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 its 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 $\mathrm{cm}^3$ in nebulae to (much) less than one molecule per $\mathrm{cm}^3$ in intergalactic space, but if we take the value of $10^4$ in the table on Wikipedia the mean free path is $100\,000\ \mathrm{km}$. 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 turbulence 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 $10^4\ \mathrm{cm^{-3}}$ 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 $v$ 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:
$$ F = \rho v^2 $$
An atom density of $10^4\ \mathrm{cm^{-3}}$ is $10^{10}\ \mathrm{m^{-3}}$ or about $1.7 \times 10^{-17}\ \mathrm{kg/m^3}$ and $0.5c$ is $1.5 \times 10^8\ \mathrm{m/s}$ so $F$ is about $0.4\ \mathrm{N/m^2}$.
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:
- the gas density goes up due to Lorentz contraction
- 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 $7.5\ \mathrm{N/m^2}$, 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.
We can make a first approximation of this situation by considering an object sliding down an incline with friction and air resistance.
$\vec{F} = m\vec{a}$
Using a coordinate system that is aligned with the incline (+x is up the incline), we can sum the forces to satisfy Newton's second law.
In the x-direction:
$D + S = mg\sin{\theta}$, where $D$ is the drag, $S$ is the surface friction, and $\theta$ is the angle of incline.
In the y-direction:
$N = mg\cos{\theta}$, where $N$ is the normal force.
Assume:
$D = \frac{1}{2}\rho V^2AC_D$
$S = \mu N$
Where $\rho$ is the density of air, $V$ is the speed of the skier, $\mu$ is the coefficient of friction, $A$ is the frontal area of the skiers, and $C_D$ is the coefficient of drag.
Now we can solve for $V$:
$V = \sqrt{\frac{2mg\left(\sin{\theta} - \mu\cos{\theta}\right)}{\rho C_D A}}$
Based on this, for the same $A$ and $C_D$, the heavier skier goes faster. If the frontal area of a skier is proportional to $(mg)^{\frac{2}{3}}$ (assuming spherical skier), then $V \propto (mg)^{\frac{1}{6}}$.
Best Answer
The answer to this question is "hard" in the sense that a general object falling from general initial conditions may or may not reorient into a lowest drag orientation.
However, one can make some educated guesses by comparing the locations of the center of drag (more generally, the center of pressure) and the center of mass of the object. Consider one commenter's example of a parachute -- nearly all the drag force is on the inner surface of the parachute -- meaning the center of drag is near the "middle" of the parachute. However, the center of mass (since parachutes are generally much lighter than people) is below the parachute, near the person. This has the effect of the person being suspended under the parachute. If the person swings a little one way or the other, the center of drag is still above the center of mass and the system does not tumble.
For a styrofoam bowl with a conical top, the styrofoam is very likely to have little impact on the center of mass of the cone, but a large effect on the center of drag. Once again, the falling preferred orientation is center of mass below center of drag. However, this can be defeated -- make the cone out of styrofoam. Another way to defeat this is to start with a very large angular velocity, so that the tendency of the mass to fall below the drag is overcome by the angular momentum. The preferred orientation will eventually prevail, but there may be many revolutions before this happens. (There is an optimization problem lurking here: Make the cone too dense and its angular momentum will be hard to dissipate. Make the cone too light and the centers of mass and drag will coincide so there will be no preferred orientation.)
An easy way to see that this should be hard is this. The drag of the system depends on the orientation of the system, so the center of drag depends on the orientation of the system. A good parachute will be designed so that a small swing by the person will lead to a small central force to bring the person back under the center of drag. A bad parachute design would lack that small central force -- it could, even worse, lower net drag so that the parachute begins to fall faster, perhaps even faster than the person. (This actually happens. It's the worst scenario in what is called "stalling a parachute.") Nevertheless, we can design for this. Recall that SpaceShipOne was designed so that in its feather configuration it would fall in a particular orientation -- i.e., small displacements from that orientation would be passively corrected by the change in the center of drag.