Newton's law of gravitational attraction
in [Relativity]
|
From: jbriggs444 on 13 Oct 2009 13:39 On Oct 13, 12:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Oct 13, 11:29 am, PD <thedraperfam...(a)gmail.com> wrote: > > PD: You are off-your-rocker. Go into a small room and talk to > yourself. No one with half a brain reads your replies. I only scan > down to your first foot-in-the-mouth nonsense. NE So do I. Usually that results in reading clear to the bottom of a PD post. > > On Oct 12, 7:32 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On Oct 10, 11:28 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > Dear Side-Stepping PD, the Parasite Dunce: WORK is NOT the same as > > > KE! > > > The contribution to the energy that changes KE is called work. Please > > consult your 7th grade science book. This one gave me pause. It's too causal for my taste. It implies that if two opposite forces act on a uniformly moving object that neither of them do any work since neither of them change KE. One can argue that this definition isn't quite right for non-rigid bodies and that it requires clarification for rotating bodies. Still, it's not a bad starting point. A simplification suitable for pedagogical purposes. Especially when the discussion doesn't involve rigidity or rotation. And if you read it just right, it is equivalent to what I consider to be a good definition. Me, I define work as the vector dot product of instantaneous point force times the incremental motion of the target object at the contact point integrated over the life of the force. If it's not a point force, you can integrate over the contact area as well. I've had discussions previously with folks who think that you're supposed to do a path integral of force times incremental movement of the force along a path. That's wrong. It's force times incremental movement of the object to which the force is applied. For point objects it's the same thing. For extended objects it's not.
From: PD on 13 Oct 2009 13:50 On Oct 13, 12:39 pm, jbriggs444 <jbriggs...(a)gmail.com> wrote: > On Oct 13, 12:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On Oct 13, 11:29 am, PD <thedraperfam...(a)gmail.com> wrote: > > > PD: You are off-your-rocker. Go into a small room and talk to > > yourself. No one with half a brain reads your replies. I only scan > > down to your first foot-in-the-mouth nonsense. NE > > So do I. Usually that results in reading clear to the bottom of a PD > post. > > > > On Oct 12, 7:32 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On Oct 10, 11:28 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > Dear Side-Stepping PD, the Parasite Dunce: WORK is NOT the same as > > > > KE! > > > > The contribution to the energy that changes KE is called work. Please > > > consult your 7th grade science book. > > This one gave me pause. It's too causal for my taste. It implies > that if two opposite forces act on a uniformly moving object that > neither of them do any work since neither of them change KE. Fair point. Net work is what is usually referenced in the Work-Energy theorem. > > One can argue that this definition isn't quite right for non-rigid > bodies and that it requires clarification for rotating bodies. Well, as for the latter, I didn't specify *linear* KE, but again the point is taken. > > Still, it's not a bad starting point. A simplification suitable for > pedagogical purposes. Especially when the discussion doesn't involve > rigidity or rotation. And if you read it just right, it is equivalent > to what I consider to be a good definition. > > Me, I define work as the vector dot product of instantaneous point > force times the incremental motion of the target object at the contact > point integrated over the life of the force. If it's not a point > force, you can integrate over the contact area as well. Agreed, but NoEinstein won't be able to look that up in his 7th grade science book. > > I've had discussions previously with folks who think that you're > supposed to do a path integral of force times incremental movement of > the force along a path. That's wrong. It's force times incremental > movement of the object to which the force is applied. For point > objects it's the same thing. For extended objects it's not. For *linear* KE, I see the distinction.
