What is the terminal velocity of the Falcon 9 at sea level (on Earth)? Has SpaceX released enough information that this could be calculated with reasonable accuracy? Assume that the Falcon 9 is completely empty of fuel. It is a Falcon 9 Block 5. It has the landing gear used for a drone ship landing, and the landing legs are extended.
$\begingroup$
$\endgroup$
6
-
1$\begingroup$ Does space.stackexchange.com/questions/49075/… help? $\endgroup$– Ryan CCommented Aug 26 at 14:55
-
$\begingroup$ @phil1008 no, terminal velocities near zero are reserved for things like feathers and leaves. this is a large, heavy object, which means its terminal velocity is going to be high. if it hits the ground at its terminal velocity, it will be destroyed. $\endgroup$– Ryan CCommented Aug 26 at 14:59
-
3$\begingroup$ @phil1008 if that were true they could stop the Falcon 9 mid-air by deploying its landing legs and running it empty of propellant. Recovery operations of such a vehicle would be much less energetic, if not exactly simpler, because at that point it's a leaf on the wind and you just have to watch it soar until you can harpoon it. $\endgroup$– Erin AnneCommented Aug 26 at 15:37
-
2$\begingroup$ @phil1008 That's not how terminal velocity works... It assumes travelling through air at the average air pressure and density for that altitude. Otherwise I could say that it's terminal velocity at 5,900m is also 0 because Licancabur lake is that high up. $\endgroup$– user53400Commented Aug 27 at 10:37
-
2$\begingroup$ Upvoting because I don't think the publicly available information is sufficient (all the unpowered flight is supersonic or transonic) though I'd like to be proven wrong $\endgroup$– Erin AnneCommented Aug 27 at 16:27
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
4
Well, we can plug the numbers into the terminal velocity formula $v_t=\sqrt{\frac{2mg}{\rho AC}}$.
- The dry mass is apparently 22 tons (although Ars Technica floated 50 t at some point)
- The air density $\rho$ at seal level is about 1.23 kg/m3
- The drag coefficient C for a long cylinder is 0.83
- The projected surface area of the cylinder front end with a diameter of 3.7 m is 10.75 m2.
With these numbers we get, ignoring legs and fins
$$v_t=\sqrt{\frac{2*22000*9.81}{1.23*10.75*0.82}}\\ \approx \sqrt{\frac{431,600}{10.8}}\\ \approx \sqrt{39,800}\\ \approx 200 m/s\ (720 km/h, 447 mph)$$.
In an earlier version of this answer I assumed a very small surface area of the landing legs; seeing now that a large surface area between the struts is covered with a white rigid sheet material, their contribution to the "projected area" A is much larger. Let's assume A=30m2, resulting in a terminal velocity of 119 m/s (429 km/h, 266 mph).
This may not be far off: Shortly before the landing burn, the deceleration slows markedly which indicates that the booster gets closer to its terminal velocity.
The main weaknesses of these calculations are:
- The projected surface area of the landing legs is just a wild guess, and of course they have a different drag coefficient, as do the grid fins (thanks, Mark).
- Because the vehicle is light, residual fuel has a significant impact. There must be at least a minimal reserve; ignoring it systematically underestimates the terminal velocity.
- The booster's base is not a smooth plane; the generic drag coefficient for a cylinder underestimates the real drag.
answered Sep 16 at 3:53
-
$\begingroup$ You can't use the "long cylinder" drag coefficient for the landing legs. They're probably somewhere between "flat plate" (1.17) and "open hemisphere" (0.38). There's also the issue of turbulent flow around various components, which greatly increases the drag. $\endgroup$– MarkCommented Sep 16 at 22:15
-
$\begingroup$ @Mark I meant to state that, got lost in the editing -- the "effective" area (which I only guesstimate based on pictures) when I consider fins and legs is meant to take the different coefficient into account. For the booster cylinder without legs it's also not perfect because the front end is not smooth. $\endgroup$ Commented Sep 16 at 23:21
-
$\begingroup$ "The projected surface area of the landing legs is just a wild guess" :( this answer could be off by quite a lot from both A and C being rough guesses. $\endgroup$ Commented Sep 17 at 1:05
-
$\begingroup$ @ErinAnne Yes, of course. This answer was not meant to produce a robust number but to "plug" plausible values into the relevant formula (which, I think, nobody had done so far here) to show how one would in principle go about it. The resulting number is within the expected order of magnitude which I found encouraging. $\endgroup$ Commented Sep 17 at 8:48