Split Thread The validity of classical physics (split from: DWFTTW)

[humber]

The key phrase is "directed toward the center of the planet": as long as we remain Earth-centred, we choose Earth as "the planet". But which planet do we choose when we are away from Earth? In particular, can you answer my previous question: what is the "real", "true" or "actual" weight of the Earth?

One for Councilor Troy, I think.

By definition there is no net force required to make an object accelerate at the rate of free fall.

Remove gravity during free-fall, and acceleration stops. It is a necessary cause.

"Key point: The gravitational force on each piece of this ‘person’ is the same, but lower parts are compressed more. In my model of a person, gravity is “sensed” by how body is compressed."

That can hardly be true (and it is) if there is no force.

I am not saying that "weight" and "apparent weight" are the same thing. We can always choose a frame where our defined weight is the same as our apparent weight, but we don't have to. If we take another frame, accelerated with respect to the first frame, our weight will be something else. That is the essential point. For instance, if we are on the surface of the Earth, we usually take the local frame defined by the surface we're standing on.

Then it is superfluous to do otherwise.

That means that our apparent weight, as measured by a simple spring balance, is the same as the weight we define with respect to the frame of reference of the Earth's surface.

The reaction from the ground is a 'spring balance'

We could stay standing on the Earth and define our weight by the local free fall frame (accelerating with respect to the frame of the surface of the earth), in which case our weight is zero. Of course, usually we don't, and the "standard" frame of reference is the surface of the Earth. This is just like measuring the speed of a car running on the surface of the Earth.

Yes.

If we say that a car is travelling at 100 km/h, usually we don't need to specify "with respect to the surface of the Earth": it's understood.

Yes.

For the person in a space station in orbit around the earth, it's no different: we may either choose to measure their weight in the freely-falling reference frame, or in one at a fixed distance from the surface of the earth. In the first case we conclude that they are "weightless", in the second case we conclude that they have weight. Both answers are right, as long as the frame of reference is made clear. It's like measuring the velocity of that same space station: it won't be the same if we measure it from a frame of reference rotating at the same speed as the Earth and from a frame of reference not rotating at the same speed as the Earth.

The effective gravitational attraction of the ISS to Earth, is the same as the Earth to the ISS. That is why it is in orbit.
Velocity is more like things in parallel, where acceleration is more like being in series.

It's OK, but he doesn't give a clear definition of weight.
It would be better if he simply insisted on the fact that your mass remains constant in all these situations. In any case, since this is the same guy who confidently predicts that a DDWFTTW vehicle can't work, calling it "free energy" and "magic" (see here and here),

He gives a good definition of both.
"Weightlessness" is the absence of reaction to the force of gravity. You can provide that reaction in free-fall by other means. Parachutes for example.
In the elevator, doing work or providing acceleration against gravity

I'm not inclined to spend too much time reading his articles.

No, you didn't.
.

 

[John Freestone]

The first option for g on that page reads thus:

"I-gravity is denoted gI and can be calculated using Newton’s law of universal gravitation, as follows: It has magnitude

|gI| = GM/r²
and is directed toward the center of the planet."

The key phrase is "directed toward the center of the planet": as long as we remain Earth-centred, we choose Earth as "the planet". But which planet do we choose when we are away from Earth? In particular, can you answer my previous question: what is the "real", "true" or "actual" weight of the Earth?

Sorry to butt in, but isn't there an 'm' missing from that law of universal gravitation? We're left with the odd question of whether the weight of the Earth should be GM^2 or Zero, I think. Usually, the small m refers to a smaller mass in the gravitational field of the earth, with mass M. To me, the beginning of any definition of weight is the understanding that it depends on a mass and the gravitational effect of it upon another mass, or more mass. How much voltage is there in one end of a 3V battery? What's a cheshire cat's smile without the cat?

By definition there is no net force required to make an object accelerate at the rate of free fall.

So what happened to F = ma ? If there is no net force, F = 0, a = 0. Curiouser and curiouser.

