quiet Posted December 20, 2018 Author Share Posted December 20, 2018 With many people saying the same thing, I tend to think the same as those people. I apologize for my ramblings. Best regards. Link to comment Share on other sites More sharing options...

Strange Posted December 20, 2018 Share Posted December 20, 2018 1 hour ago, quiet said: The main question is where come from the incorrection? You have had your misunderstandings explained several times in different ways. Here are some of the clearest: On 19/12/2018 at 12:57 AM, uncool said: You should understand it by realizing that the force is proportional to the square of the distance only if acceleration is being held constant, which is generally an unnatural assumption. Mass being held constant is a somewhat natural idea; acceleration being held constant is not. On 19/12/2018 at 1:37 AM, swansont said: Newton’s gravitational law does not include the variable a, which depends on r. So you can’t make this comparison. 3 hours ago, J.C.MacSwell said: Since acceleration in the formula is dependant on distance...it is simply incorrect to conclude that With regard to force and distance? Force is inversely proportional to distance squared, same as it was in equation (1), but now less obvious. Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted December 20, 2018 Share Posted December 20, 2018 Mass in each equation, both 1 and 2, is assumed to be the same. If you don't believe me dig Newton up and ask him. (whether they should be is debatable, but our evidence to this point in time has them as equivalent. Nothing wrong with the algebra. Nothing wrong with Equation 3. Though I don't see much use in that form, it is still valid. Given the acceleration, and distance, it will in fact give you the gravitational force between the two unknown equal masses. But that does not mean the force is proportional to distance squared. Acceleration depends on the distance (and the masses). If acceleration was independent of the distance you could conclude the force is proportional to distance squared, but acceleration is not. On 12/18/2018 at 7:57 PM, uncool said: You should understand it by realizing that the force is proportional to the square of the distance only if acceleration is being held constant, which is generally an unnatural assumption. Mass being held constant is a somewhat natural idea; acceleration being held constant is not. This is the key. Let's say we are observing two equal masses orbiting each other in space, in a perfect circle. We measure their distance apart and centripetal acceleration. The greater the distance is, the much greater the masses must be, and the much greater the force must be, to maintain that constant acceleration. So in this specific set of cases of fixed acceleration you can say the force is proportional to the distance squared. 2 Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted December 20, 2018 Share Posted December 20, 2018 1 hour ago, J.C.MacSwell said: Mass in each equation, both 1 and 2, is assumed to be the same. If you don't believe me dig Newton up and ask him. (whether they should be is debatable, but our evidence to this point in time has them as equivalent. Nothing wrong with the algebra. Nothing wrong with Equation 3. Though I don't see much use in that form, it is still valid. Given the acceleration, and distance, it will in fact give you the gravitational force between the two unknown equal masses. But that does not mean the force is proportional to distance squared. Acceleration depends on the distance (and the masses). If acceleration was independent of the distance you could conclude the force is proportional to distance squared, but acceleration is not. This is the key. Let's say we are observing two equal masses orbiting each other in space, in a perfect circle. We measure their distance apart and centripetal acceleration. The greater the distance is, the much greater the masses must be, and the much greater the force must be, to maintain that constant acceleration. So in this specific set of cases of fixed acceleration you can say the force is proportional to the distance squared. Note: this is not with respect to any one system, but with respect to the choice of system (mass pairs) required for the fixed acceleration. Link to comment Share on other sites More sharing options...

studiot Posted December 20, 2018 Share Posted December 20, 2018 1 hour ago, quiet said: With many people saying the same thing, I tend to think the same as those people. I apologize for my ramblings. Best regards. I am not saying the same thing. The way to explain it to your students is simple. F = ma is an equation of dynamics, and only meant for non equilibrium systems Newton's law of gravitation is not. So that force is only exerted if the object is accelerating. The force of gravity is exerted whether the objects are in motion or not. Physically you can't combine the equations in the way you have, although algebraically you can, the result makes no sense. Furthermore the 'force' of gravity is normally put in the frame of the larger object (eg the Earth) You are trying to work in some external frame. Any teacher should be able to guide his students in the correct direction about this. Link to comment Share on other sites More sharing options...

