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When a parachute slows the velocity of an object where does the energy go?

If it's a falling object the acceleration from gravity is roughly constant. How does air drag "dissipate" the extra energy?

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    $\begingroup$ Doesn't it just heat up and make the surrounding air turbulent? $\endgroup$
    – Geremia
    Oct 21 '14 at 2:25
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    $\begingroup$ Exactly. It warms things up. $\endgroup$ Oct 21 '14 at 2:27
  • $\begingroup$ Intuitively, that doesn't seem like enough. Can you provide a formula that shows enough energy is converted to heat to significantly slow the object down? $\endgroup$
    – CJ Dennis
    Oct 21 '14 at 2:31
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    $\begingroup$ You win again, entropy! $\endgroup$
    – Yves Klett
    Oct 21 '14 at 14:06
  • $\begingroup$ @YvesKlett I reckon Entropy has got to be Eddy Izzard's next Death Star Canteen character after Darth Vader and God. $\endgroup$ Nov 6 '14 at 6:35
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To add to HDE 226868's correct answer "heat":

  1. Even the ram drag component, arising when the parachute losslessly exchanges momentum with the relatively moving air and thus feels a Ram Pressure (see Wikipedia article of this name, and also my footnote), ends up as heat because the air eddies and currents airising from ram effect (see my footnote) then dissipate their energy by frictional drag with the surrounding air;
  2. A spacecraft re-entering the Earth's atmosphere gives a dramatic demonstration of this dissipation and this is why so much technology goes into heat shielding the spacecraft. The failure of this shielding lead to the destruction of the space shuttle Columbia in early 2003. Actually the main reason for the dramatic temperature rise here is a non dissipative (initially, at least) process like the ram pressure (see footnote).

Footnote on ram pressure: Ram drag is the (at first) non heating (non dissipative) drag exerted on a body as it losslessly exchanges linear momentum with the fluid it moves through. To understand the ram pressure, which arises particularly for supersonic objects, witness the object is just shoving fluid out of its way, and the latter flows off at some high angle to the trajectory. Think of a stationary object with a flat leading surface with a high speed flow around it. Fluid striking the flat surface gets deflected almost at right angles to the incoming flow. If you tally up the impulse per unit time that the object must be exerting on this flow to effect the change in fluid momentum, it is proportional to the flow rate (which, in turn, is proportional to the flow speed), and also proportional to the individual fluid particle momentum - also proportional to the flow speed. So the product of these two is proportional to $v^2$, where $v$ is the flow speed.

As noted by users GraphicsResearch and user121330, during a spacecraft re-entry, another, related, initially non dissipative mechanism arises. The simple ram pressure description above describes the fluid as though it were incompressible, and is a good model in water. But air can be compressed and, during re-entry is so adiabatically as the ram effect unfolds. As the spacecraft does work on the air, the latter's temperature rises - enormously. The main mechanism here seems to be non-dissipative, adiabatic work done by the spacecraft in compressing the air. As long as the spacecraft can withstand this temperature of the cushion of air squashed before it, the latter actually shields the spacecraft ensures that all the frictional dissipation (which must ultimately dissipate all the energy) happens well away from the spacecraft: it is frictional, viscous loss in the air after the spacecraft has passed through. Thanks to the cushion there is almost NO frictional drag directly between the spacecraft itself and the air. The spacecraft sits on its cushion and the flow almost stagnates in the neighbourhood of the spacecraft's surface. This is why a blunt object feels much less heat load than a streamlined one during re-entry; see the "Blunt Body Entry Vehicles" section in the "Atmospheric Entry" Wikipedia page. A highly amusing discussion of this effect is also the discussion of whether a steak can be cooked by dropping it through the atmosphere at Article 28 "Steak Drop" at What-if.xkcd.com

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  • $\begingroup$ Nice answer. As a space buff, I should have thought of the re-entry example. $\endgroup$
    – HDE 226868
    Oct 21 '14 at 2:42
  • $\begingroup$ Thank you. Although "heat" might be technically the correct answer (all waste energy eventually ends up as heat!) your answer about more energy being required to move the air out of the way (ram drag) makes more sense intuitively. $\endgroup$
    – CJ Dennis
    Oct 21 '14 at 2:47
  • $\begingroup$ How much does a parachute heat up in a typical skydive? $\endgroup$
    – AShelly
    Oct 21 '14 at 3:20
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    $\begingroup$ No. The high temperatures from reentry actually have very little to do with friction. Mostly, at these speeds, the air is compressed, which then releases heat adiabatically. The energy from spacecraft reentry is dissipated in the form of work compressing the atmosphere in front of it. This doesn't generate heat. The temperature increase is just a secondary effect. $\endgroup$
    – imallett
    Oct 21 '14 at 6:38
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    $\begingroup$ @GraphicsResearch see my last comment to user121330 - and be sure to see the Steak Drop article if you haven't already! $\endgroup$ Oct 24 '14 at 22:37
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Heat

Drag is the same thing as air resistance. It's a form of friction. Friction turns some of the kinetic energy of a moving object into heat; drag does the same thing, thus slowing a falling object down. When an object slows down due to friction, it heats up (and some of the heat dissipates to the surroundings). The same principle applies here: There will be some heating of the parachute and the air around it.

The drag equation governs the force of drag on an object. It is: $$F_D=\frac{1}{2}\rho v^2C_DA$$ where $F_D$ is the force of drag, $\rho$ is the fluid's mass density, $v$ is the relative velocity of the object (relative to the fluid), $C_D$ is the drag coefficient, and $A$ is the area. In the absence of other forces, you can use the definition of work to figure out the work done by the drag: $$W=Fs$$ $$W=\frac{1}{2}\rho v^2C_DAs$$

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