Newton's law of gravitational attraction
in [Relativity]
|
From: NoEinstein on 16 Oct 2009 13:10 On Oct 13, 2:51 pm, BURT <macromi...(a)yahoo.com> wrote: > 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- Hide quoted text - > > - Show quoted text - No. If the motion doesn't change, then, there has been no transfer of momentum. NE
From: NoEinstein on 16 Oct 2009 13:13 On Oct 13, 9:05 pm, "Autymn D. C." <lysde...(a)sbcglobal.net> wrote: > On Oct 12, 5:57 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On Oct 12, 4:12 pm, "Autymn D. C." <lysde...(a)sbcglobal.net> wrote: > > > On Oct 10, 8:04 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > will impact with a KE of 2 of its static weight units.  For each > > > > second of fall its KE will increase one additional weight unit.  All > > > > objects at rest have a 'KE' = their static weight.  The CORRECT > > > > formula is: KE = a/g (m) + v / 32.174 (m).  At the end of second one, > > > > KE = 2. > > > > That is the formula of nothing. > > > > feet per second.  FORCE OF IMPACT, i.e., KE is a function of the > > > > VELOCITY and the static weight of the object.  *** Since the velocity > > > > i.e. -> q.e. > > > energhy != forse > > > mvv/2 != ma = mjt > > > > > of all dropped objects increases uniformly, or LINEARLY with respect > > > > to time, NOT the distance traveled, KE must, therefore, be increasing > > > > LINEARLY, too.  The latter conforms to the Law of the Conservation of > > > > Energy.  But your errant "distance" notion violates the L. of the C. > > > > Momentum, cretin. > > > KE IS momentum, Jerk!  â NE â > > momentum = vis inertiæ > work = vis impressa > cinèsis = vis viva > pressure = vis mortva > > > Dear Autymn D. C.:  No, it's the formula that REPLACES E = mc^2 and KE > > = 1/2 mv^2 > > When v=0, KEâ 0: it's the formula of nothing. > > -Aut- Hide quoted text - > > - Show quoted text - At v = 0, KE = 1 weight unit. That's the constant downward force on the object not yet released to fall. â NE â
From: PD on 16 Oct 2009 14:09 On Oct 16, 11:30 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Oct 13, 12:37 pm, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD, the Parasite Dunce: "Conversations" usually have give-and- > take. In the two plus years that I've known that you... exist, not > once have you been swayed to accept the viewpoint of anyone other than > the status quo. There's a reason for that. This science. Science is the study of how nature works. Nature works only one way, and that's the one way we're trying to figure out. Scientific truths cannot be a matter of opinion, to be settled by argument. That is reserved for art and politics. Scientific truths are determined by the preponderance of data acquired from nature itself in experiment, and as that progresses, what DOES happen and SHOULD happen is that scientists all -- every one of them -- becomes convinced of the same scientific truth, because there is after all only one. If someone should be swayed to believe something new, and is swayed by an *argument*, then that person has certainly not been thinking scientifically. Now, if you want to talk about how new theories DO arise in science, I'd be happy to explain that to you. But somehow I think you are trying to see if you can do science in the manner of a Letter to the Editor on the Op-Ed page. And that's no way to do science. > Your bastion is to cling to the stupidity of the > past, rather than to question and to correct the errors of the past. > Calling you a Parasite is the highest high compliment I can make. > NE > > > > > On Oct 13, 11:08 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 > > > Interesting how you project that "no one" reads my replies. How do you > > know? > > Is this from the same pool of information that tells you that > > intelligent readers agree with your posts, even though there is not a > > one that has come out and said that? > > > As for what you read of my replies, I'm not surprised. You're not here > > to have a conversation about anything. You're here to hear yourself > > talk. And you take pride in the fact (and consider it an > > accomplishment) to do so on an unmoderated newsgroup. While you're at > > it, congratulate yourself for breathing without help today. > > > > > 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 -
From: PD on 16 Oct 2009 14:12 On Oct 16, 12:03 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Oct 13, 1:39 pm, jbriggs444 <jbriggs...(a)gmail.com> wrote: > > Dear jbriggs444: Your definition of work is like a time-motion > study. My definition is AFTER the work has been done, and its > quantity gets evaluated. My formal definition of work is: W = FD x > cos. of angle of deviation from the desired direction of motion. Good. Now, let's look at YOUR definition. Notice that term D. What do you think D stands for? It is the DISTANCE the object moves while the force is acting on the object. Now go back and look at what I told you a few days ago about why the work increases each second as an object falls, even as the force remains the same. You see? You AGREE with me, by invoking that formula above. > No > motion in desired direction would = zero work done. Effective KE is > AFTER all of the forces have been added or subtracted according to > their vectors. The resultant velocity times the mass that is in > motion yields the KE (in weight multiples, like pounds). NoEinstein > > > > > > 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 16 Oct 2009 14:15
On Oct 16, 12:06 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On Oct 13, 1:50 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > 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.- Hide quoted text - > > > - Show quoted text - > > Science is what I figured out over a lifetime, not something I needed > to 'look up' anywhere... like you do. NE Perhaps you should use a different word, other than science. Because your activity bears no resemblance to science and does not meet the criteria that is used by scientists do define that word. Just because you decide to putter with pipes and water in your backyard does not give you the right to claim what you learn from that hobby to be expertise in plumbing. PD |