geistkie Posted May 9, 2009 Posted May 9, 2009 This note does not rewrite gravity physics, nor does it attempt to -- it is a correction to Newton's Shell theorem wherein it is claimed that the force equation indicates the location of the point where all mass of the shell is concentrated;, that the total force on a test mass m from the mass on the surface of a shell located a distance d from m that the total force can be considered concentrated at a point located at the center of the sphere. This note concerns the location of the center of the force which cannot possibly be located at the sphere (center of mass) geometric center. The results of integrating the force term to get s F = GmM/d^2, where d is the distance between m and the sphere center, but the expression is only one of force and one cannot assume automatically that the center of force is located a distance d from m. I make the point the sphere half closest to m contributes a larger share of the total force than does the shell half farther from m. To avoid confusipon, I use the term 'mass-force-center' (not seen in the literature) to mean that point from which the test mass m considers the location of the source of the total [concentrated] mass bearing on it from the mass M of the shell. Newton's Shell Theorem –Bad mathematics. Bad physics, This part shows how to determine the point where m sees the force of two equal masses in a line added to provide the total force of the two masses and from this determine the mass-force-enter. Take three mass point objects m1 = m2 = m3 = 1 unit mass, G=1 unit gravitation constant, and using init distances the force of attraction between m1 and m3 separated by 10 unit distance is calculated using the universal law of gravity expression, F = Gm1m2/r^2 (minus sign omitted). F12 = (1)(1)(1)/10^2. Similarly, the force generated between m1 and m3 separated by 12 unit distance F13 = 1/144. When the masses are arranged along a common axis the total force of m2 and m3 on m1 is F123 = 1/100 + 1/144 = (1)(1)(1 + 1)/r^2 . Rearranging the terms, 2/r^2 = 244/(100)(144), or, r^2 = (2)(14400)/244 or R^2 = 118.03. The result r = 10.86 and, 10 < r < 11, where 11 is the location of the m1m2 system center of mass (COM). The combined forces' COM acting on m1 is located a distance 11 from m1. However, the center of mass-force (CMF) is located at 10.86, which is off set from the COM in the direction of m1. The total forces of the mass of this spherical shell is calculated manually (see above) using mirrored image pairs of masses on the shell where one member of the each pair is in an opposite hemispheres, one closest to m1, one farthest from m1. Clearly when all forces are calculated each and all CMFs of each calculation are located in the nearest sphere segment to m1, contrary to Newton's Shell theorem that without any physical basis, claimed that m1 may consider the mass of the sphere concentrated at the COM of the sphere. The links below are consistent examples of the rote acceptance of a developed shell theorem, referenced as an unchallenged law of physics and communicated as scientific gospel, chiseled in stone, as it were, leaving the authors and subsequent decuples immune from any heretical criticism. http://en.wikipedia.org/wiki/Shell_theorem http://www.physclips.unsw.edu.au/jw/NewtonShell.pdf http://www.absoluteastronomy.com/topics/Shell_theorem From inspection of the sphere and m1 externally located at some point r from the sphere center it is obvious that using the concept of "inverse distance squared" (and the universal law of gravity) as a starting point, the mass of M in the hemisphere closest to m1 will contribute a greater share of the total force on m1 than the mass in the farthest hemisphere, hence the CMF for the entire mass M is located on the m-M axis off set from the COM in the direction of m. Where, and how, does the Newton Shell theorem (NST) place the CMF at the sphere COM? It doesn't! The NST model begins with a ring of differential mass, dM, oriented perpendicular to and centered on r. Then, summing all forces for each dM on each ring and integrated over the surface of the sphere producing the calculated total force acting on m, which says nothing, absolutely nothing, regarding the location of the CMF. The Wikipedia model referenced above states after dF is integrated to F = GmM/r^2 that, "The shell really does act as though all the mass is concentrated at the center!" The big problem here is that the developed algorithm made no inclusion for determining the location of the CMF. The intuitive assumption that the CMF is located at the COM of the sphere was arbitrarily (instinctively) made (or so I surmise) from the 'r^2' term in the expression, that clearly is an expression for determining the total gravitational force of mass M on m, only. Another flaw is seen in the expression when obtaining the net vector force in the m-M direction derived by taking the cosine projection of force in the "r" direction only, and from this, supposedly, the inference is that 'the CMF followed the projection of the force onto the m-M axis' – the projection of force vectors is mathematically proper (forces perpendicular to the m-M axis 'cancel'', or so we are told), but to include the scalar quantity location of the CMF is just plain "bad mathematics"; not properly placing the CMF in the nearest hemisphere of M is just plain, "bad physics". Caveat emptor – beware of standing on the shoulders of giants who have been dead for 300 years.
J.C.MacSwell Posted May 9, 2009 Posted May 9, 2009 This note does not rewrite gravity physics, nor does it attempt to -- it is a correction to Newton's Shell theorem wherein it is claimed that the force equation indicates the location of the point where all mass of the shell is concentrated;, that the total force on a test mass m from the mass on the surface of a shell located a distance d from m that the total force can be considered concentrated at a point located at the center of the sphere. This note concerns the location of the center of the force which cannot possibly be located at the sphere (center of mass) geometric center. The results of integrating the force term to get s F = GmM/d^2, where d is the distance between m and the sphere center, but the expression is only one of force and one cannot assume automatically that the center of force is located a distance d from m. I make the point the sphere half closest to m contributes a larger share of the total force than does the shell half farther from m. To avoid confusipon, I use the term 'mass-force-center' (not seen in the literature) to mean that point from which the test mass m considers the location of the source of the total [concentrated] mass bearing on it from the mass M of the shell. Newton's Shell Theorem –Bad mathematics. Bad physics, This part shows how to determine the point where m sees the force of two equal masses in a line added to provide the total force of the two masses and from this determine the mass-force-enter. Take three mass point objects m1 = m2 = m3 = 1 unit mass, G=1 unit gravitation constant, and using init distances the force of attraction between m1 and m3 separated by 10 unit distance is calculated using the universal law of gravity expression, F = Gm1m2/r^2 (minus sign omitted). F12 = (1)(1)(1)/10^2. Similarly, the force generated between m1 and m3 separated by 12 unit distance F13 = 1/144. When the masses are arranged along a common axis the total force of m2 and m3 on m1 is F123 = 1/100 + 1/144 = (1)(1)(1 + 1)/r^2 . Rearranging the terms, 2/r^2 = 244/(100)(144), or, r^2 = (2)(14400)/244 or R^2 = 118.03. The result r = 10.86 and, 10 < r < 11, where 11 is the location of the m1m2 system center of mass (COM). The combined forces' COM acting on m1 is located a distance 11 from m1. However, the center of mass-force (CMF) is located at 10.86, which is off set from the COM in the direction of m1. The total forces of the mass of this spherical shell is calculated manually (see above) using mirrored image pairs of masses on the shell where one member of the each pair is in an opposite hemispheres, one closest to m1, one farthest from m1. Clearly when all forces are calculated each and all CMFs of each calculation are located in the nearest sphere segment to m1, contrary to Newton's Shell theorem that without any physical basis, claimed that m1 may consider the mass of the sphere concentrated at the COM of the sphere. The links below are consistent examples of the rote acceptance of a developed shell theorem, referenced as an unchallenged law of physics and communicated as scientific gospel, chiseled in stone, as it were, leaving the authors and subsequent decuples immune from any heretical