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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 ${2m\over A\,C_L}=3.6\,\mathrm{kg\over m^2}$. Then I can use that and the density profile of any other body to get its von 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 is has been defined at 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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