I know that if an object moves in the air, it can experience two types of drag, laminar and turbulent. For instance, I have a meteor of ideal spherical shape falling from edge of space, say, 100 km up at the Earth surface with initial velocity of 1 km per second. I would consider turbulent drag, but is it still applicable at supersonic velocities? How do I estimate meteors velocity when it hits the ground?

  • $\begingroup$ Yes, it works better the faster it moves, the effect of the turbulence is to erase the long distance correlations in the flow velocity in the fluid. The faster the object moves, the less the back reaction of the fluid matters, when the object arrives at some spot, the fluid "doesn't know" that it was coming. So, the approximation to just assume that right until the object arrives the air molecules have not yet been affected and they will then bounce off the object is going to work better and better at higher velocities. $\endgroup$ Sep 17, 2015 at 16:45
  • $\begingroup$ So, you mean, it's just turbulent drag assumption and that's it? $\endgroup$
    – Dankevich
    Sep 17, 2015 at 17:25
  • $\begingroup$ Yes, the drag formula expresses the resistance as being proportional the cross section and the square of the velocity, which is due to intercepting the flux of molecules scooped up by the meteor and letting the bounce off it. Per unit time the moment transferred to the air. And if the asteroid slams into the Earth, if that happens at velocities much faster than the speed of sound of the solid ground, the ground will behave as a fluid and you can still use that formula. $\endgroup$ Sep 17, 2015 at 17:53

1 Answer 1


The drag force, $F$, on the meteor is given by:

$$ F = C_d \frac{1}{2}\rho v^2 A$$

where $\rho$ is the fluid mass density, $v$ is the meteor speed, $A$ is the meteor cross-sectional area, and $C_d$ is the drag coefficient (discussed below).

The fluid mechanics of meteor reentry is quite complicated. Over the important ranges of altitude, the meteor is supersonic and a bow shock forms in front of the meteor. Consequently, the flow adjacent to the meteor is subsonic. The flow field is therefore non-trivial.

For a quick answer, though, meteor trajectories can be calculated by assuming $C_d=0.7$. If you want to be more accurate, a plot of $C_d$ vs far-field mach number can be found in this paper: "ESTIMATING THE DRAG COEFFICIENTS OF METEORITES FOR ALL MACH NUMBER REGIMES" by R. T. Carter, P. S. Jandir, and M. E. Kress:

enter image description here

From the plot, you can see that the rule-of-thumb value of 0.7 underestimates the drag at high speed and overestimates at low speed.

You will also need to know atmospheric density vs altitude. A NASA report from 1976 defines the standard values for this and is available online here. There are also online web calculators, such as this one, that are based on the same report. (For data above 86 km, see this report.)

  • 1
    $\begingroup$ I have been looking for that raw atmospheric data for years. Thank you. $\endgroup$
    – tyblu
    Jul 7, 2017 at 14:32
  • $\begingroup$ Just to confirm the numbers: that means that 1m^2 body at 800m/s through the atmosphere at sea level would roughly have 0.5*0.9*1.225*640000*1=352800N drag force? So for 100Kg, 350Gs of (de)acceleration? $\endgroup$ Oct 16, 2020 at 13:59

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