Arguing with physics teacher. Is there a terminal velocity(I know its a speed) ?

[StudentJoseph]

This has to do with an object falling down in a linear line threw the air towards earth.

I was arguing with my physics teacher today on whether or not terminal velocity was true. First, I understand terminal velocity refers to speed and therefore is without direction. My argument was not based on that. Its based on the idea that the equation for terminal velocity has acceleration due to gravity and that acceleration due to gravity is constantly changing as you fall( Distance or D^2 ). So the rate of acceleration changes in the formula even if its by a decimal that is extremely small. Meaning the whole equation constantly changes.

So is there a final speed for a falling object or is it a constantly changing speed and terminal velocity refers to the point in which an object's falling speed is changing so little that its not appreciable ?

 

[brno17]

I think its an assumption that gravity stays constant throughout. If not the change is SO small that it is negligable

 

[PhanthomJay]

Terminal velocity has both magnitude (speed) and direction (down in your case). Gravity can be assumed constant over the duration of the fall, since any change in g during this period is not significant. Nevertheless, an object never actually reaches terminal velocity, but it does reach about 99% of terminal velocity in a relatively short period of time. If the object, after that relatively short period when it reaches 99 % of terminal velocity , is still falling , because it was, say, dropped or thrown from a tall building, it will continue to approach the terminal velocity, reaching .995, .996, .997...etc. terminal velocity, but never actually reaching it, if that's your question. It stems from the non linear differential equation describing its motion.

 

[StudentJoseph]

PhantomHay , What I am hearing from your explination is the theory of terminal velocity is correct in the sense that an object does have an eventual terminal velocity. The required height for an object would have to be vary large for it to reach terminal velocity?

If that's true I do not understand why. If we were to have a very large height acceleration due to gravity would be less than 9.8m/s^2. It would be say 9.77 m/s ^ 2 ( Acceleration at the top of mount Everest.) however as you fall the acceleration due to gravity continues to change. There is no final acceleration do to gravity. Even the number 9.8 m/s does not account for surface level gravity all over the world because the thickness of the Earth's crust varies.

My argument was that there is an acceleration no matter how infinitesimally small it is. That would be the velocity changes by an infinitesimally small amount but there is still a change of velocity making "terminal velocity" or end velocity a misnomer. Terminal velocity is then the point at which measuring the velocity is not important in determining the final speed because the objects speed is something like 98.00000000000( and so on with zeros )01 m/s.Brno17 , Gm1m2/d^2. D being distance and m being mass. G is the gravitational constant.

P.S.
"non linear differential equation describing its motion"
I'm in high-school and have yet to take calculus. I'm still in algebra II so I'm not certain what that phrase means or that my understanding is not complete because I have not taken that course.

 

[QuantumPion]

The increase in terminal velocity due to the increase in gravitational force would be dwarfed by the decrease in terminal velocity due to the increasing density of air. In reality you would be constantly falling slightly faster than the terminal velocity.

 

[AndyNewton]

Hi Joseph. Congratulations on your probing analysis of terminal velocity. I hope your physics teacher was impressed.

Here is something else to think about that can help you resolve many technical "arguments" in a positive way, including the current one...

The physics theory regarding terminal velocity is not the same thing as a real object falling through the atmosphere of the real earth. The theory is a separate entity. It is a mathematical and mental construction. The theory exists as a "perfect" self consistent piece of thought.

The theory describes something that can never actually exist - that includes: an entirely uniform gravitational field - with air at a constant density at all heights - and a force law (F proportional to speed squared?) that is obeyed exactly ... and so forth. Within the "theory" though, all these things can be assumed to be true. And then we can explore what happens within the theoretical world we have created - and discover interesting phenomena such as a terminal velocity. That is all good, interesting, exciting thinking.

But back in the real world ... having understood the "theory" is not the real world ... we can now ask to what extent the lovely "perfect" theory of terminal velocity matches what actually happens in the real world. That question can be discussed in depth, as you have started to do yourself and other replies above have also addressed.

This critical discussion of the closeness of the match between a theory and the real world is good, and the impossible but perfect "theory" is also good - as long as we don't get confused by thinking that the theory is supposed to be the same thing as the real world.

All theries are perfect - but some are more perfect than others. :-)

 

[PhanthomJay]

PhantomHay , What I am hearing from your explination is the theory of terminal velocity is correct in the sense that an object does have an eventual terminal velocity.

