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According to Wikipedia:

[Terminal Velocity] is the velocity of the object when the sum of the drag force (Fd) and buoyancy equals the downward force of gravity (FG) acting on the object. Since the net force on the object is zero, the object has zero acceleration.

I was wondering if this can be more generalized. Assume that we replace the force of gravity with another type of force (for example, the force of a theoretical engine on a rocket whose fuel never loses mass). In this case, the force is not one of gravity (along the y-axis) but of the engine (along the x-axis). However, drag would still play a part in causing the acceleration due to the rocket's engine to reach 0.

As far as I'm concerned, these concepts are the same, but whenever one talks about "Terminal Velocity" they always discuss it in context of gravity. Can the term "Terminal Velocity" be used in the "rocket" case as described above, where the force is NOT that of gravity?

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    $\begingroup$ You could argue linguistics, but as long as the concept is the same, I don't see why not. $\endgroup$ – Pranav Hosangadi Nov 21 '13 at 22:59
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Yes, for example, an object with a constant applied force and an oppositely directed speed dependent force has a terminal speed.

The sum of forces is: $F_{NET} = F_A - kv$

Clearly, when the speed is large enough that $kv = F_A$, the net force is zero and the object stops accelerating - the object has reached terminal speed.

See this hyperphysics article:

When an object which is falling under the influence of gravity or subject to some other constant driving force is subject to a resistance or drag force which increases with velocity, it will ultimately reach a maximum velocity where the drag force equals the driving force. This final, constant velocity of motion is called a "terminal velocity", a terminology made popular by skydivers.

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  • $\begingroup$ I'm pretty sure I've heard the exact words "terminal velocity" to describe particle behaviour in electrophoresis experiments. Drift velocity in metals is another example where I've heard the term, although the derivation is a little different here: you balance the rate of energy gained by a carrier from the field with that lost through inelastic collisions. $\endgroup$ – WetSavannaAnimal Dec 22 '13 at 4:13

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