criticism. http://en.wikipedia.org/wiki/Shell_theorem http://www.physclips.unsw.edu.au/jw/NewtonShell.pdf http://www.absoluteastronomy.com/topics/Shell_theorem From inspection of the sphere and m1 externally located at some point r from the sphere center it is obvious that using the concept of "inverse distance squared" (and the universal law of gravity) as a starting point, the mass of M in the hemisphere closest to m1 will contribute a greater share of the total force on m1 than the mass in the farthest hemisphere, hence the CMF for the entire mass M is located on the m-M axis off set from the COM in the direction of m. Where, and how, does the Newton Shell theorem (NST) place the CMF at the sphere COM? It doesn't! The NST model begins with a ring of differential mass, dM, oriented perpendicular to and centered on r. Then, summing all forces for each dM on each ring and integrated over the surface of the sphere producing the calculated total force acting on m, which says nothing, absolutely nothing, regarding the location of the CMF. The Wikipedia model referenced above states after dF is integrated to F = GmM/r^2 that, "The shell really does act as though all the mass is concentrated at the center!" The big problem here is that the developed algorithm made no inclusion for determining the location of the CMF. The intuitive assumption that the CMF is located at the COM of the sphere was arbitrarily (instinctively) made (or so I surmise) from the 'r^2' term in the expression, that clearly is an expression for determining the total gravitational force of mass M on m, only. Another flaw is seen in the expression when obtaining the net vector force in the m-M direction derived by taking the cosine projection of force in the "r" direction only, and from this, supposedly, the inference is that 'the CMF followed the projection of the force onto the m-M axis' – the projection of force vectors is mathematically proper (forces perpendicular to the m-M axis 'cancel'', or so we are told), but to include the scalar quantity location of the CMF is just plain "bad mathematics"; not properly placing the CMF in the nearest hemisphere of M is just plain, "bad physics". Caveat emptor – beware of standing on the shoulders of giants who have been dead for 300 years. If you know the total net force and direction you have the resultant force. It effectively acts through the center of mass (assuming a homogenous sphere), not from that point. If you want to say the resultant acts through your CMF described that is fine, it is in the same direction, but it is not acting from that point either, the resultant simply is in that direction. The COM is simply an easier point to describe, can be used as the masses change in position, and is the point from which the inverse square is measured. Why complicate something that is being simplified? - the summing of gravitational forces into a resultant force that consistently acts through the COM. Do you think Newton didn't realize what you just described?
swansont Posted May 9, 2009 Posted May 9, 2009 The shell theorem (which is a restatement of Gauss's law) assumes spherical symmetry. You can't apply it to three objects. If you violate the assumption, the results won't hold. Put another way, you have a bunch of math in the form of the derivation of Gauss's law. If you do another bunch of math that doesn't agree, it means you did the math wrong, because math is self-consistent. The only way to show Gauss's law to be wrong is to find an error in the derivation (or show that the whole mathematical construct is wrong, i.e. show that calculus is not self-consistent)
geistkie Posted May 10, 2009 Author Posted May 10, 2009 If you know the total net force and direction you have the resultant force. It effectively acts through the center of mass (assuming a homogenous sphere), not from that point. [/Quote] Good reply by the way. Whether the system is three unique masses considered point masses relative to each other, or the mass is system is a test mass m1, the other two mass systems are the half sphere closest to m1 or the half furthest away, the systems are effectively identical in at least one respect. Using the three single masses lined on the same axis, the mass closest to the test mass will contribute a greater share of the total force acting on the test mass, than does the third mass, located farther from m. Add two forces to obtain the net force and use this value to plug into F = GmM/d^2 and solve for d, which my examples indicate that d < that distance to the center of mass of the two masses. The distance d is always measured as if the force originated at that point. ------------------------------------------------------------------------------------------------------------------------------------ The calculated net force F not cannot be said to be located at the center of mass of either of the two body system or the two spherical halves system as sensed by the test mass. The theorem says, in words, the shell really does act as if all the mass were concentrated at the center of the shell. On its face this seems a true statement in the sense that if the mass physically concentrated at the shell center m1 would surely sense the physical truth of that statement. However the shell does not act this way, and the way to prove this is to, first, measure F of the shell mass and test mass system separated by a physical distance Da. Next, calculate Dc using the force value obtained. The result, is? Dc < Da, always. Hence, the shell really acts as if the mass were concentrated at Dc, not Da, and this a correction to the shell theorem that improves the theoretical conformity to actual physical conditions. ----------------------------------------------------------------------------------------------------------------------------------------- The test mass m1 sees the shell center and concentrated mass at that center, but the force equation does not support the assumption that Da = Dc --------------------------------------------------------------------------------------------------------------------------------------- Just read between the lines. If you want to say the resultant acts through your CMF described that is fine' date=' it is in the same direction, but it is not acting [b']from[/b] that point either, the resultant simply is in that direction. [/Quote] It is acting from all the masses associated with the shell or solid sphere. There is only one way to concentrate the mass at the sphere center and that is via the mechanism of abstract mathematical calculation. The force equation tells us that the Force on m1 is the amount of measured force acting on m1 from the concentrated shell centered at Da. The shell acts as if the mass were concentrated at Dc , not as if the mass were concentrated at Da. This is another statement saying essentially the same thing as above. Why does the shell act as if the mass were concentrated at Dc? Calculate force on m1 for a point source concentration of mass at Da, call this Fp. Then calculate the force on m1 using the center of mass of ½ Mc half shell closest to m1 and ½Mf half shell furthest from m1, called Fs, S for the sum of the forces attributed to each half shell. Setting the distance from the concentrated mass at 11 unit distance, and set M = 10, m1 =1 ; Fp = (1)(10)(1)/11^2 = 10/121. Now calculate the force on m1 from 1/2Mc at a distance 10, 1/2Mf at a distance 12. The force on m1 from ½ Mc = 5, is Fc = (1)(5)/ 100, the force from the farther shell half is 5/144. The total force is 5/100 + 5/144 = 5(244)/(100x144) = 5(61)/3600 = 61/720. Ff = .0826 and Fs = .0824 where Fp/Fs = .0826/.0824 = 1.026, or Fp = Fs(1.026). Try it with any set of mass consistent numbers, Fp will always be greater than Fs, Fs represent the actual physical extension of the shell mass, while Fp located at the center of mass of the F . The COM is simply an easier point to describe' date=' can be used as the masses change in position, and is the point from which the inverse square is measured. [/Quote'] The center of mass is certainly easier to calculate, but see the response just above. If one uses the COM as the point source 'origin' of force, the calculation will always be in error similar to the simple calculation above. Calculating equal shell halves versus using one concentrated segment point s is improved by using 4, ¼ M segments, but the law of diminishing returns soon enters. I would have a slight problem of fixing on the shell theorem that is not as true, it has been falsified, a s a reflection of physical reality, but as long as one remember s that the test mass sees the shell or solid sphere mass physically located Da, but the shell acts as if the net force is located at Dc, and this