Phi for All Posted December 20, 2018 Share Posted December 20, 2018 ! Moderator Note Moved to Speculations. Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted December 20, 2018 Share Posted December 20, 2018 1 hour ago, studiot said: I am not saying the same thing. The way to explain it to your students is simple. F = ma is an equation of dynamics, and only meant for non equilibrium systems Newton's law of gravitation is not. So that force is only exerted if the object is accelerating. The force of gravity is exerted whether the objects are in motion or not. Physically you can't combine the equations in the way you have, although algebraically you can, the result makes no sense. Furthermore the 'force' of gravity is normally put in the frame of the larger object (eg the Earth) You are trying to work in some external frame. Any teacher should be able to guide his students in the correct direction about this. Sorry Studiot, but I believe this is misleading with respect to Newtonian Mechanics, where gravity is simply another force. (I realize it is now in speculations, but was moved from Classical Physics, and the equations all derived from Newtonian) The equations as combined make sense, though perhaps not good sense as the result is confusing and not very useful (I maintain it is accurate) Link to comment Share on other sites More sharing options...

Janus Posted December 20, 2018 Share Posted December 20, 2018 On 12/18/2018 at 3:40 PM, quiet said: I bring to the forum the query made by a student at the school. - Newtonian gravitation. - Simple case of two equal spheres that gravitate to each other (equal masses). Let's write the Newtonian equation of gravitational force.F=G m mr2 Both masses are the same. Then there is the following.F=G m2r2 (1) We clear the mass in the equation of Newton's second law.m=Fa We raise both members to the square.m2=F2a2 (2) In (1) we replace m2 as indicated by (2).F=G F2a2r2 We simplify and cleared F.F=a2 r2G (3) Equation (3) reports that the force is directly proportional to the square of the distance. This is not what Newton's gravitational equation informs. The gravitational equation and the second law, applied to the same system, result in something that we have not seen in school. How should we understand equation (3)? You understand equation (3) by grasping that 'a' is a value that itself depends on "r". a varies in inverse proportion to r^2, and thus a^2 varies in inverse proportion to r^4. Thus if r doubles, r^2 increases by a factor of 4, but a^2 reduces to 1/16 its value. 4*1/16 = 1/4 and the net value of F has decreased by a factor of 1/4 or 1/r^2. The equation is misleading because it contains a variable which is not independent of another variable in the equation. 3 Link to comment Share on other sites More sharing options...

studiot Posted December 20, 2018 Share Posted December 20, 2018 (edited) 6 hours ago, J.C.MacSwell said: Sorry Studiot, but I believe this is misleading with respect to Newtonian Mechanics, where gravity is simply another force. (I realize it is now in speculations, but was moved from Classical Physics, and the equations all derived from Newtonian) The equations as combined make sense, though perhaps not good sense as the result is confusing and not very useful (I maintain it is accurate) So explain to me the value of m given by quiet's equations 2, when a = 0 ? (Since the plan is to substitute this value or its square into equation 3) Or explain to me how you can validly divide by zero in Mathematics? Here is the version I'm looking at in post#1. On 18/12/2018 at 11:40 PM, quiet said: - Newtonian gravitation. - Simple case of two equal spheres that gravitate to each other (equal masses). I'm still hoping quiet will see why you cannot substitute for m in any Physics or Mathematical system Edited December 20, 2018 by studiot Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted December 20, 2018 Share Posted December 20, 2018 (edited) 1 hour ago, studiot said: So explain to me the value of m given by quiet's equations 2, when a = 0 ? (Since the plan is to substitute this value or its square into equation 3) Or explain to me how you can validly divide by zero in Mathematics? Here is the version I'm looking at in post#1. I'm still hoping quiet will see why you cannot substitute for m in any Physics or Mathematical system Where a = 0 in the described setup, F, and m are also = 0 (you could argue it is undefined, and indeed there would be no real system in that case) Why can you not substitute a/F for mass in a mechanics equation, when you do so in a correct and meaningful way? (or if by substitute mass you mean to create a massless system? quiet did not do that, the above exception where a is set at zero notwithstanding) Edited December 20, 2018 by J.C.MacSwell Link to comment Share on other sites More sharing options...

