# How fast until you feel wind in space? [duplicate]

I couldn't find this question asked anywhere else, so I thought I would ask it here: does anyone know how fast you would have to be traveling in space to feel 'wind' from the particles in front of you? Say a piece of thin, 0.05 N piece of paper traveling in perpendicular to it's width and length. How fast would it need to travel before it would experience, say, 1 N of force?

• When you say space, I presume you actually mean, say, the Earth's air at sea-level, rather than vacuum? – innisfree Sep 19 '14 at 20:49
• You can calculate the force by dimensional analysis, $F = const\times\rho v^2 A$ IIRC. But calculating the dimensionless constant probably requires an experiment...? – innisfree Sep 19 '14 at 20:53
• – Qmechanic Sep 19 '14 at 20:55
• What part of space? The particle content of intragalactic space is wildly different depending where in the galaxy you are, and intergalactic space is again a very different environment. – Kyle Oman Sep 19 '14 at 21:00
• possible duplicate of Would a fast inter-stellar spaceship benefit from an aerodynamic shape? – John Rennie Sep 20 '14 at 9:44

That's rather an unusual question though interesting, but it amuses me to make an even more unusual answer.

Disclaimer: this is in no way a rigorous answer, as many assumptions, simplifications have been made (even entire phenomenons such as gravity attraction if the velocity is not perpendicular, solar pressure from photons...) - only a reflection on drag alone. But I think you figured it out since it's Superman we are talking about here.

If you were actually flying like Superman, how strongly would you be affected by drag?

We'll see not only

1. how fast you need to go to get 1N of force from the drag, but also
2. what your top speed would be assuming we know how strong Superman's thrust is.

## Drag equation

As innisfree said in the comments, the usual formula to quickly get estimates is the drag equation: $$F_D=0.5 \rho v^2C_DA$$ Let's fill in the gaps, what do we have, what do we want?

## Cross section

Loading the above picture in an image editor, Superman's head diameter is ~200px; a lasso select yields a cross sectional area perpendicular to the velocity of ~124900px². Assuming Superman has a head 1.5 times bigger than the average male (just because), this is 725px/m, which means the area is really 0.238m².

## Coefficient of drag

This is an experimental parameter linked to the airflow around the geometry, but we'll take it as 0.82 for that of the side of a cylinder.

## Drag force / Thrust force equilibrium

When calculating Superman's top speed, we'll assume he can develop the equivalent of the thrust of a medium airplane such as the Airbus A321, or 114kN (after all, he's been seen struggling to stop a plane). No comments on how he's supposed to achieve thrust. Top speed will be reached when drag and thrust oppose eachother.

## Fluid density

As said in other comments, the density of the fluid surrounding you is not the same everywhere. In average though in the universe, it is between 0.1 and 1 $$atom/cm^3$$, or very roughly $$1.67*10^{-24}kg/m^3$$ seeing it's mostly hydrogen... For comparison, this is $$10^{17}$$ times less than at 100km of altitude (or the conventional "end" of our atmosphere), according to the simplified model (a more accurate model is available here) in SI units: $$\rho \approx 1.21e^{-x/8000}$$

## Calculations - 1N of force

First, let's answer your question. Rearranging the formula to find the speed at which 1N of drag is developed: $$v=\sqrt{\frac{2F_D}{\rho C_DA}}$$ We find that you need to fly at:

10km/h at sea level (sanity check)

19km/h at 10km (planes altitude)

68km/h at 30km (stratospheric balloons)

5427km/h at 100km ("end" of atmosphere)

Impossible in outer space ($$10^{12}$$km/h or much more than the speed of light)

## Calculations - top speed of Superman

Still considering the maximum thrust of 114kN, the same equation tells us that Superman can go as fast as:

3537km/h at sea level (or 2.9 times the speed of sound - machs)

6608km/h at 10km of altitude (or 5.4 machs)

23066km/h at 30km of altitude (or the orbital speed at 4000km of altitude)

1.8 million km/h at 100km of altitude (maybe that's why he does not age that much - starting from here numbers become irrelevant as Newtonian mechanics is no longer valid as we get closer to the speed of light)

Just under the speed of light in outer space (would have been above)

• You'd have to stop using that Newtonian pressure formula once you started approaching the speed of light -- the momentum transfer with the ISM particles would start becoming relativistic, so you'd have to calculate the 3-force. – Jerry Schirmer Sep 20 '14 at 1:13
• Yes, I very briefly mentioned relativistic effects on the top speed at 100km of altitude, but I'll make that clearer. – Mister Mystère Sep 20 '14 at 1:21
• Wow, I never expected such a comprehensive response! :) really entertaining answer. – SheerSt Sep 20 '14 at 2:17