Terminal velocity in a vacuum?

[Orodruin]

Which would conflict with "starting at rest".

This was never the main point. The main point is that the velocity at impact is going to depend on the initial conditions (it could be higher or lower than the escape velocity) while the accepted definition of terminal velocity does not. I would rather tend to agree with this:

c is a closer analogue of limiting velocity, though without the feature that it varies with the body (terminal velocity depends on size, shape and density of the object).

However, as PAllen mentions, the limiting mechanism is quite different in the two cases - force equilibrium in the case of terminal velocity and relativistic effects (without a counter force) in the case of c.

 

[sweet springs]

But you cannot reach escape velocity exactly if starting from rest unless you started infinitely far away and infinitely long ago. So it seems to me that the similarity remains -- albeit reversed in time.

Hi. We already succeeded the NASA space probe Voyager get more than escape velocity of the solar system. Voyager will get its terminal velocity at infinite future and infinite far away. Are you worrying about these infinities?

 

[jbriggs444]

I do not have a problem with those infinities. Limits work.

It had seemed to me that A.T. in post #45 was trying to draw a distinction between a limiting final behavior and an exact final behavior. He pointed to "escape velocity" measured as the final velocity of an object impacting on a planetary surface as a case of the latter. My point (more formally expressed) was that this this result is still a based on taking a limit. One is taking the limit as the initial separation increases without bound. So it is still the case that elapsed time increases without bound.

 

[Dryson]

Vacuum is space that is devoid of matter.

Gravity is gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter.

If an object is placed into a infinitely long vacuum tube that is devoid of matter where gravity would not be present based upon the attraction between all matter not being present wouldn't the object, when dropped, not achieve any type of terminal velocity because of gravity not being present to pull the object to a terminal velocity? Wouldn't the object need to have a force placed on it in order to achieve any type of velocity?

 

[phinds]

Vacuum is space that is devoid of matter.

Gravity is gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter.

If an object is placed into a infinitely long vacuum tube that is devoid of matter where gravity would not be present based upon the attraction between all matter not being present wouldn't the object, when dropped, not achieve any type of terminal velocity because of gravity not being present to pull the object to a terminal velocity? Wouldn't the object need to have a force placed on it in order to achieve any type of velocity?

You have introduced a set of conditions different than the ones being considered in this thread. If you want to start a different discussion, you should open a new thread. Or I am I misunderstanding you and you believe that you ARE describing the conditions being discussed in this thread?

 

[Drakkith]

If an object is placed into a infinitely long vacuum tube that is devoid of matter where gravity would not be present based upon the attraction between all matter not being present wouldn't the object, when dropped, not achieve any type of terminal velocity because of gravity not being present to pull the object to a terminal velocity?

There doesn't need to be matter inside the tube. You could place one end of the tube on the Earth's surface and extend the other end to infinity and gravity would indeed affect the object inside the tube. (Obviously it is impossible to extend a real vacuum tube to infinity, but the example helps us understand how gravity works)

 

[Dryson]

So basically you could use the vacuum tube to determine the strongest source of gravity then as the object inside of the vacuum tube would be attracted to the strongest source of gravity, correct?

 

[phinds]

So basically you could use the vacuum tube to determine the strongest source of gravity then as the object inside of the vacuum tube would be attracted to the strongest source of gravity, correct?

In terms of gravity, an object ANYWHERE will be attracted to the strongest source of gravity. I don't understand what that has to do with this thread, which started out as a discussion of an object in a vacuum tube being attracted to a an Earth-like body, with no consideration for other gravitationally attractive objects, which would just complicate the situation and derail the discussion.

 

[Drakkith]

So basically you could use the vacuum tube to determine the strongest source of gravity then as the object inside of the vacuum tube would be attracted to the strongest source of gravity, correct?

No, the object will be attracted to ALL sources of gravity, not just the strongest.

 

[phinds]

No, the object will be attracted to ALL sources of gravity, not just the strongest.

Right. My statement was rather sloppy. I should have said it would be attracted the MOST to the strongest gravity source, not leaving open the implication that it wouldn't be attracted to others.

 

[HallsofIvy]

If an object is dropped in a hypothetical infinitely long vacuum tube, will it reach a terminal velocity? I assume that it must because according to Einstein, no object that has mass can travel at the speed of light. My guess would be that the terminal velocity of an object in a vacuum would depend on its mass. I suggest this because I imagine some parabolic graph to denote the effect of mass on terminal velocity within a vacuum; not just a simple "does it have mass or not". I am just a junior in high school and have no great knowledge of relativity, but I post this to simply gain knowledge that I was unable to acquire from my school as my physics teacher disregarded my question as "too advanced for the class to comprehend".

This was the original post (succeeding posts have gotten away form it). Part of the confusion is that an object is "dropped" into an infinitely long vacuum tube but nothing is said about any force on the object. In air, an object moving under a force will accelerate until the friction force with the air will equal the force causing the object to accelerate, then continue at 'terminal velocity'. With no friction but some force, there will be no terminal velocity but will be a "bounding velocity". c, the speed of light is the bounding velocity but is not a "terminal velocity" because the object itself cannot reach c- its speed increases toward c as an upper bound.

