18
$\begingroup$

What was the velocity of the Command Module after it had penetrated through the Earth's atmosphere at the point where the parachutes were deployed?

$\endgroup$
  • 1
    $\begingroup$ Which parachutes? There were two drogues, followed by three pilot chutes, which pulled out the three main chutes. $\endgroup$ – Mark Adler Nov 4 '13 at 4:37
  • 1
    $\begingroup$ @Mark The wording of the question makes me believe that it's asking for when the drouges were deployed - it seems that they want the ones deployed right after it 'penetrated through the atmosphere'. $\endgroup$ – Undo Nov 4 '13 at 4:42
18
$\begingroup$

If you're asking about the deployment of the three main parachutes of C/M-ELS (Apollo Command Module Earth Landing System), then this is simple enough to answer, the pilot chutes are deployed at about 10,000 feet (3.05 km) by a barometric switch, pulling the three main parachutes from their containers. The ELS was designed so the drogue chutes slow the descent down to roughly 200 km/h (124 mi/h) before the pilot chutes pull the main chutes, eventually slowing down the CM to 22 miles per hour (35 km/h) for splashdown and to roughly 24.5 mi/h (39.5 km/h) with only two main chutes properly deployed, as it happened during the Apollo 15 splashdown.

    enter image description here

                  Earth Landing System sequence of events (Source: Project Apollo - NASSP)

For the drogue deployment though (thanks go to @MarkAdler in the comments!), we now have this diagram of the parachute deployment envelope:

    enter image description here

So in normal atmospheric entry (not launch abort), the diagram for manual deployment of the drogues describes the region in altitudes between 40,000 and 25,000 feet (12.2 - 7.6 km) and CM velocity between mach 0.7 and 0.3. Translating that to US standard atmosphere in 1962 figures, 0.7 mach at 40,000 feet equals roughly 206 m/s (743 km/h or 461 mi/h) and 0.3 mach at 25,000 feet equals roughly 94 m/s (338 km/h or 210 mi/h). That averages out at 32,500 ft (9.9 km) and velocity of 150 m/s (540 km/h or 336 mi/h).

The normal entry region for drogue chute deployment by barometric switch (such as was the case with Apollo 11) is described at altitudes between 25,000 and 20,000 feet (7.6 - 6.1 km) and CM velocity between mach 0.225 and 0.475 that come out at roughly 70.65 - 147.25 m/s (158 - 329 mi/h or 254 - 530 km/h). That averages out at 22,500 feet and 109 m/s (243.5 mi/h or 392 km/h).

$\endgroup$
9
$\begingroup$

The command module's speed is going to be terminal velocity — the speed at which the force exerted by aerodynamic drag equals the force exerted by gravity — because it entered the atmosphere going at a crazy high speed, and it's been falling for tens of thousands of meters already. So we know it's going to be terminal velocity.

But we can't stop there.

I was curious about this too, so I decided to run the numbers. The formula for terminal velocity is as follows:

$$V_T=\sqrt{\frac{2mg}{\rho AC_D}}$$

  • The mass of the command module, $m$, was 5809kg, from here.
  • The force of gravity on Earth, $g$ is right around $9.8~\textrm{m/s}^2$.
  • The drag coefficient, $C_D$, seems to be about 1.3 (judging from the graph on pg 48 of this NASA paper.
  • The projected area of the CM's heat shield, $A$, was $11.631~\textrm{m}^2$ ($125.2 ~\textrm{ft}^2$), according to page 9 of another NASA paper.

Here I'll assume you're talking about the drogue parachutes, the first ones out. They were deployed at 24,000 feet (7315.2m). The density of the air at that altitude is, according to Wolfram|Alpha, $0.57~\textrm{kg/m}^3$.

So now we have all our numbers. We just need to do the calculations — which I performed, getting a terminal velocity at that altitude of (drumroll please)… $103.71~\textrm{m/s}$ (or about 230 mph).

That is…

  • About the maximum speed of a swing by a professional golf player,
  • Almost as fast as a Ferrari F50 GT1 (about 2% slower),
  • And about a third as fast as sound in 15°C dry air @ 1 atmosphere: Mach 0.33 at that altitude.

Thanks, Wolfram|Alpha!

$\endgroup$
  • 5
    $\begingroup$ Actually, we can't be sure it's at terminal velocity. You have assumed that aerodynamic braking from reentry speed has been completed before parachute deployment. I think it's possible to deploy the drogue parachutes earlier. Haven't found a definitive answer though. $\endgroup$ – Hobbes Nov 4 '13 at 12:09
  • 2
    $\begingroup$ Since you are chasing the density (and to some extent the gravity) on the way down, you can never get to terminal velocity. However you get pretty close. For this example you get to and stay within 5% of terminal velocity below 25,000 ft. So the terminal velocity calculation is a good one. $\endgroup$ – Mark Adler Nov 4 '13 at 20:41

protected by Community Nov 24 '18 at 15:19

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.