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Why the acceleration of free-falling bodies become zero after some time? My only idea would be that the closer to the core of Earth we are, the value of "g" the lower becomes (since the mass below the body is lower and consequently, its gravitational force is also lower) but it works on large scales, like hundreds and thousands of kilometres. (and also, I'm not sure in it)

I don't count with air resistance, since an experiment with checking terminal velocity would work in vacuum too.

EDIT: I thought about enery aspects too, but obviously a body that falls with constantly increasing speed wouldn't be a perpetuum mobile, because once it's gonna reach with the larger body, and then speed becomes zero.

Zoltán Schmidt
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Terminal velocity exists because a velocity dependent force against gravity results in a net acceleration of 0.

In most cases, air resistance (drag force) is the velocity dependent force.

Out of curiosity, why does terminal velocity work in a vacuum too?

sihrc
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in the vacuum, and in absence of other frictional forces (electromagnetic, etc), you do not reach terminal velocity for linear motion. An asteroid, is always accelerated, same as a satelite orbiting around the earth. They can reach "constant" speed if they are in a circular motion, but the direction of the speed changes (it is a circular motion), due to the acceleration of gravity. So velocity, is not constant. If you could make a tunnel through the center of the earth and drop a ball, it will keep accelerating until it reaches maximum speed at the center, then it will keep going to the other side, this time decelerating, until it reaches zero speed at the other end of the tunnel and comes back, in an endless oscillatory motion

Akash
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I assume that you're asking a question about the fundamental physics involved in terminal velocity, then this is a good question.

Terminal velocity exists because at this point a falling object (e.g. rock) displaces a total mass of fluid (e.g. water or air) equal to their own mass, each second. This is similar to the concept of buoyancy for boats and balloons.

At terminal velocity:

Mass of Rock Falling = Total Mass of fluid displaced each second

which is the same as buoyancy for a boat floating in water:

Mass of Boat = Total Mass of fluid displaced each second

This is why the same object must fall a lot further through air than water, at terminal velocity for that fluid. i.e. The rock has to fall through a greater volume of air each second, before displacing the same total mass as compared to water, because air is a lot less dense than water (about 1/1000 times less dense). hence terminal velocity is a lot higher falling through air than water, as the rock falls further before achieving buoyancy. See image:

enter image description here

The 'total mass of fluid displaced' includes the fluid directly fallen through and the fluid indirectly displaced by the rock accelerating the fluid fallen through away from it.

Total mass of fluid displaced = Mass of fluid directly displaced + Mass of fluid indirectly displaced.

enter image description here

This approach allows you to calculate the downward force exerted by the rock on the air/water in terms of the mass of fluid ('m') fallen through and its acceleration ('a') i.e. Force = ma).

This is not how they explain it in school textbooks. They will blame things like viscosity of the fluid .... which is only part of the explanation. It misses the 'bigger picture' explanation of the masses involved.

Nick Landell
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