From: Y.Porat on 13 Oct 2009 14:34 On Oct 13, 5:33 pm, illed with charge. > > > > > ------------------- > > > > > common > > > > you start to amuse me: (and stun me) > > > > > empty space has the property of charge ?? (:-) > > > > ithought that empty space is by definition > > > > has nothing in it !! > > > > why not witches on brooms > > > > > does the experiment that defined that electric constant--- > > > > > -- WAS DONE IN EMPTY SPACE ?? > > > > DONT YOU THINK THAT > > > > THERE WERE OTHER PHYSICAL > > > > ENTITIES THERE THAT INFLUENCED and shaped THAT CONSTANT?? > > > > Nope. No charges need be present in the volume considered, and you > > > don't have to consider the region outside the volume either. > > > did you ever think or know > > how that electric constant was derived by an experiment?? > > A number of them, yes. > And not all of them included a charge in the region where this > property was measured! >--------------------- so it was measured in Vacum?? to measure in complete vacum should be measuring an electric charge 50 meters fron the palce in which any charge is located !! ie to measure the constant in a palce that here is no electriccharge iow anttime you measure charge all the sources of that charge are near by !! OTHA you can measure a photons one light year from the location of the entity that created it and sent it ie you measure it neutralized of its creators !!! dont you see the difference ?? ------------------ > > do you agree that it was derived by expariment ?? > > all those tools and matter and physical entities that involved in > > that experiment are meaning less for the experiment ?? > > Oh, come on. Here we go. > So for the photon, where you say Planck's constant tells you there is > mass involved and so the photon must have mass, how do you know that > the mass doesn't belong to all the tools involved in the experiment, > rather than to the photon itself? > Stop it. You twist and try to use a chain of logic one way for one see jsut above now if it does not go with you by physics arguments i have some formal physics arguments for you 1 you fell in to the trap of words interpretations not physics interpretations you heared that they call it sort of the vacuum constant'' as if it was a proerty of Vacuum but no one told you it is a property of vacuum: the constant of charge force WAS DERIVED IN VACUUM CONDITIONS IT WAS DERIVED IN SOME SPACE THAT DIDNT NOT INCLUDE SAY AIR RUBBER WATER GASOLINE ANOTHER GAS etc etc they actually told you what there ** WAS NOT IN THERE!* among posible materials that we know about them that could be but were not there !! and you wrongly understood it as a property of Net VACUUM!! -------------------------- 2 WHO PERMITTED YOU TO SEPARATE A APRT OF THE FORMULA FROM THE MAIN BODY OF THE FORMULA ANDINTERPRET IT ASD A SEPARATED ENTITY OF THE FORMULA?? iow all the componenents of the formula is ONE PACKET ONE PARCEL !! you cant trim jsut a part of it and interpret it as a separated part of the whole formula! by trimming it from the rest of the formula you actuallly castrated that formula !! from its physical meaning and went to fantastic interpretations (if you like it more abstaractly:-- half of the truth is often a lie !!) on the other hand while i examined E=hf i examined it as** a whole* i dint separate any part of it and didnt interpret-at it separately as you did !! got the difference ?? ATB Y.Porat ----------------------------------
From: Androcles on 13 Oct 2009 14:37 "jbriggs444" <jbriggs444(a)gmail.com> wrote in message news:c9b652d0-7210-43e3-8c8d-6fdb499db18e(a)m7g2000prd.googlegroups.com... On Oct 13, 12:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Oct 13, 11:29 am, PD <thedraperfam...(a)gmail.com> wrote: > > PD: You are off-your-rocker. Go into a small room and talk to > yourself. No one with half a brain reads your replies. I only scan > down to your first foot-in-the-mouth nonsense. � NE � So do I. Usually that results in reading clear to the bottom of a PD post. > > On Oct 12, 7:32 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On Oct 10, 11:28 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > Dear Side-Stepping PD, the Parasite Dunce: WORK is NOT the same as > > > KE! > > > The contribution to the energy that changes KE is called work. Please > > consult your 7th grade science book. This one gave me pause. It's too causal for my taste. It implies that if two opposite forces act on a uniformly moving object that neither of them do any work since neither of them change KE. One can argue that this definition isn't quite right for non-rigid bodies and that it requires clarification for rotating bodies. Still, it's not a bad starting point. A simplification suitable for pedagogical purposes. Especially when the discussion doesn't involve rigidity or rotation. And if you read it just right, it is equivalent to what I consider to be a good definition. Me, I define work as the vector dot product of instantaneous point force times the incremental motion of the target object at the contact point integrated over the life of the force. If it's not a point force, you can integrate over the contact area as well. ============================================= I've heard stilletto heels on women's shoes damage kitchen floors when the vector dot product of an instantaneous point force times the incremental motion of the target object at the contact point is integrated over the life of the force. Especially when the floor damage doesn't involve rigidity or rotation. If you add rotation then its called "drilling". =============================================
