Problem with UA

If the Earth is moving up to meet a skydiver, and is constantly accelerating, then how does Terminal Velocity work? Genuine question, not spam.

Short answer: Equivalence principle.
Long answer: Lurk moar.

Short answer: Equivalence principle.
Long answer: Lurk moar.

How do you get terminal velocity under constant acceleration? I'm just curious.

How do you get terminal velocity under constant acceleration? I'm just curious.

My answer to this is no different to what it was when the OP asked it, variations in wording notwithstanding.

Quote from: TheJackel on October 08, 2010, 09:00:51 PM

How do you get terminal velocity under constant acceleration? I'm just curious.

My answer to this is no different to what it was when the OP asked it, variations in wording notwithstanding.

It's a valid question Parsifal.. under UA there would be no such thing as a Terminal velocity. It seems the equivalence is subject to being finite or confined to only certain aspects. Hence a pull vs push when falling seems testable as a sensation within a vacuum correct? Hence, they aren't exactly equivalent.

So again, how does something reach terminal velocity under UA?

Quote from: TheJackel on October 08, 2010, 10:10:49 PM

It's a valid question Parsifal.. under UA there would be no such thing as a Terminal velocity.

Please justify this statement.

Quote from: TheJackel on October 08, 2010, 10:10:49 PM

It seems the equivalence is subject to being finite or confined to only certain aspects. Hence a pull vs push when falling seems testable as a sensation within a vacuum correct? Hence, they aren't exactly equivalent.

What the fuck are you even trying to say here?

Quote from: TheJackel on October 08, 2010, 10:10:49 PM

So again, how does something reach terminal velocity under UA?

Once again, my answer has not changed. This is not the first or the second or the dozenth time this question has been asked; the answer to it, which is patently obvious to anyone with an elementary understanding of physics, has been restated many times over on these forums by people who give lurkless noobs more time than they're worth. You can try searching for it, you can hope someone else comes along who feels like babysitting you, but don't expect me to play physics tutor.

Parsifal, You really like to deflect from answering a point blank question. However, I will answer it for you. In a vacuum there is no terminal velocity in either case as it is determined by resistance of air or of a fluid from which an object passes through. Watching you go into a tizzy over this was quite interesting Not every noob understands the equivalence principle. You spent more energy defending yourself than if you were to have better explained your answer in a way that most people will find easier to grasp. I don't think you realized that I wasn't disagreeing with your position.

However, I made a mistake UA Terminal velocity is possible, but not in a vacuum. A failed sentence lol Ahh well "/

Now in regards to a push vs pull.. This is where the equivalence ends. Hence, if built a room and turned it into a vacuum and performed various experiments walking, jumping, or falling off ledges, or chairs you could possibly detect the difference of the pull of gravity vs the push of acceleration. More specifically in falling or walking off a ledge. There should be some kind of experiment that can detect or at least examine push vs pull. Yes or no? And I am only discussing this out of curiosity. 

Parsifal, You really like to deflect from answering a point blank question. However, I will answer it for you. In a vacuum there is no terminal velocity in either case as it is determined by resistance of air or of a fluid from which an object passes through. Watching you go into a tizzy over this was quite interesting Not every noob understands the equivalence principle. You spent more energy defending yourself than if you were to have better explained your answer in a way that most people will find easier to grasp. I don't think you realized that I wasn't disagreeing with your position.

You stated that there would be no such thing as terminal velocity under UA. Since my position is that there would be, you have disagreed with my position.

Now in regards to a push vs pull.. This is where the equivalence ends. Hence, if built a room and turned it into a vacuum and performed various experiments walking, jumping, or falling off ledges, or chairs you could possible detect the difference of the pull of gravity vs the push of acceleration. More specifically in falling or walking off a ledge. There should be some kind of experiment that can detect or at least examine push vs pull. Yes or no? And I am only discussing this out of curiosity.

No, there is not. That is why they call it the Equivalence Principle and not the Sort Of Similar Principle.

Quote from: TheJackel on October 08, 2010, 11:56:54 PM

Parsifal, You really like to deflect from answering a point blank question. However, I will answer it for you. In a vacuum there is no terminal velocity in either case as it is determined by resistance of air or of a fluid from which an object passes through. Watching you go into a tizzy over this was quite interesting Not every noob understands the equivalence principle. You spent more energy defending yourself than if you were to have better explained your answer in a way that most people will find easier to grasp. I don't think you realized that I wasn't disagreeing with your position.

You stated that there would be no such thing as terminal velocity under UA. Since my position is that there would be, you have disagreed with my position.