No, it never reaches a specific terminal velocity, but it comes very very close to it, but as a high schooler with the (of course) absence of knowledge of the differential calculus involved, for all practical purposes, terminal velocity is reached, as long as the height of release is high enough for it to do so

The required height for an object would have to be vary large for it to reach terminal velocity?

Not necessarily. A feather or coffee filter will reach terminal velocity in a very very short distance, but a heavy small ball (like a bowling ball) would have to be dropped from a much higher distance before it reaches terminal velocity.

If that's true I do not understand why. If we were to have a very large height acceleration due to gravity would be less than 9.8m/s^2. It would be say 9.77 m/s ^ 2 ( Acceleration at the top of mount Everest.) however as you fall the acceleration due to gravity continues to change. There is no final acceleration do to gravity. Even the number 9.8 m/s does not account for surface level gravity all over the world because the thickness of the Earth's crust varies.

we're not talking dropping the thing from Mt Everest, maybe say from a 100 foot tall (30 meter) building. Here's a nice calculator, just ignore the math behind it, or you'll get really confused. But it's fun to play with.

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/fallq.html

My argument was that there is an acceleration no matter how infinitesimally small it is. That would be the velocity changes by an infinitesimally small amount but there is still a change of velocity making "terminal velocity" or end velocity a misnomer. Terminal velocity is then the point at which measuring the velocity is not important in determining the final speed because the objects speed is something like 98.00000000000( and so on with zeros )01 m/s.


Brno17 , Gm1m2/d^2. D being distance and m being mass. G is the gravitational constant.

P.S.
"non linear differential equation describing its motion"
I'm in high-school and have yet to take calculus. I'm still in algebra II so I'm not certain what that phrase means or that my understanding is not complete because I have not taken that course.

Yeah, that's sort of right , the object will never have 0 acceleration (constant speed), it will accelerate downward initially at 9.8 m/s/s, then slow due to air drag to 5, 4 3, etc m/s/s, and ultimatly within typically just a few seconds like less than 10 seconds, it's acceleration will be about 0.001 m/s, and falling to near zero acceleration as it travels further and further. Nice questions for a high schooler!

 

[StudentJoseph]

I appreciate all the help I've received with this question. I often feel as if my teacher is too analytical and because of this has a hard time communicating with us. I don't blame my conflicting understanding of the theory as her fault because I myself ( while reading andy's post) realized I had not explained it in the best way. She is still my favorite teacher because of the depth she goes into when offering an explanation some times she can be a little too black or white (Darn mathematicians and your logic :tongue2:).

Thank you phantom for sticking with my question and explaining it to me ( and the calculator ). Id also like to thank andy for helping me separate the hypothetical from reality . I think i had over lapped the two and created some sort of pseudo reality. In the future I think ill look at the problems in a better way.

Also quantumpion i appreciate your explanation but I have another question. Are you saying that as we get closer to the Earth there are more gas particles or are you saying that as the object falls it hits more gas particles ( due to the amount of distance traveled. )? ( in reference to density )I Look forward to more days in physics because its quite a lot of fun. Every explanation raises more questions and every question that has an explanation raises more questions.

 

[russ_watters]

When teaching to high school students, teachers must often choose to simplify subjects in order to move forward. If you spend too much time discussing what is wrong with the terminal velocity equation, you'll never have a chance to actually use it. As you progress with your education, you'll find much if not most of what you learn about science in high school has been similarly simplified. But you'll also learn to discard the complexities when appropriate so they don't prevent you from ever finding an answer to a problem.

 

[_Tully]

If you spend too much time discussing what is wrong with the terminal velocity equation, you'll never have a chance to actually use it.

"But Teach, all of these variables are traveling through space at a gajillion parsecs per nanoyear*..."
*not the actual speed of Earth.

 

[QuantumPion]

Also quantumpion i appreciate your explanation but I have another question. Are you saying that as we get closer to the Earth there are more gas particles or are you saying that as the object falls it hits more gas particles ( due to the amount of distance traveled. )? ( in reference to density )

The air gets thinner the higher you go in altitude. This is why jet airplanes fly at 30,000 feet, there is less air resistance and so less fuel is required to maintain a constant speed.

From the wikipedia article about terminal velocity:

Air density increases with decreasing altitude, ca. 1% per 80 metres (262 ft) (see barometric formula). For objects falling through the atmosphere, for every 160 metres (525 ft) of falling, the terminal velocity decreases 1%. After reaching the local terminal velocity, while continuing the fall, speed decreases to change with the local terminal velocity.

