Newton's law of gravitational attraction
in [Relativity]
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From: PD on 10 Oct 2009 10:50 On Oct 10, 9:27 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > On Oct 10, 3:59 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Oct 7, 12:55 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > > On Oct 7, 3:07 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Oct 7, 3:27 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > > > > On Oct 7, 2:01 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On Oct 6, 6:36 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On Oct 2, 2:07 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > PD, the Parasite Dunce, has never considered a better rationale than > > > > > > > the errant mechanics garbage that he was taught in school. Distance > > > > > > > of fall increases to the second power relative to time. But in any > > > > > > > given second the KE that's imparted depends solely on the increase in > > > > > > > VELOCITY > > > > > > > Yes, it depends on the *square* of the velocity. > > > > > > > > and that increases LINEARLY for all near-Earth falling > > > > > > > objects. > > > > > > > And because the velocity increases linearly with time, and because the > > > > > > KE depends on the square of the energy, then the KE increases as the > > > > > > square of time. You see? It's not so difficult. Even an architect > > > > > > would understand it. > > > > > > > > The force of gravity applies a uniform downward force > > > > > > > exactly equal to the static weight of the object. That downward force > > > > > > > is analogous to the uniform thrust of a rocket motor. Uniform, i. e., > > > > > > > LINEAR, > > > > > > > Nope. Uniform means CONSTANT, not linear. Constant thrust stays the > > > > > > same value. Linearly increasing things change value with time in a > > > > > > straight line. A uniform force does not increase with time. Uniform > > > > > > doesn't mean linear. You have your words mixed up. Moreover, the > > > > > > amount of energy contributed by a force is not constant, even if the > > > > > > force is constant. (This is your big mistake.) Remember that the > > > > > > energy contributed by a force is calculated as the PRODUCT of force > > > > > > times distance. So even if the force stays uniform with time, the > > > > > > distance in each second increases with time. This means that the > > > > > > product of force times distance in each second INCREASES each second. > > > > > > That means that the energy contributed by a force in the next second > > > > > > is GREATER than the energy contributed by a force in the previous > > > > > > second, even if the force remains uniform. Do you understand this now? > > > > > > Most 7th graders get it, even if you haven't. > > > > > > > > velocity increases cause a uniform increase in the KE. That > > > > > > > is saying: 'Energy IN = energy OUT." The latter conforms to the Law > > > > > > > of the Conservation of Energy! But Einstein's second power increase > > > > > > > No, this is not Einstein's increase. This was established LONG BEFORE > > > > > > Einstein. Einstein had nothing to do with it. > > > > > > > The square increase is CLASSICAL mechanics, not relativity. > > > > > > > The blame you place on Einstein has nothing to do with Einstein.. > > > > > > > And you simply don't understand classical mechanics. Which is > > > > > > remarkable for a licensed architect. > > > > > > > > in KE due to a uniform increase in velocity, violates the L. of the C. > > > > > > > of E. You have been explained these things dozens of times.. Take the > > > > > > > following Pop Quiz (and flunk) if you are to ever get basic things > > > > > > > RIGHT! NoEinstein > > > > > > > > Pop Quiz for Science Buffs!http://groups.google.com/group/sci..physics/browse_frm/thread/43f6f316... > > > > > > > > > On Oct 2, 11:13 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > > On Sep 30, 11:15 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > > Dear PD: For you, a rock that falls four times as far will hit four > > > > > > > > > times as hard. > > > > > > > > > Well, I'm not sure we're talking the same language. Physics uses quite > > > > > > > > precise language, as I'm sure architecture does to. > > > > > > > > > A rock that falls four times as far will have four times the energy > > > > > > > > when it lands. This is certainly testable by a variety of methods, > > > > > > > > including bolometers. Then there isn't any real question about it, > > > > > > > > since it's just a quantity that you measure, like the length of a > > > > > > > > carrot or the volume of syrup in a jug. > > > > > > > > > "Hitting hard" can mean how much momentum is transferred or what the > > > > > > > > force of the impact is. That strongly depends on the mass of the > > > > > > > > object it hits. > > > > > > > > > > WRONG! The KE that you seem to think accrues > > > > > > > > > proportional to distance of fall INCLUDES THE COASTING CARRYOVER > > > > > > > > > DISTANCES FROM