Quote from: TheJackel on October 08, 2010, 11:56:54 PM

Now in regards to a push vs pull.. This is where the equivalence ends. Hence, if built a room and turned it into a vacuum and performed various experiments walking, jumping, or falling off ledges, or chairs you could possible detect the difference of the pull of gravity vs the push of acceleration. More specifically in falling or walking off a ledge. There should be some kind of experiment that can detect or at least examine push vs pull. Yes or no? And I am only discussing this out of curiosity.

No, there is not. That is why they call it the Equivalence Principle and not the Sort Of Similar Principle.

Push is not equivalent to pull.. They are opposite forces. Thus I think there is possibly a means to determine how such equivalent force is being applied. I can tell when I am being pulled vs pushed. I should possibly be able to within a vacuum.


However,

You can weigh yourself or an object at the northern hemisphere, southern hemisphere, and equator to figure out why UA is false since under UA the objects wouldn't vary in weight. objects at the Equator in RE would weigh less than they would at either the southern or northern Hemispheres.. This would not be true under the UA since it would have to be uniform across it's entire surface area. And the reason why this is true on RE is because objects at the Equator are furthest from Earths center of gravity.

So the equivalence principle doesn't really apply in regards to FE vs RE. And that's because they are completely incompatible in this regard since RE is a rotating spheroid.. FE would have to match equivalence for every location on Earth to the RE model. Unfortunately for FE, that's not possible

So on RE, an object should way about 3.4% (estimated) less than it would at the poles. any object co-rotating with the Earth at the equator has its measured weight reduced by 0.34 percent, thanks to the Earth's shape and rotation.

http://en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_effect
http://en.wikipedia.org/wiki/Earth%27s_gravity#Latitude

This is the actual flaw in the UA and why you can not apply the equivalence argument. In fact this pretty much kills UA all together.


On the other thing, I had corrected myself.

Push is not equivalent to pull.. They are opposite forces. Thus I think there is possibly a means to determine how such equivalent force is being applied. I can tell when I am being pulled vs pushed. I should possibly be able to within a vacuum.

This is incorrect, and I don't feel like tutoring you on the subject. Take a physics course or even just read up on Wikipedia if you want to understand why acceleration and gravitation are equivalent.

However,

You can weigh yourself or an object at the northern hemisphere, southern hemisphere, and equator to figure out why UA is false since under UA the objects wouldn't vary in weight. objects at the Equator in RE would way less than they would at either the southern or northern Hemispheres.. This would not be true under the UA since it would have to be uniform across it's entire surface area. And the reason why this is true on RE is because objects at the Equator are furthest from Earths center of gravity.

So the equivalence principle doesn't really apply in regards to FE vs RE. And that's because they are completely incompatible in this regard since RE is a rotating spheroid.. FE would have to match equivalence for every location on Earth to the RE model. Unfortunately for FE, that's not possible

So on RE, an object should way about 3.4% (estimated) less than it would at the poles. any object co-rotating with the Earth at the equator has its measured weight reduced by 0.34 percent, thanks to the Earth's shape and rotation.

http://en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_effect
http://en.wikipedia.org/wiki/Earth%27s_gravity#Latitude

This is the actual flaw in the UA and why you can not apply the equivalence argument. In fact this pretty much kills UA all together.

This is a valid point, it has been responded to, it does not kill UA and it is off topic.

Quote from: TheJackel on October 09, 2010, 12:19:13 AM

Push is not equivalent to pull.. They are opposite forces. Thus I think there is possibly a means to determine how such equivalent force is being applied. I can tell when I am being pulled vs pushed. I should possibly be able to within a vacuum.

This is incorrect, and I don't feel like tutoring you on the subject. Take a physics course or even just read up on Wikipedia if you want to understand why acceleration and gravitation are equivalent.

Quote from: TheJackel on October 09, 2010, 12:19:13 AM

However,

You can weigh yourself or an object at the northern hemisphere, southern hemisphere, and equator to figure out why UA is false since under UA the objects wouldn't vary in weight. objects at the Equator in RE would way less than they would at either the southern or northern Hemispheres.. This would not be true under the UA since it would have to be uniform across it's entire surface area. And the reason why this is true on RE is because objects at the Equator are furthest from Earths center of gravity.