because of the asymmetrical distribution of mass relative to the m1-M1 system. In Fc + Ff systems, only the calculation that considers the asymmetric distribution of the shell mass tends to a more accurate representation of physical reality. Why complicate something that is being simplified? - the summing of gravitational forces into a resultant force that consistently acts through[/b'] the COM. See above. I see using the best representation of physics as the biggest dog on the block. The statement describing the results of the integration are facially false, the concentration of mass at Dc is where it is happening, not, as the shell theory says that "the shell acts as if all the mass were concentrated at the center of the shell. If for no other reason than to add a footnote to the progress of science, should be sufficient for some. Do you think Newton didn't realize what you just described? You are asking me to make a gross speculation. Personally' date=' I don’t care if Newton realized what I said or not. If he did realize it and kept silent, he goes down a few steps in scientific integrity measurement. If he realized it and kept silent he was robbing him self of a tad extra glory, but for what silly reason would he remain silent. It is easy to say that the shall theorem as you and I know it, is a good approximation for quick calculations, but a more complete theory 3001 years ago, who knows for sure, but at least science would have started from a higher position and science relative to the shell theorem would have started a few steps advanced. As one is said that Isaac developed calculus and the shell theorem at the same time, that Isaac was not experienced in the fullest extent in both disciplines explains to me an explanation for the math and physics shortcomings of the theory-- giants make mistakes -- so to answer your question, no I do not think IN realized what I described. However, I do appreciate the question and the form of the question that indicates me that you postulated that perhaps IN should have realized what I had just described. [/indent'] Merged post follows: Consecutive posts merged The shell theorem (which is a restatement of Gauss's law) assumes spherical symmetry. You can't apply it to three objects. If you violate the assumption, the results won't hold. Put another way, you have a bunch of math in the form of the derivation of Gauss's law. If you do another bunch of math that doesn't agree, it means you did the math wrong, because math is self-consistent. The only way to show Gauss's law to be wrong is to find an error in the derivation (or show that the whole mathematical construct is wrong, i.e. show that calculus is not self-consistent) I had always considered what I had been taught that Gauss' input here was equivalent to the results of the shell theorem, I still so consider. I briefly looked at some links re Gauss' law. To the extent that Gauss' law is based on Newton's gravitational law they are consistent and as neither predict the shell mass concentrated at the distance calculated from a proper summation of all dF centers being offset from the location of the total mass concentration at the shell center. What do whan to call the the errors, invco,mp;lete, casll it what you may, bjut the off set i have b een discussing is a fact of life, that is a fact of physics. In my posts I never said that the Sell theorem could be applied to a three body system, or any multibodied system. In the examples I offered I always made calculations of two bodies at a time, even in the modified integral over the shell surface calculating a mirror image pair, as a pair, while also providing a calculation that determined the distance x from a the combined force of two differential mirror image dM segments on the surface of the sphere. If, as in the Newton's Shell theorem, Gauss made the same assumptions that, for instance the force expression F = GmM/d^2 where d is the distance from the point mass to the shell center, no amount of math is going to escape the reality that as to the test mass m, the forces acting on m originate in a mass system that is asymmetrically distributed with respect to m, meaning that for a shell and a test mass, the half lof the shell closest to m contributes a greater share of the total mass acting on m than the shell half farther from m. So taking the summation of all the smallest segments of differential masses in pairs mkirored in the ntwo halves, or cutting the shell in half using the closest half as one spource of force, the rear half as another source of force the result is inescapable that the shell acts as if the mass were concentrated at Dc, the distance from m calculated from the net force due acting on m from the combined 1/2 of the total masses segments in the shell halves. See my post to J.C.McSwwell where I treated in some depth your concerns.
D H Posted May 10, 2009 Posted May 10, 2009 geistkie, in order to understand how Newton's shell theorem is just that -- a theorem, you first need to know how to compute the gravitational acceleration toward a thin hoop of mass. So, forget the shell for a bit. Suppose there exists a thin hoop with mass m of radius r. What is the gravitational acceleration, in math, not in words, at some point located on the hoop's axis at a distance l from the center of the hoop? Read the rules of this site. You need to answer this question if you have any hopes of keeping this thread alive.
geistkie Posted May 10, 2009 Author Posted May 10, 2009 geistkie, in order to understand how Newton's shell theorem is just that -- a theorem, you first need to know how to compute the gravitational acceleration toward a thin hoop of mass. I actually got this earlier, the understanding you refer to that is, the first time approximately 40 years ago. So' date=' forget the shell for a bit. Suppose there exists a thin hoop with mass [i']m[/i] of radius r. What is the gravitational acceleration, in math, not in words, at some point located on the hoop's axis at a distance l from the center of the hoop? Why do you need to know this? I thought you were an expert on Newton's Shell theorem. Read the rules of this site. You need to answer this question if you have any hopes of keeping this thread alive. What are you really trying to determine here, spell it out. Are you a moderator or monitor for this forum ? If so what do want from me? ----------------------------------------------------------------------- dF = Gmdm[cos(phi)]/x^2 where x is the distance of the differential segment of mass dm on the ring to the test mass m on the ring axis and phi the angle x makes with the ring axis. -------------------------------------------------------------------------- If you had read my posts you would already haveyour answe, and I repeat the process as a courtesy response to your request. I would, at some time, appreciate a direct response to any specific statement I have made, the thesis I have been discussing, mainly the correction of the statement that, "the shell really does act as if all the mass was concentrated at the center of the shell" which I have been laboring to correct.. This should read, "the shell really does not act as if all the mass was concentrated at the center of the sphere." - rather the "shell really does act as if all the mass was concentrated at a point off set from the center of the shell on the axis in the direction of m." See previous calculations conditions using three equal masses [i have discussed this process in dfetaikl.and the second using two centers of mass of halves of the shell as analogous to the three mass condition. =========================== ====================== Now a question for you; though four actual queries that are so simple counting 4 = 1 shouldn't be a problem for you. Three masses m1 = m2= m3 = Sqrt(2) unit mass, G = sqrt(2) a unit force constant. what is the force of the, a. m1- m2 system at a distance of 20 units distance? b. m1 - m3 system at a distance of 20 units distance? c.What is the total force acting on m1 for the combined forces calculated above? d. what is the calculated distance r for the combined forces Found in c? ---------------------------------------------------------------------- Your question was a tad strange, as if you had some secret agenda. I know you detest the theses I have been discussing,but I cannot understand your comittment to terminating this thread. You mention a rule for this site, so I will check it out. [/indent]