studiot Posted December 20, 2018 Share Posted December 20, 2018 (edited) 59 minutes ago, J.C.MacSwell said: Where a = 0 in the described setup, F, and m are also = 0 Oh come now, I'm sure you know better than that. setting a = 0 is nothing more than noting that the masses are not undergoing accelerating movement towards each other. This means that the net accelerating force F = 0. It does not mean that either m1 or m2 are zero. If that were the case then the bodies would exert zero gravitational effect on each other, as with photons. For massive bodies it may mean they are fixed or held as with Cavendish's balls. But even though they are not moving towards each other massive objects still experience a gravitational force in Newtonian mechanics. The bottom line is that the original diagram is incomplete since it does not explain how the masses came to be in the position shown. Any force due to gravity must be additional to any force that either drew them there or holds them there must be taken into account when calculating F. And the offered analysis does not do this. Edited December 20, 2018 by studiot Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted December 20, 2018 Share Posted December 20, 2018 5 minutes ago, studiot said: Oh come now, I'm sure you know better than that. setting a = 0 is nothing more than noting that the masses are not undergoing accelerating movement towards each other. This means that the net accelerating force F = 0. It does not mean that either m1 or m2 are zero. If that were the case then the bodies would exert zero gravitational effect on each other, as with photons. For massive bodies it may mean they are fixed or held as with Cavendish's balls. But even though they are not moving towards each other massive objects still experience a gravitational force in Newtonian mechanics. The bottom line is that the original diagram is incomplete since it does not explain how the masses came to be in the position shown. Any force due to gravity must be additional to any force that either drew them there or holds them there must be taken into account when calculating F. And the offered analysis does not do this. Sorry Studiot. I certainly do not! We are discussing a system as described, with 2 equal masses some distance apart in space. All I know of the system is what was described. two equal masses in space, some distance apart....I know of no other balls, Cavendish or otherwise. The masses of the balls are fixed by any given distance, and any given acceleration...unless you want to make stuff up...such as Cavendish balls Why not just use your hands? Why not say "no JC, the acceleration is zero because I'm holding them apart" and then ask me to guess what the masses might be? Invoking something other than what is described no longer works for the equations...why would they? Link to comment Share on other sites More sharing options...

studiot Posted December 20, 2018 Share Posted December 20, 2018 I'm glad you are not teaching me Physics. Quote Wikipedia The Cavendish experiment, performed in 1797–1798 by British scientist Henry Cavendish, was the first experiment to measure the force of gravity between masses in the laboratory and the first to yield accurate values for the gravitational constant. What was that you said about two equal masses a fixed distance apart. Google is your friend. We also had a long thread about it here recently. Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted December 20, 2018 Share Posted December 20, 2018 (edited) 7 minutes ago, studiot said: I'm glad you are not teaching me Physics. What was that you said about two equal masses a fixed distance apart. Google is your friend. We also had a long thread about it here recently. I know what the experiment is about. Anything else that is irrelevant you would like to add? There are no balls set up as described by the Cavendish experiment in quiet's description. None. Edited December 21, 2018 by J.C.MacSwell Link to comment Share on other sites More sharing options...

studiot Posted December 21, 2018 Share Posted December 21, 2018 7 minutes ago, J.C.MacSwell said: I know what the experiment is about Then you are just making trouble for reasons best known to yourself so there is no point my having further discussion with you. Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted December 21, 2018 Share Posted December 21, 2018 (edited) 19 minutes ago, studiot said: Then you are just making trouble for reasons best known to yourself so there is no point my having further discussion with you. Apparently not understood by you...but I assure you I mean no trouble. Much of your inputs here are misleading with regard to mechanics. I was simply pointing it out. By "making stuff up" with regard to Cavendish balls, I meant adding something to the system that was not included. The Cavendish balls are constrained in a manner not described in what we are discussing, and not only that but there are 4 of them constrained within an apparatus, not 2 in free space...just so you know. Edited December 21, 2018 by J.C.MacSwell Link to comment Share on other sites More sharing options...

quiet Posted December 21, 2018 Author Share Posted December 21, 2018 (edited) 1. I say that: The equation (3) is absurde into the physical context. 2. I say that: Absurde comes only from replace [math]m^2[/math] in Newton's equation of gravitatory force. 3. I say that: If we reject the idea of gravity and inertia caused by the same physical property, then, we can not put the same symbol ([math]m[/math]) in force equation and in the second law equation. 4. I say thay: We can't to put the same symbol without suppossing the same cause in gravity an in inertia. 5. I say that: The absurdity of eqution (3) requires reject the idea of the same cause. 6. I say that: You can call it: idea of the same cause. Or can choose other name. The name choosed not diminish the physical absurde. There isn't a mathematical absurde. There is an unavoidable physical absurde. 7. I say that: If someone still have doubts, try perform a proof of the items 1 to 6 are false. With one item really false, the proof is done. Remember that a proof must be not only logically valid. Must be too physically valid and referred only to well known physical principles and laws. 8. I say that: I will receive with happyness such a proof and I will be greatful. Edited December 21, 2018 by quiet Link to comment Share on other sites More sharing options...