PREVIOUS SECONDS. > > > > > > > > > Well, yes, of course. Otherwise it would just have the same value > > > > > > > > after every meter fallen, and we know THAT isn't true. > > > > > > > > > > KE, like momentum, increases in > > > > > > > > > direct proportion to the increase in VELOCITY, > > > > > > > > > No, sir, not according to experimental measurement, as I've been > > > > > > > > pointing out to you. > > > > > > > > > > NOT the increase in > > > > > > > > > distance which is mostly due to coasting carryover. > > > > > > > > > And that's simply wrong. You can test it by a number of methods, if > > > > > > > > you're careful with the method. > > > > > > > > > > Because the > > > > > > > > > velocity of falling objects increases uniformly, KE does too. > > > > > > > > > Nope. > > > > > > > > > > Read the following to better understand: Pop Quiz for Science Buffs!http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316... > > > > > > > > > > NoEinstein - Hide quoted text - > > > > > > > > > - Show quoted text -- Hide quoted text - > > > > > > > > - Show quoted text - > > > > > > ------------------- > > > > > moreover > > > > > at very high velocities > > > > > the force needed to add velocity > > > > > is not linear > > > > > it is an **exponential order** > > > > > > F/Gamma = ma !!!! > > > > > Gamma is not linear > > > > > but exponential > > > > > You and NoEinstein have the same problem understanding what > > > > "exponential" means. > > > > > > it becomes infinitely big as closing to c !!! > > > > > Actually, it gets finitely big. It never gets infinite. > > > > > > Y.P > > > > > --------------------------------------- > > > > but waht you does not understand > > > is that > > > anyphysics formula has its limits of validations > > > and a limit case mathematically > > > (and even not mathematically) > > > is th epoint at which to formula > > > stops working or being relavant > > > totake an example > > > if we say that no mass can reach c > > > it is an extrapolation beyond the > > > legitimate limits of th eformula > > > to say that a mass can or cannot reach c > > > because it is mathematically a limit case > > > so > > > you cant say that the photon cannot have mass > > > because it moves at c !! > > > it can be an exception case > > > (beyound your common paradigm) > > > of which the mass of the photon is so small > > > that it CAN move at c !! > > > even experiments can indicate that trend: > > > > as masses become smaller and smaller > > > they reach CLOSER AND CLOSER > > > to c !!! > > > But never reach it. > > Let's take a simple example. > > Let's suppose there are two physical properties F and x, and they are > > related this way: F=1 - 1/x. > > When x's value is 1, then F=1 > > When there is an object with x=2, then F for that object is 0.5. > > When there is an object with x=10, then F for that object is 0.9. > > When there is an object with x=1000, then F for that object is 0.999. > > When there is an object with x=1000000, then F for that object is > > 0.999999. > > It's plain that the larger x is, then the closer F will be to 1. > > But there will never be ANY object for which F = 1, no matter how big > > x is. > > > > that wrong paradigm that no mass canreach c > > > jsut because of expanding a formula > > > to an unknown position > > > It's not an unknown position. > > > > interpretating it arbitrarily 'as you like it' > > > was one of the disasters that happend > > > to physics at the 20 th centuries > > > leading to curved space time > > > massless particles etc etc etc > > > > ATB > > > Y.Porat > > > ------------------------- > > but still you ddint got my point about > limits of validations of a mathematical formula > > the scope of physical phenomenon is not just > always ovelaping the mathematical scope > that is why we add factors to the mathematical formula > and we add stsrt point and end limitation > to the formula > inoder to diminish the degrees of freedom > in order to fit it to reality of the physical world > as i brought the trend > of smaller and smaller masses > closing closer and closer to c > > the fact that we didnt find yet that smallmass > but onthe other hand we have photons that has > some of the mass property ie momentum Momentum is a not a property exclusive to mass. You may have been taught that momentum is *defined* as mass times velocity. If so, then you were taught incorrectly. That is not any kind of correct definition of momentum. > and if you add all of it together > it is actually duggestion and **predicting* > that there must be a smaller mass thanknown now > thatwill accommodate with al l the acumulative > experimental data > not to mention that i showed that > the > E=hf > has in it **hidden* even in the formula -- the mass entity !! And by the same argument you've used to "show" this, you can just as well "show" that empty space has electrical charge. This SHOULD be an indicator to you that something is wrong with the method used to "show". > > Y.P > -----------------