ETA: Oh, hang on, are you defining freefall as a = 0? Maybe that's it. If you're in zero G, you don't get accelerated. If you fall down an elevator shaft, you're not accelerating, the world is whizzing outwards with gravity, since gravity is an acceleration field pointing up? Somewhere near? I'm really struggling here. I can only guess we're not in Newtonian space anymore.

 

[humber]

That's incorrect. There is no way to distinguish a gravitational field from an acceleration field at a point. They are identical in every respect.

A point within a uniform gravitational field? Well of course they are indistinguishable under those circumstances.
As you say, it makes no sense. They (would be) the same thing, but from a different source.
But then again, it makes no sense to apply that to macroscopic objects, in the Earth's non-uniform gravitational field.

 

[John Freestone]

Indeed, I got "whooshed".

Too busy bolting things down and installing K filters as the harmonics in this place are causing things to float away.

Everything is all Fouriered up around here.

 

[humber]

How much voltage is there in one end of a 3V battery? What's a cheshire cat's smile without the cat?
So what happened to F = ma ? If there is no net force, F = 0, a = 0. Curiouser and curiouser.

Of course. The force is there. No gravity, no acceleration.


"I-gravity" is cool though. Goes well with an i-pod.

 

[Belz...]

I'm having trouble following this thread.

What's it about, now ?

Guess I'll never know.

 

[Clive]

The first option for g on that page reads thus:

"I-gravity is denoted gI and can be calculated using Newton’s law of universal gravitation, as follows: It has magnitude

|gI| = GM/r²
and is directed toward the center of the planet."

The key phrase is "directed toward the center of the planet": as long as we remain Earth-centred, we choose Earth as "the planet". But which planet do we choose when we are away from Earth?

Well, you could (in principle) use this formula for every other object that had mass anywhere else in the universe (or at least in your "general neighbourhood"!) and then sum all those forces to get the total gravitational force. In other words, look (through Newtonian glasses) at what would be "lost" if somehow we could remove all mass from the universe (except the object we are "weighing" I guess) or at least "turn off" gravity up in "god's control room". Now you are possibly left with an apparent force acting on the object which is presumably due to various non-gravitational accelerations.

In particular, can you answer my previous question: what is the "real", "true" or "actual" weight of the Earth?

I can answer it a whole bunch of different ways! I might just say it's a silly question, has no sensible or useful meaning and so also, there's no sensible or useful answer, or it's not clearly defined. Or, I might say I also need to know the location and mass of every other "object" in the universe that has non-zero mass so I can then calculate Gm/d² for each (with the direction towards the other mass) and sum all those vectors to get the "total acceleration due to gravity" and then multiply the earth's mass by that. Or I might decide that it's really only the earth that matters (or is at least significant enough?) and say the answer is zero (also assuming it is symmetrical in terms of density and shape) because each 'little bit' of earth is generating a gravitational force on every other little bit and vice versa. Or I could choose some completely arbitrary frame of reference, and calculate the acceleration of the earth in that frame and hence it's weight relative to that frame (this I think is your method).

By definition there is no net force required to make an object accelerate at the rate of free fall.

I think the point of that definition is that some other influences can stop a particular object from being in freefall, and so you may need to apply some "net force" to overcome those other forces and so bring your chosen object's motion into a trajectory that now matched freefall perfectly.

I am not saying that "weight" and "apparent weight" are the same thing. We can always choose a frame where our defined weight is the same as our apparent weight, but we don't have to. If we take another frame, accelerated with respect to the first frame, our weight will be something else. That is the essential point. For instance, if we are on the surface of the Earth, we usually take the local frame defined by the surface we're standing on. That means that our apparent weight, as measured by a simple spring balance, is the same as the weight we define with respect to the frame of reference of the Earth's surface. We could stay standing on the Earth and define our weight by the local free fall frame (accelerating with respect to the frame of the surface of the earth), in which case our weight is zero. Of course, usually we don't, and the "standard" frame of reference is the surface of the Earth. This is just like measuring the speed of a car running on the surface of the Earth. If we say that a car is travelling at 100 km/h, usually we don't need to specify "with respect to the surface of the Earth": it's understood.

