In Do atmospheric tides have any impact on orbiting satellites or rocket launches? we learned that the atmosphere does impact satellites in Low Earth Orbit (LEO), and that "most satellites have anything but a nice aerodynamic shape"

Why are LEO satellites not aerodynamically shaped?

  • $\begingroup$ I'm wondering if it would be a good idea to preface the title of your question with 'since LEO satellites are affected by drag', or something like that. For viewers with a more basic knowledge of space, if that isn't said they might undervalue the question, thinking there is no drag in space. $\endgroup$
    – kim holder
    Nov 29, 2014 at 15:32
  • $\begingroup$ @briligg That would be a rather long title. The question itself is fairly self-explanatory. $\endgroup$ Nov 29, 2014 at 17:08
  • $\begingroup$ Heh. Reminds me of "why are SciFi interstellar vehicles almost always supplied with rudders and ailerons?" $\endgroup$ Nov 29, 2014 at 19:17
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    $\begingroup$ Shape doesn't matter an awful lot, but you should read up about "Night Glider" mode on the ISS. They actually rotate the solar panels at night and can reduce drag by up to 30% $\endgroup$
    – Dylan
    Dec 1, 2014 at 1:40

1 Answer 1


Why are LEO satellites not aerodynamically shaped?

The need for electrical power overwhelms the need to reduce drag. That means a sizable cross sectional area to incoming solar radiation. Sometimes that cross section to solar radiation corresponds nicely (or not so nicely) with cross section to drag.

What's worse, it's hard to claim that any shape is "aerodynamic" in low Earth orbit. In the troposphere, the coefficient of drag of an streamlined object can be less than a tenth that of a spherical body, which in turn is about a quarter that of a well-designed parachute. On orbit, the standard value for the coefficient of drag is 2.2, regardless of shape. Keep in mind that parachutes typically have a drag coefficient of 1.75. Per this standard view, shape doesn't matter, and whatever shape an object has, it is less aerodynamic than a parachute. All that matters is cross sectional area.

A somewhat recent paper by Kenneth and Mildred Moe, Moe & Moe (2005), "Gas–surface interactions and satellite drag coefficients," Planetary and Space Science 53.8:793-801 puts some doubt to this standard drag coefficient of 2.2. Above 200 km altitude, most shapes have a drag coefficient that exceeds 2.2. Objects in space are not "aerodynamic".

Update: Why shape doesn't matter (at least not that much)
For objects moving through the troposphere, the shape of the object has a dramatic effect on drag. The coefficient of drag can vary by factor of forty depending on shape. Shape is much less important in the thermosphere. There, coefficient of drag varies by perhaps a factor of two instead of forty. Moreover, the very shapes that are considered to be aerodynamically-shaped in the troposphere can have a very high coefficient of drag in the thermosphere.

For example, a flat plate oriented normal to the wind flow has close to the worst shape an object can have in the troposphere with regard to drag. (A parachute is of course even worse.) An arrow with a nicely shaped arrowhead has a significantly smaller coefficient of drag than does a flat plate. The situation is reversed in the thermosphere. It's the flat plate that has a lower coefficient of drag.

The reason for this reversal is the way drag works in the lower atmosphere versus the upper atmosphere. The mean free path between collisions of atmospheric molecules is extremely short in the troposphere. On the other hand, mean free path ranges from about a quarter of a kilometer at 200 kilometers altitude to 2.6 kilometers at 300 kilometers to hundreds of kilometers at 600 kilometers altitude. This long mean free path means that drag operates very differently in the thermosphere than in the troposphere. Viscosity is the dominant force in the troposphere. Viscosity is essentially zero in the thermosphere.

Instead, drag in the thermosphere is described by free molecular flow. Atmospheric molecules in the thermosphere don't "know" about the existence of the object that is subject to drag unless they collide with it. The object is long gone by the time surrounding molecules do interact with the molecules that collided with the object.

  • $\begingroup$ Can you say why drag coefficients are so high when atmosphere is so thin? It seems like the fewer molecules there are, the greater effect the impact of each one has. $\endgroup$
    – kim holder
    Nov 29, 2014 at 15:03
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    $\begingroup$ @briligg Too high Reynolds number in exospheres. Basically, because viscosity is non-existent and the drag of non-streamlined bodies isn't dominated by the pressure component in the wake region at extremely high Reynolds numbers any more. And the other way around, streamlining doesn't help by creating a boundary layer effectively further extending this streamlining by local air pressure at higher viscosity atmospheric pressure. You have to be in extremely low LEO for streamlining to make much of a difference, but some satellites were/are streamlined, e.g. GOCE and Swarm trio (all ESA). $\endgroup$
    – TildalWave
    Nov 29, 2014 at 15:51
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    $\begingroup$ @TildalWave - I think you meant extremely high Reynolds numbers. Perhaps too simplistically, the Reynolds number is the ratio of inertial forces to drag forces. Reynolds number doesn't even make sense above the mesosphere because drag changes from that due to viscosity to that due to free molecular flow. $\endgroup$ Nov 29, 2014 at 18:31
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    $\begingroup$ @DavidHammen Yes, that's it, I always mess it up with this LOL Basically ${{\rho {\mathbf v} L} \over {\mu}}$ where $\mu \to 0$ (pretty much non-existent viscous forces). So yes, $\mathrm{Re} \to \infty$. Cheers for correction! I've abused mod powers and reflected that in initial comment. ;) $\endgroup$
    – TildalWave
    Nov 29, 2014 at 19:08
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    $\begingroup$ @Rob - An arrow (or any longish cylinder flying with the long axis parallel to the velocity vector) has two terms that lead to drag: a term proportional to the area of the cross section normal to the flow, plus another term proportional to the area of the cross section parallel to the flow. This second term can dominate for a sufficiently long cylinder. The second term is due to collisions along the cylinder wall. As the constant of proportionality for the first term varies by at most a factor of two in low earth orbit, arrows and other long skinny objects are not aerodynamic in space. $\endgroup$ Dec 17, 2019 at 2:50

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