at the tippy-top of Earth's atmosphere a few things are different. Atoms can be much hotter than 300 Kelvin and the distributions of kinetic energy can be non-thermal in nature, and the densities are so low that once an atom achieves escape velocity there's a good chance it will escape without hitting anything further.

At lower altitude the atmosphere's mixture has relatively constant ratios, but (as the plot shows) once you get above 90 to 100 km the density is so low that the gases are no longer automatically mixed. At this point each gas can have its own scale height such that the exponential rate of drop off is faster for the heavier gases.

This produces a sequence of primary gas component as a function of altitude.

See concepts of scale height and turbulent mixing in answers to Why does Earth's atmospheric density have a big “knee” around 100 km? Is there a good analytical approximation?

As shown in the plot below, above about 190 km monatomic oxygen (O, not O2) is the primary component of Earth's atmosphere. At 450 km helium is the primary component and above roughly 1250 km hydrogen becomes the primary component.

Source

This answer to Is Earth getting heavier or lighter? says:

It seems a similar question was asked in Astronomy Stack Exchange, and this short answer links to the BBC News article Who, What, Why: Is the Earth getting lighter? which answers this question.

[...]

"Physicists have shown that the Earth is losing about three kilograms of hydrogen gas every second. It's about 95,000 tonnes of hydrogen that the planet is losing every year.

"The other very light gas this is happening to is helium and there is much less of that around, so it's about 1,600 tonnes a year of helium that we lose."

I thought that most light gas atoms (or maybe molecules) are blown away by the solar wind, so they don't (initially) need escape velocity. That's why the magnetic field is essential for retaining an atmosphere. What happens if it's missing can be seen on Mars. — Peter - Reinstate Monica, Aug 09 '21 at 14:03
@Peter-ReinstateMonica now that's an interesting point, but the situation for helium is special; even though it is constantly being released into the atmosphere and there is no sink for it (nothing else removes it, it's not chemically active) it never builds up, unlike N2, O2, CO2, H2O. There's much less release of Ar than He from Earth and yet Ar is 1% and He is only 0.0005 %. It's the mass that's different; He escapes, Ar stays put. — uhoh, Aug 09 '21 at 16:37
Thanks for those details at the top atmosphere, it makes sense for the gaz up there to occasionally leave when there is few gaz resistance. — KeitelDOG, Aug 09 '21 at 17:23
@uhoh The mass will also influence the impact of the solar wind. — Peter - Reinstate Monica, Aug 09 '21 at 17:59
@Peter-ReinstateMonica that can be, but as you can see from the edit I've made one above about 200 km mixing stops and each gas has a very different height profile. — uhoh, Aug 09 '21 at 23:14

score 8 · Answer 3 · answered Aug 09 '21 at 14:01

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I'd be surprised if the event that created the Moon didn't scatter some debris beyond the Earth's Hill sphere (in order to choose one meaning of "escape Earth gravitation").

answered Aug 09 '21 at 14:01

Peter - Reinstate Monica

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Can an object from a natural process escape earth gravitation?

3 Answers3