D H Posted May 10, 2009 Posted May 10, 2009 (edited) What are you really trying to determine here, spell it out. Are you a moderator or monitor for this forum ? I am one of this site's physics experts, which means I have limited moderation capability. By taking active part in this thread I can no longer apply these capabilities to this thread. I can however ask moderators who are not taking part to put this thread on a suicide watch pending your failure to answer the question posed in post #6. If so what do want from me? I want you to answer the question posed in post #6, So, forget the shell for a bit. Suppose there exists a thin hoop with mass m of radius r. What is the gravitational acceleration, in math, not in words, at some point located on the hoop's axis at a distance l from the center of the hoop? ----------------------------------------------------------------------- dF = Gmdm[cos(phi)]/x^2 where x is the distance of the differential segment of mass dm on the ring to the test mass m on the ring axis and phi the angle x makes with the ring axis. -------------------------------------------------------------------------- That is not the answer. m is the mass of the ring, not the test particle. I asked for acceleration, not force. If you want to use force rather than acceleration (which is what I asked for), you need to introduce the mass of the test particle. Don't call that mass m because that is the given mass of the ring. That is a differential force, not the total force. You have the magnitude of the differential force correct, but in which direction does that differential force point, and how does this differential force relate to the total force? It's not right, even as a differential force. You don't have the magnitude right, and you don't even have the force represented as a vector. Hint: Do the math. The answer I am looking for will say two things: The direction of the total gravitational acceleration (or force) induced by the ring. The magnitude of this total acceleration (or force), in terms of the given quantities m, the mass of the ring; r, the radius of the ring; and l, the distance between the test point and the center of the ring. Edited May 10, 2009 by D H
swansont Posted May 10, 2009 Posted May 10, 2009 Now a question for you Let us be very clear that you are making claims here, and must address questions put to you. Asking questions of others is OK as long as you are doing that, but if not, it is simply shifting the argument, and will be considered trolling.
J.C.MacSwell Posted May 10, 2009 Posted May 10, 2009 (edited) The gravitational net force towards a hoop does not act as if the mass was concentrated at the COM. In this particular case what Geistke describes makes sense (it would act as if the mass was closer to the object than the COM when outside the hoop but in the plane of the hoop) and I think he is confused with regard to the sphere being similar to that or to a group of masses. Edit: I thought Geistke's point was that the closer half of the sphere contributed to the gravitational force more than the further half, which is true and of course obvious to Newton. The resultant force acting as if at the COM is pretty much unique to the sphere, or shell of the sphere. It is exact and not a close approximation. Edited May 10, 2009 by J.C.MacSwell
D H Posted May 10, 2009 Posted May 10, 2009 The reason I asked the question is because knowing the answer to that question is central to knowing how to calculate the total force exerted by a spherical shell. You have to integrate over a surface. In short, you have to do a double integral. By far the easiest way to do this double integral is to slice the spherical shell into infinitesimally small circular hoops.
J.C.MacSwell Posted May 10, 2009 Posted May 10, 2009 (edited) The reason I asked the question is because knowing the answer to that question is central to knowing how to calculate the total force exerted by a spherical shell. You have to integrate over a surface. In short, you have to do a double integral. By far the easiest way to do this double integral is to slice the spherical shell into infinitesimally small circular hoops. When you integrate for a hoop though, say for a test mass inside the hoop, you would get a resultant force towards the nearest point on the hoop, where inside a spherical shell you would have no net force at all. A single hoop could be modeled as a point mass at the COM only as an approximation at a distance. It is the geometry of a sphere that conspires with the inverse square law to get the shell theorem. It doesn't work with any other shapes or groups of masses unless you carefully "tune" them. It is amazing that with everything else he did, Newton practically invented calculus to do this. Edited May 10, 2009 by J.C.MacSwell
D H Posted May 10, 2009 Posted May 10, 2009 J.C., I never said that the single hoop can be modeled as a single point mass. The force can nonetheless be calculated. Calculating the force for a point along the hoop's axis is not that hard. Knowing how to do that is central to understanding the shell theorem. I am asking geistkei, and not you, to provide this calculation.
J.C.MacSwell Posted May 10, 2009 Posted May 10, 2009 J.C., I never said that the single hoop can be modeled as a single point mass. The force can nonetheless be calculated. Calculating the force for a point along the hoop's axis is not that hard. Knowing how to do that is central to understanding the shell theorem. I am asking geistkei, and not you, to provide this calculation. I thought your post #11 was in part replying to my post #10. Sorry about that.
geistkie Posted May 11, 2009 Author Posted May 11, 2009 The answer I am looking for will say two things: The direction of the total gravitational acceleration (or force) induced by the ring. [/Quote] The direction of the total gravitational acceleration induced by the ring is straight forward' date=' yet I distinguish between net force and total force. I assume the question goes to the net force. The forces on the test mass located on the axis of the ring is in the shape of a cone with the base of the cone defined by the ring, the apex of the cone located at the test mass, the origin of the coordinate system. The length of each side of the cone is determined by the particular units chosen and the location of the ring on the sphere. Note, that for a single ring, the total force is confined to the sides of the cone and only when each differential force is projected onto the axis does the force appear in a mode different than that observed at the ring-mass site – the physical forces making the cone are reduced to a sum of vectors along the ring axis. I note that you asked for the "total gravitational acceleration (or force induced by the ring", yet methinks the net force is what you seek. The net force is a vector on the ring axis pointing from the origin along the ring and projected as indicated below by the cos(phi) phi being the angle between the axis and the line to the mass on the ring. The magnitude of this total acceleration (or force), in terms of the given quantities o m, the mass of the ring; o r, the radius of the ring; and o l, the distance between the test point and the center of the ring. [/Quote] Fr (of the ring) is found from the universal law gravitation, F = GmM/r^2, Int(GmdM(cos(phi)/x^2), m the test mass, and dM the differential mass on the ring a distance x from m, the cos function being the vector projection of the differential force onto the mM axis where phi is the angle between the mM-axis and x. The mass of the ring is found from density, D, of the mass on the shell and the Volume, Vr, of the ring, which is Mr = DV. The volume is, found from the circumference of the ring, 2(pi)R times the thickness t times the differential width, Rd(theta). Or Mr = D2(pi)RtRd(theta). The mass of the ring therefore, is, Mr = 2(pi)R si n(theta)tRd(theta) or, Mr = D2(pi)R^2sin(theta)d(theta). Force from the ring is Fr = Gmcos(phi)dM/x^2 and substituting for dM, Fr = Gm cos(phi)2 (pi)R^2sin(theta)d(theta). Fr =2(pi)tDR^2Gm[int]cos(phi)sin(theta)d(theta)/x^2 Using the law of cosines and l the distance of the shell center to the mass is, (after some algebra, cos(phi)sin(theta)/x^2 is, {l^2 - R^2 + x^2}dx/x^2 the integral is modified with r the unit vector along l as, Fr = [(pi)tDR^2Gm][int]{(d^2 – R^2/x^2 ) +1}dx[r] This is the total force of the ring acting on m. This is what was asked for. Then evaluated at d –R the closest point on the shell to d + R, the farthest point on the shell to m, the total force of the shell or Fs = GmM/d^2 . -------------------------------------------------------------------------------------------- Using the universal law of gravity there is nothing objectionable about this expression. The error needing correcting is the statement that, "the shell acts as though all the mass was located at the center of the shell". The results of he integration produces only the force on m from the shell