swansont Posted December 21, 2018 Share Posted December 21, 2018 1 hour ago, quiet said: 3. I say that: If we reject the idea of gravity and inertia caused by the same physical property, then, we can not put the same symbol (m ) in force equation and in the second law equation. There are other tests, though. If M>>m, you can look at e.g. a circular orbit. m will cancel, leaving a result that works. i.e. the cancellation was valid. So we have confirmation that inertial and gravitational mass are the same. It works for elliptical orbits, and works (with some more effort) even when you don’t have M>>m 1 Link to comment Share on other sites More sharing options...

quiet Posted December 21, 2018 Author Share Posted December 21, 2018 40 minutes ago, swansont said: There are other tests, though. If M>>m, you can look at e.g. a circular orbit. m will cancel, leaving a result that works. i.e. the cancellation was valid. So we have confirmation that inertial and gravitational mass are the same. It works for elliptical orbits, and works (with some more effort) even when you don’t have M>>m Hello swansont. I liked the argument. And I would like it to stay firm. I'm thinking of a kind of chess between gentlemen. A is your argument and B is mine. I will try to look for a weak point in A and you can, if you decide to play, do the same with B. --------- In the case of the satellite orbit you mentioned, as in all applications of Newton's gravitational equation, G is calibrated precisely so that it is possible to use the same term [math]m[/math], with the same value, in the gravitation and inertia. This is a conventional decision, it is not a physical law. The number of convetional decisions that could be proposed is unlimited and all would work without contradiction. So the satellite case does not show that the same physical property is responsible for gravity and inertia. Link to comment Share on other sites More sharing options...

Strange Posted December 21, 2018 Share Posted December 21, 2018 3 hours ago, quiet said: 2. I say that: Absurde comes only from replace m2 in Newton's equation of gravitatory force. m is an independent variable. You have replaced it with something that is, not only not independent, but also may be dependent on m. 3 hours ago, quiet said: 3. I say that: If we reject the idea of gravity and inertia caused by the same physical property, then, we can not put the same symbol (m ) in force equation and in the second law equation. except we know that we can use the same m in both and we get the results that are consistent with experiment (eg. Galileo's famous demonstration, pendulum period, etc). And there is no evidence that gravitational and inertial mass are different. 3 hours ago, quiet said: 5. I say that: The absurdity of eqution (3) requires reject the idea of the same cause. The "absurdity" of (3) is just a demonstration of your wilful ignorance; after all, several people have explained why there is no problem and what the equation represents. 1 hour ago, quiet said: In the case of the satellite orbit you mentioned, as in all applications of Newton's gravitational equation, G is calibrated precisely so that it is possible to use the same term m , with the same value, in the gravitation and inertia. The value of G is irrelevant. You can set it to 1 or 42 and you will still find that you can use the same value for mass in both equations. (After all, G only appears in one of the equations.) As the is in the speculations forum, it is up to you to provide some evidence that inertial and gravitational mass are not the same. (Meaningless and childish tricks with algebra are not evidence.) Link to comment Share on other sites More sharing options...

swansont Posted December 21, 2018 Share Posted December 21, 2018 2 hours ago, quiet said: Hello swansont. I liked the argument. And I would like it to stay firm. I'm thinking of a kind of chess between gentlemen. A is your argument and B is mine. I will try to look for a weak point in A and you can, if you decide to play, do the same with B. --------- In the case of the satellite orbit you mentioned, as in all applications of Newton's gravitational equation, G is calibrated precisely so that it is possible to use the same term m , with the same value, in the gravitation and inertia. This is a conventional decision, it is not a physical law. The number of convetional decisions that could be proposed is unlimited and all would work without contradiction. So the satellite case does not show that the same physical property is responsible for gravity and inertia. Different orbits, with multiple satellites, and it still works. You could pick a value of G to make it work for a single case, but not with the hundreds of them we have, with widely varying parameters. plus the high-precision tests that show that gravitational and inertial mass are the same. Link to comment Share on other sites More sharing options...

