If you were standing on a planetesimal in the planet forming disk of a new solar system (or our own, billions of years ago), would you be able to feel "interplanetary wind"? Would it be physically possible to "fly" from one planetismal to another or change your solar orbit using an airplane, rotorcraft, or balloon instead of a rocket?

(Obviously you'd need a space suit in the first case, and something other than an air breathing engine in the other - perhaps assume the propeller is spun using electricity and/or nuclear power.)

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    $\begingroup$ what do you mean by "interplanetary wind" ? $\endgroup$ Jan 27, 2022 at 4:15
  • $\begingroup$ Most of the matter in the Solar System is in the Sun (it's >1000× more massive than Jupiter). If you spread the Sun into a uniform cylinder with a radius of 30 AU (Neptune's orbit) with the same density as air, it'd be about 60 km thick. $\endgroup$
    – PM 2Ring
    Jan 27, 2022 at 11:01
  • $\begingroup$ I'd recommend rewording this question to something quantitative like "what is the gas density as a function of time in the last billion years before a solid planet emerges. $\endgroup$ Jan 27, 2022 at 13:47
  • $\begingroup$ The surface density of gas (and dust) in a protoplanetary disk decreases with distance from the host star. Most models use a standard equation from Lyndon Bell & Pringle (1974), which is a power law with an exponential taper. The precise density at a given point will depend on the mass/size of the disk that is being modelled. $\endgroup$
    – lucas
    Jan 27, 2022 at 22:26
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    $\begingroup$ Why have you accepted an answer which does not suggest any value for the density of gas in a protoplanetary disk? $\endgroup$
    – ProfRob
    Jan 28, 2022 at 17:01

2 Answers 2


The total mass of all the planets is about $3\cdot10^{27} kg$. If you assume the proto-planetary disk to have a radius of $10^9 km$ (somewhat beyond the orbit of Jupiter) and a thickness of $10^7 km$ (roughly the thickness the planetary system has today) this results in a volume of about $3\cdot10^{34} m^3$. This means the matter density in the proto-planetary disk would have been about $10^{-7} kg/m^3$ , which is 7 orders of magnitude smaller than the air pressure on earth at seal level.

Interestingly, this estimate is close in order of magnitude to the density you get for the sun at a stage where its radius was also $10^9 km$: the average density of the sun at its present radius ($7\cdot 10^5 km$) is $1.4\cdot10^3 kg/m^3$, so at $10^9 km$ it would have been about $5\cdot 10^{-7} kg/m^3$. A factor 2 error in either of these estimates would practically make the figures equal (and they should really be equal as it would not be physically plausible if the gas component with a high angular momentum (the proto-planetary disk) had a different local pressure/density to the component with a low angular momentum (the proto-sun).

So the gas densities in the proto-planetary disk would have been on average similar to that in the earth's atmosphere at a height of about $100 km$. And the practical use of aerodynamics in the usual sense (e.g. as with airplanes) is not possible anymore at this height. Even though you have still aerodynamic drag and lift (after all, satellite orbits are unstable in the long term because of drag even at greater heights), the air has practically no viscosity anymore as molecules do practically not collide with each other anymore. Their mean free path with regard to mutual collisions between molecules is about $100m$ (at sea level it is $0.01 mm$), which is larger than the dimensions of the aerodynamical structure, so the latter would not behave in the usual way even if you would fly them at much higher speed to compensate for the smaller density.

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    $\begingroup$ This is probably good enough to answer the question, but the amount of mass in a protoplanetary disk can be more like 10% of the stellar mass. On the other hand, disks are a lot bigger than the orbit of Jupiter - by an order of magnitude, so these possibly cancel. $\endgroup$
    – ProfRob
    Jan 28, 2022 at 21:09
  • $\begingroup$ @ProfRob Well, I assumed for simplicity a constant density, so I had to take a cut-off point somewhere. In reality, there will of course be a radial dependence of the density and the disk will extend further. See also my edited answer. $\endgroup$
    – Thomas
    Jan 30, 2022 at 10:29

I imagine that the density of the gas in the protoplanetary disc around a young star would vary with distance from the star.