So the equivalence principle doesn't really apply in regards to FE vs RE. And that's because they are completely incompatible in this regard since RE is a rotating spheroid.. FE would have to match equivalence for every location on Earth to the RE model. Unfortunately for FE, that's not possible

So on RE, an object should way about 3.4% (estimated) less than it would at the poles. any object co-rotating with the Earth at the equator has its measured weight reduced by 0.34 percent, thanks to the Earth's shape and rotation.

http://en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_effect
http://en.wikipedia.org/wiki/Earth%27s_gravity#Latitude

This is the actual flaw in the UA and why you can not apply the equivalence argument. In fact this pretty much kills UA all together.

This is a valid point, it has been responded to, it does not kill UA and it is off topic.

Actually it's not off topic. And yes, it does kill UA. Do tell us how UA allows for various parts of Earth to accelerate upwards at different rates. Sorry, but the model can only work on a Round Earth to which rotates. But you can feel free to link me to where they explain a flat Earth uniformly accelerating. And then provide me the backup that supports it. Sorry but this will not be equivalent to RE. Your argument is like trying to state that coordinates can be equal between a flat circle and a Sphere. They are incompatible.  Your argument makes no sense here Parsifal :/

This would require you to a best show FE accelerating faster at the northern and southern hemispheres than it would at the equator. That would be impossible. It would rip the Planet apart. There is now way you can argue that it does. And we will just forget for a moment that pendulums exist because that would be unfair right? Nah!

Actually it's not off topic. And yes, it does kill UA. Do tell us how UA allows for various parts of Earth to accelerate upwards at different rates. Sorry, but the model can only work on a Round Earth to which rotates. But you can feel free to link me to where they explain a flat Earth uniformly accelerating. Your argument makes no sense here Parsifal :/

Make a new thread if you'd like to discuss this. This thread is about terminal velocity, not variation in g.

Quote from: TheJackel on October 09, 2010, 12:57:31 AM

Actually it's not off topic. And yes, it does kill UA. Do tell us how UA allows for various parts of Earth to accelerate upwards at different rates. Sorry, but the model can only work on a Round Earth to which rotates. But you can feel free to link me to where they explain a flat Earth uniformly accelerating. Your argument makes no sense here Parsifal :/

Make a new thread if you'd like to discuss this. This thread is about terminal velocity, not variation in g.

That would effect terminal velocity. Hence, it's on topic. Terminal velocity would equally be different due to this very fact. So there is no need to open a new thread. Your equivalence problem fails because of this. Since when did Fer's think RE would have equivalence with FE? Tsk Tsk Tsk.. :/

That would effect terminal velocity. Hence, it's on topic. Terminal velocity would equally be different due to this very fact. So there is no need to open a new thread. Your equivalence problem fails because of this. Since when did Fer's think RE would have equivalence with FE? Tsk Tsk Tsk.. :/

I am not going to discuss this with you here because it is unrelated to the question the OP asked. Make another thread, and I will gladly talk with you about it.

Parsifal, if he opened a thread you would just not answer to it. And you wouldn't re-appear in this one either, we all know you.

TheJackel has beated you to pieces, deal with it.

Parsifal, if he opened a thread you would just not answer to it. And you wouldn't re-appear in this one either, we all know you.

TheJackel has beated you to pieces, deal with it.

Ah... actually Parsafil is more right than wrong on this point. But he could explain it a few sentences rather than being a troll:

The EP provides that the difference between a uniform acceleration, g, and a uniform gravity field with acceleration, g, cannot be distinguished locally. Since we can and do detect that the Earth gravity field is not uniform, EP does not apply here.

If we ignore that fact that the UA is false, it can be argued, though only academically, that the upwards force of the column of air forced up by the Earth forced up by the UA could balance against the psuedoforce pulling the object down and thus create a terminal velocity.

we can and do detect that the Earth gravity field is not uniform

Please justify this statement.

Quote from: ClockTower on October 09, 2010, 02:02:15 AM

we can and do detect that the Earth gravity field is not uniform

Please justify this statement.

I already have in the appropriately titled thread. Search More

Quote from: ClockTower on October 09, 2010, 02:02:15 AM

we can and do detect that the Earth gravity field is not uniform

Please justify this statement.

That's why things weigh differently at the poles vs the equator. Do learn to keep up.. That's non-local comparison you are going to have to deal with. That effects Terminal velocity when the object weighs less at the equator or when acceleration is less due to the distance from its gravitational center, shape of the Earth, and how it rotates. And this is also why P-waves, and S-waves show a spherical Earth as well.. You can feel free to visit the Earthquakes thread for a good schooling on that subject as well. UA simply can not account for this, so it will have to simply attempt to ignore it.