quiet Posted December 21, 2018 Author Share Posted December 21, 2018 (edited) Let me go step by step. I have read all the posts, but first I analyze the main point of all this, to submit it to your criticism. --------- - Free fall. [math][/math] [math]m_{g_1} \ \rightarrow[/math] gravitational mass of the planet [math]m_{g_2} \ \rightarrow[/math] gravitational mass of the falling object With this symbols, we write the force as: [math]F = G \ \dfrac{m_{g_1} \ m_{g_2}}{r^2} \ \ \ \ \ \ \ \ \ (A)[/math] The falling object is accelerating, and we write Newton's second law: [math]F = m_{i_2} \ a \ \ \ \ \ \ \ \ \ (B)[/math] [math]m_{i_2} \ \rightarrow[/math] inertial mass of the falling object We equal (A) and (B). [math]G \ \dfrac{m_{g_1} \ m_{g_2}}{r^2}= m_{i_2} \ a [/math] We clear [math]m_{i_2}[/math] [math]G \ \dfrac{m_{g_1} \ m_{g_2}}{a \ r^2}= m_{i_2} \ \ \ \ \ \ \ \ \ (C)[/math] - Acceleration and distance are measurables with rule and clock. - Both gravitatory masses, of the planet and of the object, have the same physical nature and are measured and expressed in the same way. - Inertial mass of the object is the term under discussion. We can stablish conventionally the value of G, without destroy the equality of the forces neither alter the equation (C). - We can, conventionally, choose a value of G that makes [math]m_{g_2}= m_{i_2}[/math]. This simplify many calculus, but there is no physical law, nor physical principle, that necessarily requires [math]m_{g_2}= m_{i2}[/math]. Edited December 21, 2018 by quiet Link to comment Share on other sites More sharing options...

studiot Posted December 21, 2018 Share Posted December 21, 2018 Quiet. Ask yourself the following questions. 1) What system does Newton's 2nd Law apply to or represent.? Answer it represent a single body or system expressed as a free body diagram. Note the crucial part. There is only one mass. 2) What system does Newton's Law of gravitation apply to or represent? Answer it represents two bodies that interact with each other. Note the crucial part. There are two masses. This is the first place your student is going wrong. Link to comment Share on other sites More sharing options...

quiet Posted December 22, 2018 Author Share Posted December 22, 2018 (edited) Hi, studiot. Since the gravitational equation and the second law are applied together, the military can, unfortunately, accurately calculate ballistic missile launches. And long before the cannon shots. In those cases the vertical component of the ballistic movement works like a vertical shot, which reaches its apogee and initiates the free fall. In order not to result in verbal arguments I have put the same development that the texts present to teach that the equality of both types of masses is conventional. Einstein himself emphasizes that in the physics of Newton nothing does impose equality. And he emphasizes that, since equality is not the consequence of principles or of physical laws prior to GR, the hypothesis of equality is a new element introduced to give a basis to GR. Einstein calls it a principle, as befits a new element that is introduced as the basis of a theory. You do not need me to name it, but I will do it to avoid verbal scuffles because of the name. He called it the equivalence principle of the gravitational and inertial masses. This principle is alien to Newtonian physics, to the theories of Lagrange and Hamilton, to thermodynamics, to statistical mechanics, to electrodynamics, to special relativity and to quantum theory. Only GR is based on that kind of hypothesis. If I, with seriousness, take the task of accompanying my statements with mathematical developments and demonstrate that I affirm something coherent and in accordance with physical laws, it is only fair that my affirmations are refuted in the same way. In the last mathematical post I presented there are no terms in absurd places of some equation. It is only a free fall, because thus the texts teach that the equality of both types of mass is not a consequence of the well-known physical laws, nor of the principles that are the basis of all that is not GR. Tired of the sterile verbal scuffle, in another post I put, ironically, that many people saying the same thing made me want to join them and think the same. I had really decided to leave the thread served to those who prefer the verbal fray, without writing me any new post. When swansont calmly resumed the dialogue, I thought: nobility obliges. That's why I wrote again and showed the corresponding mathematics. The tranquility was short lived. Now again the thread was thrown into empty verbality. What I do ? I follow ? I give up? They say that I must notice this, that, the other, take into account this and that detail and many more etceteras. Discounting me, did someone take paper and pencil, to see how the subject remains mathematically, before filling a post with phrases? I have patience, all the patience that a teacher exercises in the classroom, but that patience is finite. And it has reached the end. Edited December 22, 2018 by quiet Link to comment Share on other sites More sharing options...

swansont Posted December 22, 2018 Share Posted December 22, 2018 52 minutes ago, quiet said: This principle is alien to Newtonian physics People were doing calculation based on the concept, for many years before GR, so clearly it is not. Einstein included it in GR, but that does not mean the idea was new. Link to comment Share on other sites More sharing options...

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