So as I said, if you are falling from a high altitude e.g. skydiving, you will be constantly decelerating as the air gets thicker as you decrease in altitude. This effect is far greater than the change in gravitational force, which is negligible, unless you are free falling from a REALLY high altitude, like geosynchronous orbit. And in that case, you wouldn't have a terminal velocity anyway since there is no atmosphere to speak of at that altitude. By the time you hit the atmosphere, you would be close enough to the Earth where there would no longer be much more change in gravity.

 

[brainstorm]

My impression was that for skydivers drag from air-resistance reaches a point where it prevents further acceleration of the person. I assumed that this would have the effect of neutralizing the effect of weightlessness from free-fall in that it would be like lying on a bed of air (while the ground approaches at high speed).

Surely meteors that burn up as shooting-stars have a "terminal velocity" if only because they evaporate before accelerating to their maximum potential speed.

 

[litup]

My impression was that for skydivers drag from air-resistance reaches a point where it prevents further acceleration of the person. I assumed that this would have the effect of neutralizing the effect of weightlessness from free-fall in that it would be like lying on a bed of air (while the ground approaches at high speed).

Surely meteors that burn up as shooting-stars have a "terminal velocity" if only because they evaporate before accelerating to their maximum potential speed.

Meteors can have widely varying incoming velocities, some having to catch up with Earth and some come slamming in headfirst, Earth's orbital velocity added to the incoming velocity rather than subtracting from it, so some meteors can hit the ground with way higher than terminal velocity.

Think about how meteors hit the moon, no atmosphere to muck things up:)
if the velocities add up to a larger # the damage caused is a lot greater, depth of the crater, and amount of ejecta widely varies because of that difference in velocity for two meteors of the exact same mass.

If a mile wide meteor comes down to Earth like the one 65 million years ago, the atmosphere would not have much effect on its terminal velocity because our atmosphere would just not be able to greatly effect it's incredible kinetic energy.

 

[Dr Lots-o'watts]

Terminal velocity for a human being falling horizontally through air is the speed of the air that is coming out from this machine (wind tunnel):
http://www.skyventuremontreal.com/?lang=en

 

[Born2bwire]

PhantomHay , What I am hearing from your explination is the theory of terminal velocity is correct in the sense that an object does have an eventual terminal velocity. The required height for an object would have to be vary large for it to reach terminal velocity?

If that's true I do not understand why. If we were to have a very large height acceleration due to gravity would be less than 9.8m/s^2. It would be say 9.77 m/s ^ 2 ( Acceleration at the top of mount Everest.) however as you fall the acceleration due to gravity continues to change. There is no final acceleration do to gravity. Even the number 9.8 m/s does not account for surface level gravity all over the world because the thickness of the Earth's crust varies.

My argument was that there is an acceleration no matter how infinitesimally small it is. That would be the velocity changes by an infinitesimally small amount but there is still a change of velocity making "terminal velocity" or end velocity a misnomer. Terminal velocity is then the point at which measuring the velocity is not important in determining the final speed because the objects speed is something like 98.00000000000( and so on with zeros )01 m/s.


Brno17 , Gm1m2/d^2. D being distance and m being mass. G is the gravitational constant.

P.S.
"non linear differential equation describing its motion"
I'm in high-school and have yet to take calculus. I'm still in algebra II so I'm not certain what that phrase means or that my understanding is not complete because I have not taken that course.

One of the basic skills in physics is the ability to make appropriate approximations. As you go through your studies, you'll be surprised about the level of assumptions and approximations that physicists make even on very difficult and complex theories. It really has to come down to the fact that most of the exact rules of physics make a problem far too complex to easily solve. Instead, we find that an appropriate set of assumptions, approximations, and constraints can greatly simplify the problem with only causing a minimal deviation in the solution. So maintaining a high level of pedantry is going to cause problems like this. If you want, you should sit down and work out an actual case yourself. Use the actual definition for the force of gravity and an appropriate drag force and find the equations of motion. Then, go back and use a constant force of gravity and find the equations of motion. Take a look at the velocities and compare and also see how they compare to the idea of a terminal velocity. You should find that the results differ negligably. Of course, this also requires that you know calculus, which you have stated you do not, and that should also indicate to you the value of such an approximation. That is, it's now accessible to a high school student where before it would require a higher level of mathematical knowledge to work out.