If an object loses mgh when travelling at terminal velocity, where's the energy now?

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  • #1
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This is a simple problem.

Picture a ball dropping from great height such that it reaches terminal velocity.
Its KE will not increase anymore, but mgh is still being lost. That energy has to go somewhere.

1) My friend and I believe the energy is lost to heating the surrounding fluid (what a convenient answer. Always blaming friction or heating...). Do you guys think this is correct?

2) Assuming the above is correct, then I have an extended problem. An object travelling at terminal velocity v will cause heating to the surrounding, but another object (which has a higher terminal velocity threshold) passing through velocity v,v+Δv, etc. will see a region of 'cooler' fluid around it because the energy goes to KE instead of heating?

It seems non-intuitive that a faster object will cause less heating of the surrounding fluid.

Hope you can understand what I mean.
Any comments?
 

Answers and Replies

  • #2
boneh3ad
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Yes the culprit here is heat. Drag is a dissipative process, do it does account for your missing energy.

As for your second question, the amount of heat generated in the air in the way you describe is minimal. Most aerodynamic heating problems, such as on a re-entering spacecraft, are do to more complicated processes like shocks.
 
  • #3
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thanks for the answer.
but there's no shock here!
just terminal velocity...
 
  • #4
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Let us analyze your statements in a little more detail. I took the liberty of editing your post so that only the important sentences are retained:

...
A) Assuming the above (...the energy is lost to heating the surrounding fluid...[my addition from above]) is correct, then, an object traveling at terminal velocity v will cause heating to the surrounding.

B) Another object (which has a higher terminal velocity threshold) passing through velocity v,v+Δv, etc. will see a region of 'cooler' fluid around it, because the energy goes to KE instead of heating?

C) A faster object will cause less heating of the surrounding fluid.
...
First of all, you seem claim that if an object has a larger terminal velocity, then a larger part of the potential energy decrease ([itex]m \, g \, \vert \Delta h \vert[/itex]) should go to the change of kinetic energy, then to heating.

This is wrong! When an object reaches terminal velocity, its velocity remains just that - terminal, i.e. it does not change. If the velocity does not change, then the kinetic energy does not change. Therefore, ALL of the potential energy loss goes to heating.

Then, you made a change of terms. In sentence B), the blue text says one thing, but then you change the term in sentence C) (green text). This is a logical fallacy. Therefore, sentence C does not follow from the previous two statements.

Your problem stems from the fact that you do not understand terminal velocity thoroughly, and that you performed at least one logical fallacy.
 
  • #5
boneh3ad
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thanks for the answer.
but there's no shock here!
just terminal velocity...
I never said there was a shock. I said that when you see the intense heating such as on a re-entry vehicle, that is a result of things like shocks.

In the case you described, the change in temperature of the air will be nearly negligible, especially since the object is moving. Also keep in mind that the air is taking energy from the object even before it reaches terminal velocity so the lighter object doesn't really have a head start. Furthermore, even though it takes longer for the heavier object to reach terminal velocity, since m is larger, that means each [itex]\Delta h[/itex] it falls transfers more energy to the air than the same [itex]\Delta h[/itex] by the lighter object, so the heavier object (with higher terminal velocity) is actually heating the air more anyway. It will still be a negligible amount of heat though.
 
  • #6
rcgldr
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My friend and I believe the energy is lost to heating.
A lot of the energy goes into KE of the air. The air affected by the falling object is accelerated downwards to some "exit velocity", which is the velocity of the affected air at the moment it's pressure returns to ambient. I'm not sure how much of the power of a falling object goes into heat versus KE of the air.

another object (which has a higher terminal velocity threshold) ...
Assume "another object" is the same size but with more weight (higher density), so the force, m g is higher for this object. The power from gravity at any moment in time at some velocity v is m g v = m g Δh / Δt. The power and drag related to the air is the same for both objects if the velocity is the same, but at terminal velocity for the "original object", the "other object" is receivng more power and force from gravity due to it's greater weight, so it continues to accelerate.
 
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