2
$\begingroup$

In this answer I've estimated the peak rate of energy loss of a particularly unwise re-entry as 3 to 4 gigawatts. Never mind that it's 15-20 gees of acceleration, that's a lot of power injected into the gas and plasma directly in front of the heat-shield.

My guess is that for survivable capsule re-entries with existing heat shields, the power generation is way lower than that, even on a log scale.

Are there any ballpark, order of magnitude-type numbers out there for the rate of energy loss to the atmosphere in manned capsule reentries?

$\endgroup$
2
$\begingroup$

Here is the descent profile for Soyuz.

At 08:53:30 the speed is 7.62km/s and touchdown is at 09:14:39.

Over 1269 seconds the object sheds 7.62 km/s. Kinetic energy is $.5mv^2$. So that's 29,032,200 joules per kilogram.

29,032,200 joules/1269 seconds = 22878 watts. Over that 21 minute interval I get about 23 kilowatts per kilogram.

According to the descent profile, Soyuz descent maximum g load is around 4 g's.

I am trying to find the descent profiles of the Apollo capsules. They would enter the earth's atmosphere at almost 11 km/s. But so far I haven't been able to find descent profiles that give altitudes and speeds at different times.

In the scenario you link to I get max speed of 4.52 km/s at about 71 km altitude. Impact is 117 seconds later. For this I get 87 kilowatts per kilogram. More than triple of the Soyuz capsule. This is with a 1.85 meter radius, 6500 kilograms and a drag coefficient of .5

Increasing the radius to 2.9 meters, max speed of 4.5 km/s is reached at about 79 km altitude. Impact is 185 seconds later. Over that 185 seconds the capsule endures 54 kilowatts per kilogram. More than double the Soyuz.

$\endgroup$
  • $\begingroup$ I think $dE/dt$ can be calculated directly from the spacecraft mass and deceleration in g's, right? All you need is the spacecraft's velocity at the moment of maximum g-force. Isn't the power just $P=Fv=mav$ where $a=4g$? $\endgroup$ – uhoh Sep 2 '17 at 15:09
  • $\begingroup$ The majority of the kinetic energy loss happens in a narrow time window, somewhere between tens of seconds and a few minutes depending on the vehicle, anywhere between 20 and 200 seconds. This time is so short that heat is not distributed uniformly, so kW/kg is not a relevant concept. Capsules that have survivably re-entered from orbit with several humans on board are all about the same size, (around the 2 to 2.5 meters diameter ballpark) so let's just talk about the peak power heating them; maximum of mass x deceleration x velocity. $\endgroup$ – uhoh Sep 2 '17 at 18:11
  • 2
    $\begingroup$ Definitely don't count the parachute descent towards the duration. Since 08:53:30 +03:01:00 -0:21:09 Entry Guidance enabled (80.4km, 7.62km/s) ) until parachute opening (09:00:18 +03:07:48 -0:14:31 Parachute Opening (10.8km, 217m/s)) you have 462 seconds, and average power output of 0.18 gigawatt. If you narrow it down to neighborhood of peak g-load, you'll likely exceed a gigawatt for a short time. $\endgroup$ – SF. Sep 5 '17 at 11:57
  • 1
    $\begingroup$ Also, for the Space Shuttle profile, during the blackout you're getting about 1.7 GW. $\endgroup$ – SF. Sep 5 '17 at 12:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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