and does not prove that the shell acts on m as if the mass were concentrated at the shell center. There is a fairly simple way to test the 'physical' accuracy of the statement. First, make two calculations, the first with G = 1, m = 1, M = 10 and d = 11 arbitrarily, using Fs above. Then, calculate F with M1 = ½M = 5 located at 10 and M2 = ½ M = 5 located at 12, which uses the same total mass of M and the same center of mass as the first example , which is the same total mass as the former and with the same center of mass at 11. First: The concentrated mass force calculation Fc is (1)(1)(10)/121 or Fc = 10/121 =.0826. Second: The force on the half nearest m, F1 = (1)(1)(5)/10^2 = 1/20 and for the half farthest from m, F2 = 5/144, for a total of force for the 'half mass' duo , Fh = 1/20 + 5/144 = .05 + .0347 = .0847 . Fh/Fc = .0847/.0826 = 1.025, or Fh = 1.025Fc. Using Fh to find d of this force, Fh = 10/d^2 = .0847 or d^2 = 10/.0847 = 118.063 to obtain d = 10.866. In other words, the shell acts as if the total mass of the shell was located at 10.866 which is only a first order correction. Calculating the force center by splitting the sphere into ¼ segments of equal volume (equal mass), the calculated distance d, which is the distance to the center of mass, creeps up toward 11, but will never arrive at the center of mass of the system. All calculations using mirror image mass segments (or sub-segments) in the near and far segments are necessary in order to determine where m sees the mass of the shell concentrated. If the shell theorem were valid as stated, that the shell acts as if all the mass were concentrated at the center of the shell it stands to reason that by calculating the point by incrementally as above, the same location should result. How can these competing results be resolved? One can start by understanding that, like noticing the mass of the shell closest to m1 must contribute a greater share of the total force on m than the half shell farthest from m. If we use the results as stated for the shell theorem, adding the force of the nearest half shell to the force attributed to the farthest half of the shell should produce the same results. The development of the Shell theorem did not provide a calculation to determine the center of force. How is this established? There is no interpretation of the algorithm arriving at F = GmM/d^2 where the centers of force are considered in the slightest. As was mentioned in a previous post, the authors of the theorem that also developed the calculus that was fundamental to the resulting theorem and virtually all contemporaneous reviewers were novices, at best, in the mathematical and physical details that only scrutiny by Sunday morning analysis could uncover. This writer required 40 years of Sunday mornings to detect the flaw. The projection of the force on the ring to the ring axis included the projection of the scalar location of the center of force in a way that is not clear, there being no discussion of this process as seen in the mathematical expressions. One question I have mentioned with little fan fare and no response is under what basis is the net projected force used in the result? Looking at vectors plotted on the coordinate system and using only the net force along the circle axis seems proper. However, can the force perpendicular to the axis force be ignored completely and in all cases? Two ropes held by a physicist in each hand when pulled in equal force in opposite directions will result in the forces cancelling using vector analysis. The physicist holding the ropes will still feel two forces, one this way, the other the opposite way, she and does not count the net forces as zero. Also, those results a indicating that the force between stellar bodies in instantaneous, that the speed of the force V > 10^10© requires a serious revaluation of gravity physics as this phenomenon states a space absent force fields existing between the stellar bodies. Google on 'speed of gravity'. I haven't any speciofic questions but i do note that no specific error or flaw to anything I have posted has been criticized by yourself. Merged post follows: Consecutive posts merged Let us be very clear that you are making claims here, and must address questions put to you. Asking questions of others is OK as long as you are doing that, but if not, it is simply shifting the argument, and will be considered trolling. I don't get your meaning clearly, " as long as you are doing that" - do you mean as long as I am answering questions and asking questions is the restriction? if so i have no problem. However, and just a question, take the question by DH which was answered, yet nothing in the question or the answer goes to the thesis of this thread. I guess I am saying, in a sense, so what if I can or cannot reproduce the mathematics and derivations of the Shell theorem that exists in thousands of tomes, what has this to say regarding the four corners of my thread ? If I don't know all the words to "La Marseillaise" " am I anti-French? Merged post follows: Consecutive posts merged The gravitational net force towards a hoop does not act as if the mass was concentrated at the COM. In this particular case what Geistke describes makes sense (it would act as if the mass was closer to the object than the COM when outside the hoop but in the plane of the hoop) and I think he is confused with regard to the sphere being similar to that or to a group of masses. I don't get this. Edit: I thought Geistke's point was that the closer half of the sphere contributed to the gravitational force more than the further half' date=' which is true and of course obvious to Newton. The resultant force acting as if at the COM is pretty much unique to the sphere, or shell of the sphere. It is [b']exact[/b] and not a close approximation. J.ClMcSwell, You said, "I thought Geistke's point was that the closer half of the sphere contributed to the gravitational force more than the further half, which is true and of course obvious to Newton." Where did Newtoin say this was obvious? If he did say it he was conscious of the error in the shell conclusion when claiming 'that the shell acts as if all the mass was concentrated at the center aof the shell'. If he had made a calculation using the half shell nearest m and the ferthewr mate to this half, he would get a different answer. See my post to DH where I do the math in detail. GThe shgell theorem can be corrected to, the shell acts as if the mass of the shell was concentrated at a point off set from the shell center in the direction of m. Merged post follows: Consecutive posts merged The reason I asked the question is because knowing the answer to that question is central to knowing how to calculate the total force exerted by a spherical shell. You have to integrate over a surface. In short, you have to do a double integral. By far the easiest way to do this double integral is to slice the spherical shell into infinitesimally small circular hoops. Edit: I thought Geistke's point was that the closer half of the sphere contributed to the gravitational force more than the further half' date=' which is true and of course obvious to Newton. The resultant force acting as if at the COM is pretty much unique to the sphere, or shell of the sphere. It is exact and not a close approximation. [/Quote'] Are you correct about what was obvious to Newton? The first ring closest to m area dM of thickness t is not constant from ring to ring. Projected on m1 is an area of dM where the center of mass of this dM of thickness t is at ½ t and the location of the center of mass of each dM relative to the coordinate system local to each dM is (½t, 1/2 Rd(theta)). This last () locates the center of mass of each dM. Using an example of one dM that is located on the vertical axis closest to m, simply rotate this dM way from you as the dM creeps up the shell surface. The effect of rotation changes the distance of the mass in ½ dM forward of the center of this mass relative to m1. If the dM is square initially; relative to m1, then we have a mini problem of an asymmetric distribution of mass relative to m1. The mass on the outer edges of the dM are farther from m1 than the mass toward the center of the dM. The mathematics of taking dM to approach zero cannot obliterate this physical fact. If as, JCS believed, was obvious to Newton that the mass in the shell closest to m1 contributes a greater share of the total force on m1 than the mass farther from m1 then this must extend down to all the dM on all the rings and where no dM on succeeding rings will expose the same distance of mass to the center of mass on the dM relative to m1 . If all this was obvious to Newton he was quiet about it.
