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Some (generous) assumptions:

  • We have a spaceship that can reach a reasonable fraction of light speed.
  • The ship is able to withstand the high energies of matter impacting at that speed.

Given the amount of matter in inter-stellar space, at high speed, would it encounter enough of it and frequently enough that an aerodynamic shape would significantly reduce its drag (and thus save fuel)?

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3 Answers 3

up vote 17 down vote accepted

For the sorts of vehicles we're used to, like cars and aeroplanes, there are two contributions to drag. There's the drag caused by turbulence, and the drag caused by the effort of pushing the air out of the way. The streamlining in cars and aeroplanes is designed to reduce the drag due to turbulence. The effort of pushing the air out of the way is basically down to the cross sectional area of whatever is pushing it's way through the air.

Turbulence requires energy transfer between gas molecules, so you can't get turbulence on length scales shorter than the mean free path of the gas molecules. The Wikipedia article on mean free paths helpfully lists values of the mean free path for the sort of gas densities you get in space. The gas density is very variable, ranging from $10^6$ molecules per cm$^3$ in nebulae to (much) less than one molecule per cm$^3$ in intergalactic space, bt if we take the value of $10^4$ in the table on Wikipedia the mean free path is 100,000km. So unless your spaceship is very big indeed we can ignore drag due to turbulence.

A sidenote: turbulence is extremely important in nebulae, and a quick glance at any of the Hubble pictures of nebulae shows turbulent motion. However the length scale of the turblence is of the order of light years, so it's nothing to worry a spaceship.

So your spaceship designer doesn't have to worry about the sort of streamlining used in aeroplanes, but what about the drag due to hitting gas molecules? Let's start with a non-relativistic calculation, say at 0.5c, and use the density of $10^4$ I mentioned above, and let's suppose that the gas is atomic hydrogen. If the mass per cubic metre is $\rho$ and you're travelling at a speed $v$ m/sec then the mass you hit per second is:

$$ m = \rho v $$

Suppose when you hit the gas molecules you accelerate them to match your speed, then the rate of change of momentum is this mass times your speed, $v$, and the rate of change of momentum is just the force so:

$$ F = \rho v^2 $$

A density of $10^4$ atoms/cm$^3$ is $10^8$ per m$^3$ or about $1.7 \times 10^{-19}$kg and 0.5c is $1.5 \times 10^8$m/sec so $F$ is about 0.004N per square metre.

So unless your spaceship is very big the drag from hitting atoms is insignificant as well, so not only do you not worry about streamlining, you don't have to worry about the cross section either. However so far I've only talked about non-relativistic speeds, and at relativistic speeds you get two effects:

  1. the gas density goes up due to Lorentz contraction
  2. the relativistic mass of the hydrogen atoms goes up so it gets increasingly harder to accelerate them to match your speed

These two effects add a factor of $\gamma^2$ to the equation for the force:

$$ F = \rho v^2 \gamma^2 $$

so if you take v = 0.999c then you get $F$ is about 7.5N/m$^2$, which is still pretty small. However $\gamma$ increases without limit as you approach the speed of light so eventually the drag will be enough to stop you accelerating any more.

Incidentally, if you have a friendly university library to hand have a look at Powell, C. (1975) Heating and Drag at Relativistic Speeds. J. British Interplanetary Soc., 28, 546-552. Annoyingly, I have Googled in vain for an online copy.

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so extremely fast spaceships should be needle-shaped to minimize cross section? – endolith Jul 17 '13 at 14:33
Unless you're travelling very, very close to the speed of light the shape of the spaceship makes little difference. – John Rennie Jul 17 '13 at 14:40
Very interesting answer -- you mention the gas density increasing due to Lorentz contraction. Is it possible that the density increase (and hence the mean free path decrease) makes the turbulent length scales small enough to matter? Or does the Knudsen number stay constant regardless of velocity? – tpg2114 Dec 31 '13 at 1:29

At that speed aerodynamic shape would be unimportant since almost all particles would penetrate the hall. So more important would be total cross-sectional area perpendicular to velocity (how many particles per second collide with ship). So I assume that ship should have torpedo like shape to reduce total cross-sectional area.

More probable is use of some kind of electromagnetic shielding (to protect crew) but in such case aerodynamics of ship is also unimportant.

Don't forget that dimension in direction of movement is compress in such speed according to Special Relativity so aerodynamic shape becomes less effective (harder to achieve).

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Yes, this shape would be good, but not for aerodynamic reasons. As the others have commented, there's not much matter in your path. However, the matter that is in your way really hits your hull hard.

At that point, look at tank designs. Since WW2, tanks have sloped armor: armor which is at a angle to incoming shells. This increases the effective armor thickness by 1/cos(θ). An aerodynamic shape of a spaceship achieves a similar effect.

So, essentially your second assumption is dependent on this shape.

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