From: BURT on 13 Oct 2009 14:51
On Oct 13, 8:08 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Oct 13, 11:29 am, PD <thedraperfam...(a)gmail.com> wrote: > > PD: You are off-your-rocker. Go into a small room and talk to > yourself. No one with half a brain reads your replies. I only scan > down to your first foot-in-the-mouth nonsense. NE > > > > > > > On Oct 12, 7:32 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On Oct 10, 11:28 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > Dear Side-Stepping PD, the Parasite Dunce: WORK is NOT the same as > > > KE! > > > The contribution to the energy that changes KE is called work. Please > > consult your 7th grade science book. > > > > Distance of fall always includes a COASTING component. If the > > > fall distance in one second, 16.087 feet, is taken as a datum and > > > called 'd', the coasting distances (hidden) within the parabolic time > > > vs. distance curve are as follows: Second 1, coasting distance 'zero'; > > > Second 2, coasting distance 2d; Second 3, coasting distance 4d; and > > > Second 4, coasting distance, 6d. The distances 'left' over are > > > increasing one 'd', or 16.087 feet, each secondand that is a LINEAR > > > increase in NON COASTING distance of fall, consistent with KE being a > > > LINEARELY increasing quantity. > > > Sorry, but no. The contribution to energy as it increases each second > > is the product of a force times the distance covered in that second. > > This is known by every 7th grader but it still seems to elude you. Why > > didn't you ask a question about this in the 7th grade? > > > > It would be possible to convert KE to useful work done at impact or > > > slow-down. But such would be a tacked-on, different physics problem > > > concerned with the weight of the object(s) impacted. Knowing you, you > > > will weasel out by insisting that the discussion be on work, not KE. > > > Like I've told you, the existing definition of work is the only > > > correct term in the chapters on mechanics, > > > Good, and then you'll also notice right next to the definition of work > > something called the Work-Energy theorem. > > > > which I have otherwise > > > correctly re written for the benefit of science. NoEinstein > > > > > On Oct 10, 10:21 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On Oct 8, 1:01 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > Dear PD, the Parasite Dunce: An object's "distance of fall" isn't an > > > > > "energy" component! > > > > > Of course it is. Please consult on the definition of work, which is > > > > how a force makes an energy contribution. > > > > This contribution is the force times the distance the force acts > > > > through. > > > > The distance is indeed a factor in the energy contribution. > > > > > I'm astounded -- ASTOUNDED, I tell you! -- that you have forgotten > > > > this basic fact that 7th graders know. > > > > It is in fact the basis for the playground see-saw, not to mention the > > > > block-and-tackle pulley system. > > > > > > The only force acting on a dropped object is the > > > > > uniform (per unit weight) force of gravity. An object in space that > > > > > is traveling 1,000 miles per hour, after traveling 1,000 miles, will > > > > > impact with the identical KE as a like-size object that traveled one > > > > > million miles at 1,000 miles per hour. > > > > > But if it is traveling at a constant 1000 miles per hour, then there > > > > is obviously no force acting on the object. > > > > If there were a force acting on it, it would continue to accelerate.. > > > > Since there is no force acting on it, then there is no contribution to > > > > the energy whether it is traveling a thousand miles or a million > > > > miles. > > > > > If you want to know how a force contributes to the energy, then > > > > consider cases where the force is present. Looking at cases where > > > > there is no force acting on the object any more will not help you > > > > understand the basics. > > > > > > Distance of travel has NO > > > > > direct influence on the object's KE. Try to get that through your > > > > > hard head! Noeinstein > > > > > I'm sorry, NoEinstein, but this is REALLY basic, 7th grade stuff. > > > > Nobody that failed to learn the material from the 7th grade should be > > > > allowed to be licensed as an architect, in my opinion. > > > > > > > On Oct 8, 10:46 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On Oct 7, 7:48 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > Dear PD, the Parasite Dunce: The INPUT energy is the UNIFORM force of > > > > > > > gravity. > > > > > > > No sir. The input energy is the PRODUCT of the force of gravity and > > > > > > the distance the force acts through. > > > > > > This is a simple fact, verified by experiment. > > > > > > And as an object falls, though the force is uniform, the distance per > > > > > > second increases each second, and so the product is not uniform though > > > > > > the force is uniform. > > > > > > I've explained this to you a half-dozen times, and you still don't > > > > > > seem to understand this. Do you have a reading comprehension problem? > > > > > > > > To have the "output" KE be the square of the time, > > > > > > > immediately, violates the LAW OF THE CONSERVATION OF ENERGY: energy IN > > > > > > > must = energy OUT! NoEinstein > > > > > > > Sorry, no, the INPUT energy is also increasing as the square of time, > > > > > > as I've explained several times. Energy is conserved.- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text - Can motion be conserved in exchange of momentum? Bodies exchanging their states of motion? MItch Raemsch |