I note that the atmosphere on Mars is quite thin compared to that on Earth:

The average surface pressure is only about 610 pascals (0.088 psi) which is less than 1% of the Earth's value


But there are sometimes sandstorms on Mars, and airborne sand and dust does settle on the solar panels of Martian rovers. So the winds on Mars might be felt through a spacesuit.

The atmosphere on Triton, the moon of Neptune, is much thinner than the atmosphere on Mars.

Triton's surface atmospheric pressure is only about 1.4–1.9 Pa (0.014–0.019 mbar).[7]


The surface pressure is only 14 microbars (1.4 Pa or 0.0105mmHg), 1⁄70000 of the surface pressure on Earth,1


In 1989 Voyager 2 discovered that near the surface there are winds blowing to the east or north-east with a speed of about 5–15 m/s.[9] Their direction was determined by observations of dark streaks located over the southern polar cap, which generally extend from the south-west to north-east. These winds are thought to be related to the sublimation of nitrogen ice from the southern cap as there was summer in the southern hemisphere in 1989.[9] The gaseous nitrogen moves northward and is deflected by the Coriolis force to the east forming an anticyclone near the surface. The tropospheric winds are capable of moving material of over a micrometre in size thus forming the streaks.[9]

The atmosphere is dense enough to allow the formation of dunes.[17]


So it is possible that even an interplanetary atmosphere as thin as the atmosphere of Triton would have visible effects.

A meteoroid is a small rocky or metallic object in outer space, ranging in size from a grain of dust to about a meter in diameter.

When a meteoroid, comet, or asteroid enters Earth's atmosphere at a speed typically in excess of 20 km/s (72,000 km/h; 45,000 mph), aerodynamic heating of that object produces a streak of light, both from the glowing object and the trail of glowing particles that it leaves in its wake. This phenomenon is called a meteor or "shooting star". Meteors typically become visible when they are about 100 km above sea level.


Meteors become visible between about 75 to 120 km (250,000 to 390,000 ft) above Earth. They usually disintegrate at altitudes of 50 to 95 km (160,000 to 310,000 ft).3


So gas as thin as Earth's atmosphere at altitudes of about 5 to 120 kilometers would cause meteoroid's passing thorugh it to glow with the heat of friction and even disintigrate.

If you were standing in such a dense interplanetary gas you would see meteors as passing meteorids glowed with heat, especially since the protoplanetary disc should have contained many times as many meteroids as interplanetary space now has.

The interplanetary dust grains would be slowed down by collisions with gas particles and would spiral in toward the star. And they would be broken down into smaller particles, and so would have a hard time clumping together to form larger particles and eventually planets.

So if the gas in the protoplanetary disc was even as dense as Earth's atmosphere at altitudes of 50 to 120 kilometers, astronomers would have to include that in their computer simulations of the formation of the solar system.

So how dense or thin is Earth's atmosphere at altitudes of 50 to 120 kilometers?

The scale height of Earth's atmosphere is the height difference where the density of the atmosphere decreases by a factor of e, about 2.718. It isabout 8.5 kilometers.

50 kilometers is a littles less than 51 kilometers, 6 times 8.5, while 120 kilometers is a little more than 119 kilometers, 14 times 8.5.

Accordign to my rough calculations at 51 kilmeters altitude the density of Earth's atmosphere compared to sea level is about 1 divided by 403.17787, or 0.0024802, and the density of Earth atmosphere at 119 kilometers compared to sea level is one divided by 1,201,004.7, or 0.0000008.

And i expect that a much thinner interplanetary gas would have significant effects on the orbits of interplanetary dust particles over thousands and millions of years as a solar system forms.

So theories and calculations and simulations of planetary formation should probably include estimates of the density of interplanetary gas in those eras and its effects on dust particles.

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    $\begingroup$ What is the density in the protoplanetary disk is the central question. $\endgroup$
    – ProfRob
    Jan 27, 2022 at 19:43

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