From: NoEinstein on 10 Oct 2009 11:04 On Oct 8, 12:21 pm, Jerry <Cephalobus_alie...(a)comcast.net> wrote: > Dear Jerry ("another" hard-head): You need to take the following Pop Quiz for Science Buffs. If you doand will learn from the rationale (Though that's doubtful for you...)you will realize THIS basic fact: A compact object is traveling 32.174 feet per second, as at the end of one second of free-fall, and will have traveled 16.087 feet. Such 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. Now, suppose that a second same-size-and-weight object is allowed to fall under the influence of gravity for that same one second, but by some magic process, the force of gravity stops after one second. That object will be traveling 32.174 feet per second, and after two more seconds will have fallen 80.435 feet, mostly, by virtue of COASTING. Its total distance of fall is 5 times greater than the first object dropped, but its KE will remain just 2 static weight units! The reason? Both objects will impact at the identical velocity of 32.174 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 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. E. NoEinstein Pop Quiz for Science Buffs! http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316... Dropping Einstein Like a Stone http://groups.google.com/group/sci.physics/browse_thread/thread/989e16c59967db2b?hl=en# > > On Oct 7, 6:27 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > Falling objects are accruing energy in direct proportion to how long > > the object is subjected to the action of gravity. > > No. Falling objects accrue energy in direct proportion to HOW FAR > THEY FALL under the influence of gravity. > > After the first second, an object has fallen about 16 feet. > After the second second, an object has fallen about 64 feet. > After the third second, an object has fallen...guess what??? > > 1) Each foot that an object falls represents the same amount of > energy accrued. Each foot that one lifts an object represents the > same amount of energy expended in lifting. > 2) The DISTANCE an object falls is proportional to the square of > the time that it has spent falling. > 3) The SPEED of a falling object is directly proportional to the > time that it has spent falling. > > Do you have any problem with statements (1), (2), or (3)? > Together, they explain why there is a square in the kinetic > energy equation. > > Jerry
From: NoEinstein on 10 Oct 2009 11:21 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! 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. Distance of travel has NO direct influence on the object's KE. Try to get that through your hard head! Noeinstein > > 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.
From: PD on 10 Oct 2009 11:28 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.
From: NoEinstein on 10 Oct 2009 11:40
On Oct 8, 1:04 pm, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > Dear Y.Porat: The failure of scientists to make charged particles reach 'c' was due to the presence of ether inside the vacuum tubes. The ridiculous notion behind SR resulted from extrapolation of the particle experiments. In space travel, the density of ether drops off the further away from massive objects you stay. In the 'Swiss Cheese' voids between the galaxies, there is little or no ether. Plot your intra-universe courses of travel to stay in those voids, and your maximum velocity is limitless. Extra-terrestrials, and yours truly, know that, because ether is polar, it can be magnetized and made to RIP apart so that the spaceships don't ever impact the ether. I respect you for rejecting warped space-time. But you should realize that when warped space-time goes, SR goes with it! NoEinstein > > On Oct 8, 1:32 am, NoEinstein <noeinst...(a)bellsouth.net> wrote:> On Oct 7, 4:27 am, "Y.Porat" <y.y.po...(a)gmail.com> wrote: > > > Dear Y. P.: I've disproved SR; remember? Nothing 'special' happens > > close to 'c' other than simply reaching a Universal measuring datum of > > much larger potential velocities. NE > > --------------------- > > you forgot that i agree with you only about the nonsense of curved > space time > i do not agree with you about SR > i accept SR ! > > it is actually all based on the physical fact that > it becomes more and more difficult to accelerate a big mass > closer and closer to c !! > > ATB > Y.Porat > -------------- > > > > > > > > On Oct 7, 2:01 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On Oct 6, 6:36 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On Oct 2, 2:07 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > PD, the Parasite Dunce, has never considered a better rationale than > > > > > the errant mechanics garbage that he was taught in school. Distance > > > > > of fall increases to the second power relative to time. But in any > > > > > given second the KE that's imparted depends solely on the increase in > > > > > VELOCITY > > > > > Yes, it depends on the *square* of the velocity. > > > > > > and that increases LINEARLY for all near-Earth falling > > > > > objects. > > > > > And because the velocity increases