So please justify your statement to support a non-uniform acceleration of a FE all the way to the north pole and southern edge expanding from the equator in order for your equivalence argument to be in equivalent to the RE model. This is in regards to the OP btw.

That's why things weigh differently at the poles vs the equator.

This is just a restatement of what CrackToter said. I asked for justification.

@Parsifal: how about you answer questions instead of complaining about how they were posted. Also, I just realized how it would work. If air is moving up with the Earth, and we must assume it is, then it moving against the skydiver would make him/her move up at a certain speed.

@Parsifal: how about you answer questions instead of complaining about how they were posted.

How about you don't bother us with questions that have been asked hundreds of times before? We aren't here to cater to your whims, although if you'd like to pay us money that may change.

@Parsifal: how about you answer questions instead of complaining about how they were posted. Also, I just realized how it would work. If air is moving up with the Earth, and we must assume it is, then it moving against the skydiver would make him/her move up at a certain speed.

Correct. I don't see how TheJackle didn't understand this given the amount of time he has been here. In RE you are being pull through air, slowing you down, and on FE air is being pushed into you slowing the amount of time it takes for Earth to rise up to you.

Quote from: doyh on October 09, 2010, 02:15:22 PM

@Parsifal: how about you answer questions instead of complaining about how they were posted. Also, I just realized how it would work. If air is moving up with the Earth, and we must assume it is, then it moving against the skydiver would make him/her move up at a certain speed.

Correct. I don't see how TheJackle didn't understand this given the amount of time he has been here. In RE you are being pull through air, slowing you down, and on FE air is being pushed into you slowing the amount of time it takes for Earth to rise up to you.

I never argued otherwise.. Your problem is that acceleration is not uniform across the entire globe due to it's rotation and the fact that it's a sphere. FE can't explain this problem because FE is trying to say that it's equivalent when it's not. Hence, FE would have to be equivalent to every coordinate location on Earth. UA has a weight problem it can't deal with without ripping itself apart.

http://en.wikipedia.org/wiki/Gravity_of_Earth

I never argued otherwise.. Your problem is that acceleration is not uniform across the entire globe due to it's rotation and the fact that it's a sphere. FE can't explain this problem because FE is trying to say that it's equivalent when it's not. Hence, FE would have to be equivalent to every coordinate location on Earth. UA has a weight problem it can't deal with without ripping itself apart.

http://en.wikipedia.org/wiki/Gravity_of_Earth

Again, this is irrelevant to the question asked by the OP. Stop derailing the thread and discuss it elsewhere.

Quote from: TheJackel on October 09, 2010, 07:27:37 PM

I never argued otherwise.. Your problem is that acceleration is not uniform across the entire globe due to it's rotation and the fact that it's a sphere. FE can't explain this problem because FE is trying to say that it's equivalent when it's not. Hence, FE would have to be equivalent to every coordinate location on Earth. UA has a weight problem it can't deal with without ripping itself apart.

http://en.wikipedia.org/wiki/Gravity_of_Earth

Again, this is irrelevant to the question asked by the OP. Stop derailing the thread and discuss it elsewhere.

it's not irrelevant to the OP.. It states that terminal velocity is a problem with UA.. In this sense it most certainly is! I simply expanded the OP's premise to account for the differences between RE and FE.  If you can't handle that Parsifal, then please leave the discussion.

So since things weigh differently by 3.4 % at the poles vs the equator, this becomes a problem for UA. Do note the OP's title "Problem for UA".. In order for UA to make any sense Parsifal it's going to have to match up to the real world. To bad it doesn't.

it's not irrelevant to the OP.. It states that terminal velocity is a problem with UA.. In this sense it most certainly is! I simply expanded the OP's premise to account for the differences between RE and FE.  If you can't handle that Parsifal, then please leave the discussion.

The OP asks how terminal velocity would work under UA. This is irrelevant to whether or not terminal velocity varies over the surface of the Earth -- indeed, terminal velocity is also influenced far more strongly by various other factors, so the differences with location may not even be detectable. Just because you can relate your point to terminal velocity doesn't make it automatically relevant to anything else that might be said about terminal velocity.

If you post about this again in this thread, I will report you.

Quote from: TheJackel on October 09, 2010, 07:49:24 PM

it's not irrelevant to the OP.. It states that terminal velocity is a problem with UA.. In this sense it most certainly is! I simply expanded the OP's premise to account for the differences between RE and FE.  If you can't handle that Parsifal, then please leave the discussion.

