Terminal velocity in a vacuum?

[TumblingDice]

And just to clarify my scenario, the gravitational force reacting upon the accelerating object is constant throughout the entire tube.

The setup you're describing has some issues that lead to more issues. oo) It sounds like the target of your OP is about the speed limit of c in a scenario that provides constant acceleration (no atmosphere to introduce terminal velocity).

Jimmy's example of the rocket in post #25 is a bit more straight-forward to picture and discuss. The constant acceleration of the rocket parallels the constant acceleration you're looking at with gravity. Is this at all where you were trying to go with the OP? If so, that's where relativity can come into explain how you can keep accelerating in your frame of reference, never reach a 'terminal velocity', and yet, you will alway measure light to travel at c.

 

[PAllen]

I agree with Jimmy. Terminal velocity is typically used to refer to the limiting velocity when drag and accelerating force cancel exactly. If an object is falling faster it will slow down, if it is falling slower it will speed up. This is not the case with escape velocity, if I give the object an initial velocity it will have a velocity which is larger than the escape velocity when (and if) it hits the surface.

This is the correct answer. Further, the only limit to such velocity directly observed by some static observer for an object moving in a vacuum is c, and this limit has nothing to do with force balance as is normally understood for terminal velocity.

 

[Simon Bridge]

What sort of answer we can give depends on how we read the intent behind the original question.

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.

@kubaanglin: "terminal velocity" is a technical term in physics, and I suspect that you are not using it in technical sense.

I'm guessing that you are asking if a falling mass has a definite final speed less than the speed of light.
i.e. if massless objects travel at the speed of light, then perhaps massive objects have a slower "top speed".

That guess would seem to be borne out by:

My guess would be that the terminal velocity of an object in a vacuum would depend on its mass.

... that and your education level.

The thing to understand about the speed of light is that this is the "invarient speed" - it is the same to all observers. It is the only speed which has this property. Light must travel at this speed, and may not travel at any other speed, because it has no mass. However, any massive object may get arbitrarily close to it - in principle. The graph of a massive object's speed vs time does not flatten out like it does for an object falling with air resistance.

As others have pointed out, an object falling, from rest, under gravity can never exceed it's escape velocity ... so if you are actually looking for a limiting speed rather than a terminal one, then that is it. There are some wrinkles due to having to talk in relativity-speak though - i.e. we have to say who is measuring the speed.

But we'd really appreciate some feedback from you so we can be sure we are answering the right question. The trouble with lots of education and experience, some of which are the same thing, is that it makes you aware of many more possibilities so it's sometimes hard to be sure what someone with less of both is trying to say. Help us out eh?
Has any of this been of any use.

 

[BobG]

I now understand that there is no truly confirmed answer to my question as the scenario I described is completely hypothetical and will never exist in the real world to be observed and analysed. I definitely have a better understanding of the rules and possibilities concerning my question, but understand that an evidence-based answer is impossible.

And just to clarify my scenario, the gravitational force reacting upon the accelerating object is constant throughout the entire tube.

Part of the problem is that you need a certain level of understanding to even pose the question correctly.

Russ is correct - specifically because acceleration due to gravity depends on the distance between the two objects (force of gravity and acceleration due to gravity is inversely proportional to the square of the distance).

There is no "universal" maximum escape velocity (except your escape velocity can't be greater than the speed of light). The force of gravity is also directly proportional to the product of the two masses involved. Force also equals mass times acceleration. Once you divide out the mass of the object you're observing, acceleration is directly proportional to the mass of the object you're accelerating towards.

You have two variables - mass of the object you're accelerating towards and the distance you are from the object you're falling towards. Escape velocity depends on both of those variables. But the maximum velocity an object could achieve (assuming no atmosphere) would be escape velocity at the surface of the black hole/star/planet/etc the object is falling towards.

(Once below the surface, the situation gets a little more complicated since you have some mass above you and some mass below you.)

Escape velocity:
$$v = \sqrt{\frac{2GM}{r}}$$

G = gravitational constant
M = mass of body you're falling towards
r = distance from body you're falling towards

 

[PAllen]

But as olorudin pointed out, you don't have to start out at rest. Then, even for a small planet, the maximum velocity is limited only by c. This is distinct from terminal velocity in that you slow down to terminal velocity if you start out faster (and the atmosphere is thick enough).

 

[sweet springs]

Hi.

A vacuum tube is set at a diameter of the Earth. A ball is dropped in the tube at surface. The ball keeps falling and reach the maximum speed at the center then decreases its speed and stops at the opposite Earth surface. The ball then start falling back thus keeps vibrating motion forever. There is no terminal velocity.

You pop up a ball in the vacuum tube of infinite length from surface then it reaches maximum height and starts falling. But in case we increase pop up speed to the escape velocity, the ball keep　moving out and stops at infinite distance. In this sense terminal velocity realized at infinite distance is written as
pop up initial velocity - escape velocity > 0

How is it? Relativistic cases, for example falling into BH or velocity change by expansion of Universe, have other interesting features but not the topics here.

 

[Doug Huffman]

QED Better a good error than a bad question.

 

[anorlunda]

The answer is deceptively simple. As the OP phrased it, the length of the tube is arbitrarily long. Impact with a surface can be arbitrarily far away and the elapsed time is arbitrarily large also.

The terminal velocity (as defined by the OP, meaning speed tapering off in some roughly parabolic curve approaching terminal velocity asymtotically) is the speed of light. One thing the OP left out was the reference frame. He should have said speed and time observed from his own reference frame at rest.

Good for you kubaanglin. That is a very astute question from a junior, and you phrased it well. You deserve an A for the question.

 

[sweet springs]

Hi.

Linear accelerator consists of a vacuum tube and surrounding power supply. Produced electromagnetic field accelerate charged particles in the tube. Accelelator needs more and more power supply to accelerate heavier particles to velocity of light. We do not have theoretical (not practical or technical ) limit in power supply. In this sense maximum accelerated speed do not rely on mass and is velocity of light c as special theory of relativity (SR) says.

How is it?

 

[A.T.]

Further, the only limit to such velocity directly observed by some static observer for an object moving in a vacuum is c, and this limit has nothing to do with force balance as is normally understood for terminal velocity.

It is somewhat different, but the c limit seems to me much more like terminal velocity than escape velocity. You approach both, the c limit and terminal velocity, without ever reaching them, no matter how long you continue to fall and accelerate. Escape velocity, on the other hand, is what you actually do reach, when the accelerating fall suddenly ends.

 

[jbriggs444]

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.

 

[PAllen]

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.

No, you can trivially exceed escape velocity by starting from some modest initial speed at some finite height. This is completely different from terminal velocity, where you would still reach the same constant speed for a wide range of initial conditions. 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). Also, of course, terminal velocity has the feature that you can start faster and then you slow down to it.

 

[jbriggs444]

No, you can trivially exceed escape velocity by starting from some modest initial speed

Which would conflict with "starting at rest".

 

[A.T.]

Which would conflict with "starting at rest".

The OP seems mainly interested in the ultimate speed limit that you will approach regardless of how you vary the initial conditions.

 

[jbriggs444]

It seemed to me that you were trying to draw a distinction between asymptotically approaching a speed and and exactly hitting a speed. That's a fine line. And a distinction that is spoiled if it depends on exact and unobtainable initial conditions.

 

[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.