linearly with time, and because the > > > > KE depends on the square of the energy, then the KE increases as the > > > > square of time. You see? It's not so difficult. Even an architect > > > > would understand it. > > > > > > The force of gravity applies a uniform downward force > > > > > exactly equal to the static weight of the object. That downward force > > > > > is analogous to the uniform thrust of a rocket motor. Uniform, i. e., > > > > > LINEAR, > > > > > Nope. Uniform means CONSTANT, not linear. Constant thrust stays the > > > > same value. Linearly increasing things change value with time in a > > > > straight line. A uniform force does not increase with time. Uniform > > > > doesn't mean linear. You have your words mixed up. Moreover, the > > > > amount of energy contributed by a force is not constant, even if the > > > > force is constant. (This is your big mistake.) Remember that the > > > > energy contributed by a force is calculated as the PRODUCT of force > > > > times distance. So even if the force stays uniform with time, the > > > > distance in each second increases with time. This means that the > > > > product of force times distance in each second INCREASES each second. > > > > That means that the energy contributed by a force in the next second > > > > is GREATER than the energy contributed by a force in the previous > > > > second, even if the force remains uniform. Do you understand this now? > > > > Most 7th graders get it, even if you haven't. > > > > > > velocity increases cause a uniform increase in the KE. That > > > > > is saying: 'Energy IN = energy OUT." The latter conforms to the Law > > > > > of the Conservation of Energy! But Einstein's second power increase > > > > > No, this is not Einstein's increase. This was established LONG BEFORE > > > > Einstein. Einstein had nothing to do with it. > > > > > The square increase is CLASSICAL mechanics, not relativity. > > > > > The blame you place on Einstein has nothing to do with Einstein. > > > > > And you simply don't understand classical mechanics. Which is > > > > remarkable for a licensed architect. > > > > > > in KE due to a uniform increase in velocity, violates the L. of the C. > > > > > of E. You have been explained these things dozens of times. Take the > > > > > following Pop Quiz (and flunk) if you are to ever get basic things > > > > > RIGHT! NoEinstein > > > > > > Pop Quiz for Science Buffs!http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316... > > > > > > > On Oct 2, 11:13 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On Sep 30, 11:15 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > Dear PD: For you, a rock that falls four times as far will hit four > > > > > > > times as hard. > > > > > > > Well, I'm not sure we're talking the same language. Physics uses quite > > > > > > precise language, as I'm sure architecture does to. > > > > > > > A rock that falls four times as far will have four times the energy > > > > > > when it lands. This is certainly testable by a variety of methods, > > > > > > including bolometers. Then there isn't any real question about it, > > > > > > since it's just a quantity that you measure, like the length of a > > > > > > carrot or the volume of syrup in a jug. > > > > > > > "Hitting hard" can mean how much momentum is transferred or what the > > > > > > force of the impact is. That strongly depends on the mass of the > > > > > > object it hits. > > > > > > > > WRONG! The KE that you seem to think accrues > > > > > > > proportional to distance of fall INCLUDES THE COASTING CARRYOVER > > > > > > > DISTANCES FROM PREVIOUS SECONDS. > > > > > > > Well, yes, of course. Otherwise it would just have the same value > > > > > > after every meter fallen, and we know THAT isn't true. > > > > > > > > KE, like momentum, increases in > > > > > > > direct proportion to the increase in VELOCITY, > > > > > > > No, sir, not according to experimental measurement, as I've been > > > > > > pointing out to you. > > > > > > > > NOT the increase in > > > > > > > distance which is mostly due to coasting carryover. > > > > > > > And that's simply wrong. You can test it by a number of methods, if > > > > > > you're careful with the method. > > > > > > > > Because the > > > > > > > velocity of falling objects increases uniformly, KE does too. > > > > > > > Nope. > > > > > > > > Read the following to better understand: Pop Quiz for Science Buffs!http://groups.google.com/group/sci.physics/browse_frm/thread/43f6f316... > > > > > > > > NoEinstein - Hide quoted text - > > > > > > > - Show quoted text -- Hide quoted text - > > > > > > - Show quoted text - > > > > ------------------- > > > moreover > > > at very high velocities > > > the force needed to add velocity > > > is not linear > > > it is an **exponential order** > > > > F/Gamma = ma !!!! > > > Gamma is not linear > > > but exponential > > > it becomes infinitely big as closing to c !!! > > > > Y.P > > > ---------------------------------------- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > - Show quoted text - |