D H Posted May 11, 2009 Posted May 11, 2009 The direction of the total gravitational acceleration induced by the ring is straight forward, yet I distinguish between net force and total force. I assume the question goes to the net force. That you think there is any distinction between the two terms is part of your problem. It is not a good idea to invent your own jargon that is contradictory to conventional notation. Doing so puts you dangerously close to the realm of crackpots. Total force and net force are synonyms. Fr (of the ring) is found from the universal law gravitation, F = GmM/r^2, Int(GmdM(cos(phi)/x^2), m the test mass, and dM the differential mass on the ring a distance x from m, the cos function being the vector projection of the differential force onto the mM axis where phi is the angle between the mM-axis and x. Good, but what happens to the components of the force normal to the ring axis? (Hint: It vanishes upon integration. However, since you have a hard time comprehending the vectorial nature of forces, it would be good for you to show this is the case.) The mass of the ring is found from density, D, of the mass on the shell and the Volume, Vr, of the ring, which is Mr = DV. The volume is, found from the circumference of the ring, 2(pi)R times the thickness t times the differential width, Rd(theta). Or Mr = D2(pi)RtRd(theta).The mass of the ring therefore, is, Mr = 2(pi)R sin(theta)tRd(theta) or, Mr = D2(pi)R^2sin(theta)d(theta). Nope. First off, the mass of the ring is given. You should have computed the density in terms of the mass of the ring. More importantly, your equation is incorrect. In terms of the given mass m of the ring, the mass of the infinitesimally small portion of the ring within some infinitesimally small angle subtended from the center of the ring is [math]\frac m{2\pi}d\theta[/math]. If instead you want to treat the ring as a torus of density ρ with radii R and r, where R is the distance from the center of the torus to the center of the tube and r is the radius of the tube, then the mass of the torus is [math]m=\rho(\pi r^2)(2\pi R)[/math]. The mass of an infinitesimally small portion of the torus within some infinitesimally small angle subtended from the center of the torus is [math]dm=\rho(\pi r^2)Rd\theta[/math] Or, in terms of the total mass, [math]dm=\frac m{2\pi}d\theta[/math]. There is no sin(theta) term. Using the law of cosines ... Your mistake in the calculation of the differential force made you calculation of the net force incorrect. ...Fr = [(pi)tDR^2Gm][int]{(d^2 – R^2/x^2 ) +1}dx[r] When you correct your errors, finish off the math. In other words, I want you to evaluate the integral. The rest of your post is rambling gobbledygook.
geistkie Posted May 11, 2009 Author Posted May 11, 2009 This note is in anticipation of past, present and future question regarding the physical integrity of the shell theorem. This pre-emption should cut a lot of distortion in the communications. Construct the dM volumes on the rings to spheres where the x axis terminates on the center of mass of each sphere. The plane described by the unit vector along the x axis bisects the dM thus illustrating a contradiction in the present matter. The claim that the half hemisphere closest to m1 requires this half to contribute a greater share of the total force of this dm on m1, than that contributed by the sphere half on the adjacent side of the plane. This is the geometric condition from which the shell theorem was and is constructed. As J.C. Swell commented that the closest spherical half produce the greater share of the force must have been obvious to Newton, then this being the case requires the center of the force on m1 be located in the nearest half of the dm sphere relative to m1 and along the mdM axis. One instinct may be to use the stated claims of the shell theorem that the shell acts as if the mass was concentrated at the center of the shell to dispel the scientific heresy. It is a known fact the art, practice and effect of the power associated with circuitous reasoning as a mechanism for establishing logical truth is enormous, well in some industries, I am sure. I am however, comforted with unambiguous certainty of the acutely monitored stare of awareness of the progress of these discussions will continue to buffer this thread from such imposing nonsense. What has been historically and tacitly assumed is that the center of mass always coincides with the center of force. The asymmetrical distribution of mass results in a shift of the mass force center along the axis in the direction of m1 . The term 'the center of mass force', a term somewhat 'coined', no pun intended, if you will, by this writer, turns out to be an unintended disguise of the physical concept of 'the center of gravity, AKA 'the cg'.
J.C.MacSwell Posted May 11, 2009 Posted May 11, 2009 (edited) I think it was obvious to Newton, no calculation required, that the closer half sphere contributed more to the gravitational force on an external mass, than the further half. Every point on the closer half has a symmetrically opposite point. Each of those on the closer half contributes more gravitational force than the corresponding point on the further half. Not sure if Newton would have needed to give it even that much thought, to draw the above conclusion. Was it not obvious to you, or did you have to think about it? (I assume you didn't have to calculate it). Edit: Adding after seeing your last post: Geistke, there exists a point closer than the COM where half of the force contribution lies further away and half lies closer. However, for the purposes of modeling the gravitational force using the inverse square law the COM is exactly where to place a point mass for a sphere, not closer. Edited May 11, 2009 by J.C.MacSwell
swansont Posted May 11, 2009 Posted May 11, 2009 I don't get your meaning clearly, " as long as you are doing that" - do you mean as long as I am answering questions and asking questions is the restriction? if so i have no problem. You have an obligation to answer questions put to you. There will be little tolerance for shifting of the burden of proof. That's all that I was trying to say. Merged post follows: Consecutive posts merged Using the universal law of gravity there is nothing objectionable about this expression. The error needing correcting is the statement that, "the shell acts as though all the mass was located at the center of the shell". The results of he integration produces only the force on m from the shell and does not prove that the shell acts on m as if the mass were concentrated at the shell center. There is a fairly simple way to test the 'physical' accuracy of the statement. First, make two calculations, the first with G = 1, m = 1, M = 10 and d = 11 arbitrarily, using Fs above. Then, calculate F with M1 = ½M = 5 located at 10 and M2 = ½ M = 5 located at 12, which uses the same total mass of M and the same center of mass as the first example , which is the same total mass as the former and with the same center of mass at 11. First: The concentrated mass force calculation Fc is (1)(1)(10)/121 or Fc = 10/121 =.0826. Second: The force on the half nearest m, F1 = (1)(1)(5)/10^2 = 1/20 and for the half farthest from m, F2 = 5/144, for a total of force for the 'half mass' duo , Fh = 1/20 + 5/144 = .05 + .0347 = .0847 . Fh/Fc = .0847/.0826 = 1.025, or Fh = 1.025Fc. Using Fh to find d of this force, Fh = 10/d^2 = .0847 or d^2 = 10/.0847 = 118.063 to obtain d = 10.866. In other words, the shell acts as if the total mass of the shell was located at 10.866 which is only a first order correction. Calculating the force center by splitting the sphere into ¼ segments of equal volume (equal mass), the calculated distance d, which is the distance to the center of mass, creeps up toward 11, but will never arrive at the center of mass of the system. This is not functionally equivalent to the shells. What allows you to say that a half-shell should act the same as a point mass located at its center-of-mass?