The OP asks how terminal velocity would work under UA. This is irrelevant to whether or not terminal velocity varies over the surface of the Earth -- indeed, terminal velocity is also influenced far more strongly by various other factors, so the differences with location may not even be detectable. Just because you can relate your point to terminal velocity doesn't make it automatically relevant to anything else that might be said about terminal velocity.

If you post about this again in this thread, I will report you.

It says "Problem for UA" in the title.. Do try harder in order to understand why expanding on this subject is on topic. Crying a river over this just means you can't handle the subject. And yes, other factors can and do play a part..Your problem is that weight also plays a part. It's entirely relevant btw since there are differences in gravity at different latitudes. This automatically rules UA out of the picture on this subject. Which brings us to also realize the equilateral bulge. You can find all kinds of data on Earths gravity not being uniform by Googling "Gravimeter Data".. UA would be uniform across the entire board and I have yet to see anything show it as such.

This is also a fun read:

http://rst.gsfc.nasa.gov/Intro/Part2_1b.html


And btw, you could very well detect the differences in a controlled environment such as a non-vacuum tube, or even a vacuum tube. Measurements can then be taken to detect the differences at different latitudes. Even in a vacuum tube there would be a difference on acceleration between the North pole, South Pole, and the Equator. The acceleration due to gravity is thus slightly greater at the poles. It is ~9.832 ms^-2 at the North Pole and ~ 9.782 near the equator. So yes Parsifal the effect is noticeable because the force of gravitation is inversely proportional to the radius squared.

Quote from: Parsifal on October 09, 2010, 07:50:55 PM

Quote from: TheJackel on October 09, 2010, 07:49:24 PM

it's not irrelevant to the OP.. It states that terminal velocity is a problem with UA.. In this sense it most certainly is! I simply expanded the OP's premise to account for the differences between RE and FE.  If you can't handle that Parsifal, then please leave the discussion.

The OP asks how terminal velocity would work under UA. This is irrelevant to whether or not terminal velocity varies over the surface of the Earth -- indeed, terminal velocity is also influenced far more strongly by various other factors, so the differences with location may not even be detectable. Just because you can relate your point to terminal velocity doesn't make it automatically relevant to anything else that might be said about terminal velocity.

If you post about this again in this thread, I will report you.

It says "Problem for UA" in the title. Therefore I can post whatever I want as long as I put UA in it somewhere.

Since the OP says, "skydiver", lets talk about skydiving! Amiright?

Just because you can relate your point to terminal velocity doesn't make it automatically relevant to anything else that might be said about terminal velocity.

Quote from: TheJackel on October 09, 2010, 08:19:08 PM

Quote from: Parsifal on October 09, 2010, 07:50:55 PM

Quote from: TheJackel on October 09, 2010, 07:49:24 PM

it's not irrelevant to the OP.. It states that terminal velocity is a problem with UA.. In this sense it most certainly is! I simply expanded the OP's premise to account for the differences between RE and FE.  If you can't handle that Parsifal, then please leave the discussion.

The OP asks how terminal velocity would work under UA. This is irrelevant to whether or not terminal velocity varies over the surface of the Earth -- indeed, terminal velocity is also influenced far more strongly by various other factors, so the differences with location may not even be detectable. Just because you can relate your point to terminal velocity doesn't make it automatically relevant to anything else that might be said about terminal velocity.

If you post about this again in this thread, I will report you.

It says "Problem for UA" in the title. Therefore I can post whatever I want as long as I put UA in it somewhere.

Since the OP says, "skydiver", lets talk about skydiving! Amiright?

Quote

Just because you can relate your point to terminal velocity doesn't make it automatically relevant to anything else that might be said about terminal velocity.

All of this will effect the skydiver. If you assumed identical factors in resistance you can determine the difference. All you need to know is if objects accelerate at different rates at different latitudes. This can be done with vacuum tubes. And I do believe he used a skydiver as an example object to where any object is applicable to the discussion on terminal velocity.

And yes he's correct that other factors could make it undetectable..So that's kind of why we wouldn't use a skydiver to determine this vs object being dropped in a vacuum tube or even in a non-vacuum tube. UA would have to provide uniform acceleration at all locations. Unfortunately this is a problem for UA.

Quote from: doyh on October 09, 2010, 02:15:22 PM

@Parsifal: how about you answer questions instead of complaining about how they were posted.

How about you don't bother us with questions that have been asked hundreds of times before? We aren't here to cater to your whims, although if you'd like to pay us money that may change.

I wasn't specifically speaking of this question. I've been lurking here a bit, and I've found that you never answer questions, you just complain about how they're worded. Why?