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Is there a common definition of where the space begins on other planets (and other celestial bodies)? An equivalent of the Karman line.

How far one should go from the martian surface to get to the outer space?

Am I already in space jumping 2 meters above the Moon surface?

(please don't answer that anything beyond the earthly Karman line (including any other planets' surfaces) is in space by definition).

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    $\begingroup$ I imagine there isn't one common definition, since the Karman line is defined atmospherically. If you use a purely atmospheric definition, then you are in outer space if you are standing on the moon. Maybe that definition is useful, maybe not. $\endgroup$ – called2voyage Sep 23 '13 at 14:14
  • $\begingroup$ @called2voyage you could easily turn that into the answer. Give it a shot! $\endgroup$ – Stu Sep 23 '13 at 14:50
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    $\begingroup$ @Stu It doesn't answer "Is there a common definition" definitively because there could be an official standard that indicates how to determine the outer space boundary. It also doesn't answer his question as to whether there is any sort of analog for any celestial bodies, which there may be. $\endgroup$ – called2voyage Sep 23 '13 at 14:55
  • $\begingroup$ Exactly. There can exist a common definition I hope to get my hands on. Or, there could exist separate well established definitions for several well known objects such as Mars, Venus, Moon and Jupiter moons, used in martiology, venology, lunology and whateverlogy. $\endgroup$ – horsh Sep 23 '13 at 15:45
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    $\begingroup$ @horsh The terms are actually areology (Mars), aphrology (Venus), and selenology (Moon/Luna). $\endgroup$ – called2voyage Sep 23 '13 at 17:01
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I am not aware of anyone using a Kármán line on another planet, but it's not hard to calculate. I get about 88 km for Mars.

The equation is $r\,\rho(r)={2m\over A\,C_L}$, where $r$ is the radius of the Kármán line from the center of the body, $\rho$ is the density as a function of the radius, $m$ is the mass of the aircraft, $A$ is the planform area, and $C_L$ is the coefficient of lift. von Kármán picked some representative values for $m\over A$ and $C_L$, which I don't know. But I don't need to know. I can calculate ${2m\over A\,C_L}$ by plugging the Kármán line for Earth into $r\,\rho(r)$. I get that ${2m\over A\,C_L}=3.6\,\mathrm{kg\over m^2}$. Use that and the density profile of any other body to get its Kármán line.

By this definition, you are already in space lying on your back on the Moon.

Though I don't know of anyone using the Kármán line outside of Earth, what is in use is the entry interface altitude. By definition, the entry phase has begun when this altitude is crossed. It is sufficiently high to have no perceptible accelerations from the atmosphere, but low enough that you don't have to wait very long to see such accelerations.

At Earth for Apollo it was defined as 400,000 feet, or about 122 km. At Mars it has been defined as 125 km. That is when the $n$ minutes of terror begins.

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Kármán himself would have defined it for nonterrestrial bodies just like he did for Earth:

The height where the atmosphere becomes so thin that aerodynamic lift strong enough to support an aircraft happens only at airspeeds faster than orbital velocity.

So you might as well call that the boundary between atmosphere and space, as far as propulsion is concerned. http://books.google.com/books?id=dTwIDun4MroC&pg=PA84#v=onepage&q&f=false

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    $\begingroup$ +1 This answer really nails in my point in the first comment to the question. Let's call this answer the "Maybe...useful" case, for propulsion. $\endgroup$ – called2voyage Dec 10 '13 at 19:59
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I will elaborate a bit on Mark's answer starting from the basic formula $R\rho = 2m/(A*Cl)$. Above the troposphere, the density falls exponentially as $rho = rho_0 \exp(-h/H_0)$ where $H_0$ is the scale height. This means you have to solve an equation like: $$rhs = (R+h)*\exp(-h/H_0), rhs = 2m/(A*Cl*\rho_0)$$ This can be solved analytically using the Lambert W function, which is the inverse $w(x)$ of $x*\exp(x)$. The answer turns out to be $r = -H_0W(-1,y)-R, y=-\exp(-R/H_0)*rhs/H_0$. The notation $W(-1,x)$ indicate that the -1 branch of the Lambert W function is used. Using the standard ISA atmosphere model above the troposphere, m~10000 kg, Cl~ 0.5, A=30 (typical small fighter aircraft) the answer is about 93 km. Using the atmospheric composition of other planets it should be relatively easy to calculate this.

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The defined "Kármán line" at 100 km (62.14 mi) is actually not the true one. The true line Theodore of Kármán mentioned was at 57 mi / 91.5 km. Eventually, Kármán and the FAI just took the nextmost double-0-value in metric units which is 100 km and defined that as the "Kármán line". However, 57 miles are also a bit too high. Uhoh determined what is considered the Kármán line should actually be at 50 mi (80.5 km) which is the U.S.-defined space border and around the mesopause (meso-/thermosphere border).

So on Earth, outer space begins at the mesopause at around 50 mi (80.5 km).

As for other planets and moons, it depends on how high the atmosphere reaches which is not necessarily conclusive from the surface air pressure. Martian surface pressure is analogous to that of the Earth's stratosphere, the locations that have the highest air pressure (0.012 atm) are like about 19 mi (30 km) above the Earth's surface. I don't know well how high it reaches but Mark Adler already told you that it are obviously 88 km (54.7 mi).

So on Mars it's 54.7 mi (88 km) and thus higher than on Earth despite the surface pressure being lower than the Earth's.

The atmosphere of Venus (surface pressure: 91.8 atm) reaches very high, up to 200-250 mi (320-400 km).

And the atmosphere of Titan (surface pressure: 1.45 atm) reaches even higher, up to 300-400 mi (480-640 km).

The four gas planets consist only of gas except a likely core (Saturn however has no solid core) so you can only consider an altitude of the atmosphere either from the core's surfaces (the cores are about as big as the entire planet Earth) or from what is defined as Earth-like atmospheric pressure level so the altitude at which space begins may be similar to that on the Earth.

Neptunian moon Triton and the "dwarf planets" Pluto and Eris have a surface air pressure similar to that in the Earth's mesosphere (around 1 Pa) from which we may conclude that it's not very high, however Pluto's atmosphere indeed is, I however dunno how high exactly. Perhaps Earth-like.

Jupiter's moon Io has has an atmospheric pressure of up to 40 nbar in certain locations close to volcanoes. All of Io's collisional atmosphere reaches quite high too (perhaps as high as the Earth's space border) but such high pressure rather is found at those locations only.

Question on space border and uhoh's answer to it

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