geistkie Posted May 11, 2009 Author Posted May 11, 2009 (edited) That you think there is any distinction between the two terms is part of your problem. It is not a good idea to invent your own jargon that is contradictory to conventional notation. Doing so puts you dangerously close to the realm of crackpots. Total force and net force are synonyms. The net force as I made clear was the force projected onto the mM axis. Whether my language was a step poutside the familiar, the idea was communicated. Your protection of 'coventional notation' is to be commended, but is of limited value in the present matter and it is distracting, only, to the contention that the half of the shell closest to m contributes the greater share of the total force of the two shell halves, a slight ommission fom the results of the shell theorem. Good' date=' but what happens to the components of the force normal to the ring axis? (Hint: It vanishes upon integration. However, since you have a hard time comprehending the vectorial nature of forces, it would be good for you to show this is the case.) [/quote'] If you make a public statement that I have a 'hard time comprerhending the vector nature oif forces' point to the specoifoc lack if you please. I add to your question and respose that vector analysis becomes useless in the physical world where the speed of gravity forces has been demonstrated as greater than 10^10c, wwithin experimental error, or simply instantaineous. The forces on m and M just are. Likewise, it wasn't necesary to the the completion of the prioject of answering questionsw that are distractions from attacks on the thesis if this thread. If I had anever heard of vector analysis the only way tom reviver the shell theorem is to expain away the fact that the nearest shell half to m contributes the greater share of the total force on m. Nope. First off' date=' the mass of the ring is given. You should have computed the density in terms of the mass of the ring. More importantly, your equation is incorrect. In terms of the given mass [i']m[/i] of the ring, the mass of the infinitesimally small portion of the ring within some infinitesimally small angle subtended from the center of the ring is [math]\frac m{2\pi}d\theta[/math]. If instead you want to treat the ring as a torus of density ρ with radii R and r, where R is the distance from the center of the torus to the center of the tube and r is the radius of the tube, then the mass of the torus is [math]m=\rho(\pi r^2)(2\pi R)[/math]. The mass of an infinitesimally small portion of the torus within some infinitesimally small angle subtended from the center of the torus is [math]dm=\rho(\pi r^2)Rd\theta[/math] Or, in terms of the total mass, [math]dm=\frac m{2\pi}d\theta[/math]. There is no sin(theta) term. Interesting. The theta is the angle from the COM of the sphere to dm. The circumference of the tube is 2piR, of thickness t and Rd(theta) high and when multiplied together give the volume of the rectangular tube. There is a sin(theta) in my version as the ring creeps up the shell surface.. Your mistake in the calculation of the differential force made you calculation of the net force incorrect. I used a version on the internet as a reference' date=' one of four consistent models I reviewed. Fight them not me. You are trying a straw man opposition. You first establish that yours is different than mine, then you say that my calculation of the net force, the one that's different than yours, must be incorrect and hence, my answer is wrong even though it leads to the correct answer. When you correct your errors, finish off the math. In other words, I want you to evaluate the integral. The rest of your post is rambling gobbledygook. What are you talking about? You express an opinion regarding gobbledygook which has the effect of preventing a rational response. A.re you referring to the experimental resjult proving the speed of gravity force is infinite, iunstantaneous, or most weak, Vg > 10^10c. What is the vector notation for forces that are 'just there'? Google on speed of gravity - the physics indutry knows about it, the astronomy indiustry knows about it. Some scientists actrually talk about it openly. If you have already pointed out what you see as errors I see nothing to correct. I did evaluate my integral for the total force, and I came up with the same result that Newton arrived at. I would like you to find any flaws in the specifics of my theses. I have other posts in this thread than the ones reponding to you. I cannot forward all my communications and copy you with same. I only ask that you point to the errors in the thesis; A problem for you - Assuming the differential volumes on the ring are spherical, with the x axis running from m to the center of mass of dM and unit vector r points along the x axis thereby defining the plane that cuts the dM in two equal mass segments, for all dM on the ring, and where one segment is closest to m. Can you tell what is coming? This configuration is an exact model as that used in the development of the shell thelorem -- the very model used in the questions and response in this post. The problem to sort out here is, the shell half closest to m contributes the greater share of the total force of the dM than does the farthest shell half from m. This observation shifts the center of the mass forces from the center of the mass of the dM. The universal law of gravitation tells us that the mass half nearest m contributes greater share of the total force due to the inverse distance squared law which says that of two equal masses in line with the test mass, the closer of the two equal masses to m contributes more force on m than the farthest of the equal masses. How do you resolve the contradiction that the gravity law pulls the location of the center of gravity toward m along xr and hence the shell does nlot act as if all the mass were concentrated at the center of mass of the shell? The balance of forces require the m to see the center of gravity, AKA the cg of dM off set toward m on xr. I'll reciprocate in your offer of a clue to me above - you cannot solve the foregoing with the use of circuitous reasoning, that is by invoking the proof of the shell theorem on the shell dM problem presented here. Edited May 11, 2009 by swansont turned volume down
swansont Posted May 11, 2009 Posted May 11, 2009 This configuration is an exact model as that used in the development of the shell thelorem -- the very model used in the questions and response in this post. No, it's not. You have assumed that a half-shell behaves this way, ironically, to show that shells do not behave this way. The problem to sort out here is, the shell half closest to m contributes the greater share of the total force of the dM than does the farthest shell half from m. This observation shifts the center of the mass forces from the center of the mass of the dM. The universal law of gravitation tells us that the mass half nearest m contributes greater share of the total force due to the inverse distance squared law which says that of two equal masses in line with the test mass, the closer of the two equal masses to m contributes more force on m than the farthest of the equal masses. How do you resolve the contradiction that the gravity law pulls the location of the center of gravity toward m along xr and hence the shell does nlot act as if all the mass were concentrated at the center of mass of the shell? The balance of forces require the m to see the center of gravity, AKA the cg of dM off set toward m on xr. All you've shown is that a half shell does not behave as if all the mass is located at its COM. You are taking a weighted average (as it were) dependent on r^2, so this isn't going to work. The theorem only works for spherical symmetry.