 

[anorlunda]

The OP is a junior in high school. Must we be so unforgiving about him using scientifically precise language phrasing his question? I do not want to drive away students from PF by seemingly hostile answers.

All that the student needed was a simple answer and perhaps a suggestion for how to better phrase his question in a scientifically correct way.

 

[BobG]

If an object is dropped in a hypothetical infinitely long vacuum tube, will it reach a terminal velocity? I assume that it must because according to Einstein, no object that has mass can travel at the speed of light. My guess would be that the terminal velocity of an object in a vacuum would depend on its mass. I suggest this because I imagine some parabolic graph to denote the effect of mass on terminal velocity within a vacuum; not just a simple "does it have mass or not". I am just a junior in high school and have no great knowledge of relativity, but I post this to simply gain knowledge that I was unable to acquire from my school as my physics teacher disregarded my question as "too advanced for the class to comprehend".

(bolding mine)

The question isn't "perfectly" phrased, which I think has led people off on a tangent that doesn't address what I think (admitting there is some room for seeing a different main point) is the key question asked in this post:
"My guess would be that the terminal velocity of an object in a vacuum would depend on its mass."

No, that isn't correct. Yes, "terminal velocity" isn't used correctly, but I don't think that's the point. The maximum velocity the object will reach does not depend on the mass of the falling object. The maximum velocity depends on the mass of the object it's falling towards and on how far away the object was when it was "dropped" (which I assume to mean starting from rest).

And, yes, if you're talking about all possible celestial bodies the object could be falling towards, the maximum maximum velocity the object could reach would be the speed of light, but I don't think that was the point of the post.

 

[TumblingDice]

This was the original post (succeeding posts have gotten away form it).

Ageed - kudos for your efforts to get the thread on (any) track!

...there will be no terminal velocity but will be a "bounding velocity". c, the speed of light is the bounding velocity but is not a "terminal velocity" because the object itself cannot reach c- its speed increases toward c as an upper bound.

Not wanting to go off on a tangent, but adding these comments because "terminal velocity" was referred to in the OP:

I argued the 'terminal' side years ago with a friend who took a jump for his birthday. He said "terminal velocity" was a misnomer because it could never be reached. It wasn't until I discovered PF that my exploration to learn about thread topics lead me to discover that "terminal velocity" is one of those labels that, unfortunately, is intuitively misleading. Maximum velocity is reached asymptotically, so I think it can also be viewed as an "upper bound" just like c, rather than a final value.

It isn't easy to find this detail with web searches - unfortunately most explanations adhere to the 'terminal' aspect. <sigh> I hope I'm choosing the right leaders to follow! Here's a clip to illustrate:

Using the standard equations of motion and assuming that the air resistance force is proportional to the velocity squared then you can solve for the velocity and distance. There are two parameters in the solution in addition to V0: the characteristic time, T0 = V0/g = 5.6 sec, and the characteristic distance, X0 = V0T0 = 315 m.
The full solution is V = V0 tanh(T/T0) and X = X0 log( cosh(T/T0) ) .
Notice that V only approaches V0 asymptotically, it never really gets there.

 

[TumblingDice]

No, that isn't correct. Yes, "terminal velocity" isn't used correctly, but I don't think that's the point. The maximum velocity the object will reach does not depend on the mass of the falling object. The maximum velocity depends on the mass of the object it's falling towards and on how far away the object was when it was "dropped" (which I assume to mean starting from rest).

No, that isn't correct. Gravity depends on all mass. e.g., The mass of the Earth attracts you and your mass attracts the Earth. Just because one is larger doesn't mean it's the only contribution to the system.

 

[BobG]

No, that isn't correct. Gravity depends on all mass. e.g., The mass of the Earth attracts you and your mass attracts the Earth. Just because one is larger doesn't mean it's the only contribution to the system.

The force of gravity depends on both masses. Acceleration depends on the mass of the object you're accelerating towards (Force = mass * acceleration).

 

[TumblingDice]

The force of gravity depends on both masses. Acceleration depends on the mass of the object you're accelerating towards (Force = mass * acceleration).

Well you've got Force on the LHS and acceleration sitting there on the RHS. If the force of gravity increases, doesn't that imply acceleration will, too? That's how I was looking at this, from the standpoint of Newton's Law of Universal Gravitation and: F = G (m1m2/r)

Just asking, wouldn't an object the size of the moon, for instance, collide with the Earth more rapidly than a golf ball released from the same distance? (Don't need to involve CoM - I'm willing to give the golf ball a head start.)

 

[jbriggs444]

$F = Gm_1m_2/r^2$ is only half of the story. $F = m_1a$ is the other half. Solve for a.

 

[TumblingDice]

Solve for a.

Here it comes... Wait for it... ... !

"I see," said the blind man.

Apologies @BobG, and thanks too, to both you and @jbriggs444

 

[mfb]

Just asking, wouldn't an object the size of the moon, for instance, collide with the Earth more rapidly than a golf ball released from the same distance?

Yes, but just because Earth would gain more speed.

This thread left the original topic very long ago, and then went in circles. Please open a new thread with a clear topic, if you think something needs more discussion.