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?


1 Answer 1


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.

  • $\begingroup$ 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, 2017 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, 2017 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, 2017 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, 2017 at 12:00

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