Yes, sorry, I realised when heading off to sleep after my previous post that your definition only matched "apparent weight" if the frame you wanted to use was that of the object itself.

It's OK, but he doesn't give a clear definition of weight. It would be better if he simply insisted on the fact that your mass remains constant in all these situations. In any case, since this is the same guy who confidently predicts that a DDWFTTW vehicle can't work, calling it "free energy" and "magic" (see here and here), I'm not inclined to spend too much time reading his articles.

Yeah - I knew that already. I just found this particular post as it was given as a reference in the Wikipedia page on apparent weight. It seems typical of how most people want to talk about weight and apparent weight. I still haven't found any other places on the web that have explicitly given a formal definition of weight along the lines that John S. Denker has done. Do you have any other references for me? I did find some other comments by him in another forum where there was more disucssion amongst a group of lecturers on the best way to define weight. (https://carnot.physics.buffalo.edu/archives/2001/08_2001/msg00213.html)

 

[fredriks]

Some people are discussing gravity.

Humber are talking nonsense.

Thats the thread right now.

 

[Furcifer]

This is very of topic, but why are you using the very large spaces in the equations?



instead of

[latex] \[ E = mc^2 ( 1 - v^2/ c^2 ) ^ {- 1/2}\] [/latex]

Latex is a new language to me

 

[fredriks]

Latex is a new language to me

ok, sorry that I was nitpicking. Align is very nice to use.

[latex]
\begin{align}
\dot{x}&=Ax+Bu\\
y&=Cx+Du \label{sys}
\end{align}
\textmf{see} \eqref{sys}
[/latex]

This is an excellent document about latex if you are interested
http://tobi.oetiker.ch/lshort/lshort.pdf

The reference didn't work...

 

[Furcifer]

I'm having trouble following this thread.

What's it about, now ?

Anubis flying a kite on a treadmill (the belt of which has harmonics akin to chewing gum on the road) in an elevator, in absence of gravity, accelerating at much less than the speed of light, the speed of light being refered to as stiff.

Pick yur poison.

 

[Furcifer]

ok, sorry that I was nitpicking. Align is very nice to use.

[latex]
\begin{align}
\dot{x}&=Ax+Bu\\
y&=Cx+Du \label{sys}
\end{align}
\textmf{see} \eqref{sys}
[/latex]

This is an excellent document about latex if you are interested
http://tobi.oetiker.ch/lshort/lshort.pdf

The reference didn't work...

Oh, that's a better link. The tutorial here is a link to some posts. I just copied and pasted what I could. [latex] doh! [/latex]

 

[recursive prophet]

I hate to be 'pushy,' but....

It seems obvious to me that I am the least qualified person here wrt physics, yet you left me all by my lonesome for two days countering Humbers rubbish about freefall!

HA! Gotcha Ross! You're not even close to 'least qualified,' unless you don't count 'recluses.' Now to prove my point, 2 simple questions.

I recall several instances in the thousands of replies I've read here the issue of pushing the cart to get it started-as they did in all the videos I've seen-come up. Can anyone post a link showing one of these carts going forward on it's own with no push? Also, if you put the cart on a very long treadmill that was turned off and then turned it on, could the cart go forward without someone holding it down against the belt to get it started? Sorry if these are really lame questions, but I really did look for an answer before posting them.

@Belz: Dan O explains the topic of this thread on page 1. As the rather biased title he chose indicates, most here believe that many of the arguments put forth by humber claiming you can’t go DDWFTTW violate classical physics. Just the humor alone makes it a very worthwhile read.

 

[CriticalThanking]

What's this thread about?

About 1993 posts too long at last count...

CT

 

[ThinAirDesigns]

Let the air out of your tires, JB. Flat tires improve fuel consumption?

Non sequitur.

Then you also won't mind demonstrating what happens if you increase the friction of the cart to the belt. Let's say, a 1kg mass over the wheel.

