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.

  • 2
    $\begingroup$ What makes you think "an experiment with checking terminal velocity would work in vacuum too"? Or have you managed to drill a hole to the centre of the earth and sustained a vacuum in such a hole? $\endgroup$ – Henry Jul 29 '13 at 18:44
  • $\begingroup$ You may be interested by the paragraph b) of this answer $\endgroup$ – Trimok Jul 29 '13 at 18:47
  • $\begingroup$ @Henry check my comment below the answer of sihrc. $\endgroup$ – Zoltán Schmidt Jul 29 '13 at 18:53

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?

  • $\begingroup$ Honestly, I'm not a physicist, but I'm enthuastic at it, and with my logic, if it wouldn't work in vacuum, that'd mean that the speed of asteroids would accelate constantly, and their speed may even come near to the speed of light. But because of the terminal velocity, they can't approach this level of speed. The question is that what is the velocity dependent force that results in a net acceleration of 0, if it's not the air resistance? $\endgroup$ – Zoltán Schmidt Jul 29 '13 at 18:52
  • $\begingroup$ +1 for the answer. And it works in vacuum, because if the thing is pulled down, at one point it make Boom! :) $\endgroup$ – Nikolaj-K Jul 29 '13 at 18:54
  • $\begingroup$ @ZoltánSchmidt In a vacuum, there is little acceleration to pull the asteroid to begin with. You wouldn't need a resistive force for "terminal velocity" because there isn't acceleration (very little due to gravity over great distances). In this scenario, the acceleration is little, if not none, so the velocity is constant (rather than terminal). If it came close enough to a planet (earth) for gravitational force to "change" the velocity, then it will accelerate more and keep going until BOOM. $\endgroup$ – sihrc Jul 29 '13 at 18:58
  • $\begingroup$ But if the gravitational field of the Sun (among enourmous amount of other stars) has so strong gravity field that it can make giant planets orbit around itself, why it can't make either asteroids of these planets accelerate? $\endgroup$ – Zoltán Schmidt Jul 29 '13 at 19:13
  • $\begingroup$ @ZoltánSchmidt Large masses do accelerate comets. If there is only one significantly large mass near it, the comet will accelerate towards that mass and probably crash/burn. Comets don't accelerate indefinitely because there are also large masses pulling them in other directions. And as julian frenandez posted, it is also possible that they reach circular motion around large masses. $\endgroup$ – sihrc Jul 29 '13 at 19:33

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


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