D H Posted May 11, 2009 Posted May 11, 2009 This note is in anticipation of past, present and future question regarding the physical integrity of the shell theorem. This pre-emption should cut a lot of distortion in the communications. Construct the dM volumes on the rings to spheres where the x axis terminates on the center of mass of each sphere. The plane described by the unit vector along the x axis bisects the dM thus illustrating a contradiction in the present matter. This is gobbledygook. For starters, there is only one sphere involved here. Next, before you advance to looking at the sphere as a whole you really need to understand the gravitational field induced by a ring of mass. You do not understand that result. The claim that the half hemisphere closest to m1 requires this half to contribute a greater share of the total force of this dm on m1, than that contributed by the sphere half on the adjacent side of the plane. Who made that claim? If anyone here made it, its wrong. Split a sphere in half. In fact, split a sphere any way you want into two separate parts. Now compute the gravitational force induced by the two parts of the sphere at some test point inside the sphere. No matter where you place the test point and no matter how you split the sphere into two parts, the forces from these two parts on the test point will be equal in magnitude and opposite in sign. Interesting. The theta is the angle from the COM of the sphere to dm. The circumference of the tube is 2piR, of thickness t and Rd(theta) high and when multiplied together give the volume of the rectangular tube. There is a sin(theta) in my version as the ring creeps up the shell surface.. Think of it this way. If a sin(theta) term does creep in, that means half the ring has negative mass. There is no sin(theta) term. I have asked you many times now to compute the gravitational force exerted by a thin hoop of mass on a test point mass. You have yet to do this.
J.C.MacSwell Posted May 11, 2009 Posted May 11, 2009 . Who made that claim? If anyone here made it, its wrong. Split a sphere in half. In fact, split a sphere any way you want into two separate parts. Now compute the gravitational force induced by the two parts of the sphere at some test point inside the sphere. No matter where you place the test point and no matter how you split the sphere into two parts, the forces from these two parts on the test point will be equal in magnitude and opposite in sign. Just to be clear, any time I referred to this ,it was with regard to a mass outside the sphere. I know I mentioned it with regard to a mass inside a hoop.
geistkie Posted May 15, 2009 Author Posted May 15, 2009 Let us be very clear that you are making claims here, and must address questions put to you. Asking questions of others is OK as long as you are doing that, but if not, it is simply shifting the argument, and will be considered trolling. Swansont, as the monitor bearing careful attention on this problem I would like your detailed or cursory assessment of the following to date. 1. From observation and the effects of the inverse square distance law for gravity the shell (or solid shell) half closest to the test mass m1 contributes the greater share of the total force on m than the half shell farthest from m. 2. From the above alone, the center of the force, or 'cg', cannot be located at the COM of the shell. 3. The shell does not act as if all the mass was concentrated at the shell COM; it acts as if the mass were concentrated at the cg of M which is off set from the COM in the direction of m along the mM axis. This is the correction to the shell theorem that is offered, 4. The cg of the shell and test mass system can be determined by modifying the shell integral to calculate the force on m of dm mirror image pairs where the force Fc of the dm is on a ring in the closest shell half, the force Ff from the mirror image dm on a ring in the farthest shell half. For G = 1 and m = 1. We see, the forces associated with each dM are, Fc(dM) + Ff(dM) = Fcf(2dM) = (2dM)/r^2 or r^2 = 2/Fcf, or, r = sqrt(2/Fcf) which is the location of the cg of the pair of calculated forces. The resultant cg may then be determined from the weighted average of the cg determined from each pair calculated. The projection of the resulting force onto the mM axis does not locate any concentration of mass; the projection determines the incremental force along the mM axis and for sure also points to the COM and the cg which is also on the mM axis offset in the shell half closest to m. DH means well but his exercises he had me engaged in are of no significance to the claims of this thread. The hoops I was jumping in for DH were intended, or so I surmise, to determine if I could recreate the shell maths from scratch. I used models from the Internet, but the model I focused on was different than that used by DH. My answer was the same as any other consistent model published over the years, such that F = G mM/d^2 , that says to me, the force of a shell of total mass M on a test mass m for a shell located a distance d from m". This as I have said is a correct statement and is incorrect if stated, "the force of a shell of total mass M on a test mass for the concentrated mass of the shell located a distance d from m". Merged post follows: Consecutive posts merged No, it's not. You have assumed that a half-shell behaves this way, ironically, to show that shells do not behave this way.- In developing the shgell theorem use differential spherical volumes of constant radius R" located a distance x to m with r being a unit vector along x defining the plane through the 1/2 the volume of the differential shell; Here, each differential shell segment also has the same effect on m as the total integrated effect. How is the asymmetrical distribution of mass on the shell relative to the the test mass different than the integrated distribution ? They are equivalent in form and differ by total amount..es And how do you explain the differentrial mass spheres without running into circuitous logic? What I did was to take the mass in the closest snhell half and concentrate this mass at the COM of the shell half. Then I calculated the total force on m from each shell half leaving the COM of the undivided shell fixed. The resulots of these calculations produces the contradiction. So when taking the force of a sphere of mass 1/2 M located at some point equivalent to the COM of the shell half nearest mand then calculated the force of a shell located equivaslent to the COM of the undivided shell [of mass M] and use the shell theorem concentrating the masses appropriatrely the reuslting calculation is contradictory as I have shown. All you've shown is that a half shell does not behave as if all the mass is located at its COM. You are taking a weighted average (as it were) dependent on r^2' date=' so this isn't going to work. The theorem only works for spherical symmetry.[/quote'] The spherical symmetry is only symmetric with respect to the shell mass and the COM of the shell. Regarding the force on m the hass is asymmetric as the forward half. shell must contribute the greater of the total force on m; Do this - calculate the force on m from the closest shell half using the shell integral evaluated from d - R to d. aue down. d - R to d, write this value. The n calculate the force from d to d + R and compare the two results. As each integration is using the inverse distance squared for calculating the incremental foirces the two calculations should be different. Take the weighted average of all the calculayed centers oif frce for each mirror image pair and then take the weighted average of these locations. Remember I am no quarelling with the final statement from the integration arriving at F = GmM/d^2. ae onlym saying that this statement locates the sphere only. Dh could probably coem up with a model for a thin shell cube of side 1 with the center located at d as in the sphere. I say the result will be equivalent to the shell model of the sphere. Or use cylinders, cubes, or any mass configuration rotated around the mM axis and the results of the shell theorem will be reproduced.
D H Posted May 15, 2009 Posted May 15, 2009 1. From observation and the effects of the inverse square distance law for gravity the shell (or solid shell) half closest to the test mass m1 contributes the greater share of the total force on m than the half shell farthest from m. Prove this claim, with math. Hint: You can't, because its wrong. The forces exerted by each of the two shells are equal in magnitude but have opposite signs. In fact, if you split the spherical shell into two measurable parts, the forces exerted by the two parts will be equal in magnitude and opposite in sign. Your entire argument against the shell theorem is based on this faulty assumption.
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