Been there done that. When looking for the best advance ratio of the cart, we used many different diameter wheels. "Many different diameters" meant many different designs -- some with hard and shiny surfaces and some with soft and sticky surfaces. The soft tires weren't slipping, but the harder one's did so we had to add weight to get them to drive the prop.

We went through this with you about 5 thousand posts ago Humber -- you're just cracked on this one. Performance goes up on the cart as the slip against the belt goes down.

If friction increases performance, then why not glue it to the belt?

For the same reason that McClaren doesn't glue their cars tires to their starting grid spot. (you Moron!).

JB

 

[ThinAirDesigns]

I recall several instances in the thousands of replies I've read here the issue of pushing the cart to get it started-as they did in all the videos I've seen-come up. Can anyone post a link showing one of these carts going forward on it's own with no push?

http://www.youtube.com/watch?v=kWSan2CMgos

http://www.youtube.com/watch?v=QTAd891IpRs

http://www.youtube.com/watch?v=MCB1Jczysrk&feature=related

Also, if you put the cart on a very long treadmill that was turned off and then turned it on, could the cart go forward without someone holding it down against the belt to get it started?

Yes.

JB

 

[recursive prophet]

Thanks for the links TAD. I looked at quite a few on Youtube, but of course there are now literally hundreds. Another theory shot down.

 

[Furcifer]

Also, if you put the cart on a very long treadmill that was turned off and then turned it on, could the cart go forward without someone holding it down against the belt to get it started?

TAD says yes, I've never seen any video to support this. ???

 

[ThinAirDesigns]

TAD says yes, I've never seen any video to support this. ???

Then you didn't look at the videos I posted. Those video show that with wind at the tail, the device self-starts.

It doesn't take a lot of CPU cycles to realize that on a long treadmill, there would be increasing wind at the devices tail as the speed of the treadmill is increased.

JB

 

[Michael C]

I can answer it a whole bunch of different ways! I might just say it's a silly question, has no sensible or useful meaning and so also, there's no sensible or useful answer, or it's not clearly defined. Or, I might say I also need to know the location and mass of every other "object" in the universe that has non-zero mass so I can then calculate Gm/d² for each (with the direction towards the other mass) and sum all those vectors to get the "total acceleration due to gravity" and then multiply the earth's mass by that. Or I might decide that it's really only the earth that matters (or is at least significant enough?) and say the answer is zero (also assuming it is symmetrical in terms of density and shape) because each 'little bit' of earth is generating a gravitational force on every other little bit and vice versa. Or I could choose some completely arbitrary frame of reference, and calculate the acceleration of the earth in that frame and hence it's weight relative to that frame (this I think is your method).)

This illustrates exactly my point: the term "real weight" has no clear definition. If we are to talk of weight at all, the only really clear definition I can imagine must include the frame of reference.

As I said before, in situations on Earth, the frame of reference is implied and we don't need to specify it. If we find ourselves on the surface of the Moon, it's usually accepted that the implied frame of reference is that of the Moon's surface; ditto for Mars or any other planet. So we can talk about our "weight" on the Moon, Jupiter or Mars without much confusion. However, if we start talking about "real weight" and "apparent weight" in free fall situations or other situations where we aren't just standing or walking around on the ground, we start having problems. Some people will say, as you did, that the "real weight" of the astronaut in the space station orbiting the earth is defined by the gravitational acceleration as measured relative to the Earth at this point above the Earth's surface. Others, like those who define "weight" as "the reaction to the normal force a spring scale would exert on an object" (taken from the first post in the discussion you linked to) or, even more simply "what the spring scale reads", will say that the astronaut is weightless. In both cases, a frame of reference is in fact being used: those who take your point of view are using a frame of reference stationary relative to the Earth's surface, while those who take the second view are using the free-falling frame of reference defined by the space station. Here's a nice lecture from Walter Lewin of MIT, who takes the second view.

So maybe I can best express my point thus: when somebody talks about the "weight" of an object, a certain frame of reference is always being used, even if the person is not stating the fact, nor even aware of it.