geistkie

When the spherically symmetric mass is located near m1 this mass sees an asymmetric distribution of mass on the shell. What an observer considers is irrelevant to the matter being considered by us. Surely you can prove that a spherically symmetric mass behaves as claimed. Be as if the mass was concentrated at the center of the shell Originally Posted by geistkie For G = 1, m = M = 1 and the shell center located at d = 10. Therefore F = 1/100 = .010. No one can complain if I see what occurs when place .5M at 9 and .5M at 11 units distances from m1 Yes. Geistkie was input in error tlo which I informed this forum' Oh so now you argue with psychology. We are wasting each others time with this kind of exchange, but you are from the Wizards's Office so you win, right? I use the two (or more_ points) to give extension to the sphere that remains unchecked because it is a single entity and is assumed to act like a concentrated mass at a point . If I used 100 points 50 in the closest ½ shell the accuracy of my calculation would be effectively perfect..

I looked at the problem and i will produce an answer. However, you must be aware that I did not intend to shift the buirden onto DH. I was attempti ng to get him to see the real effect of the calculating the forces in pairs in order to see the effect of distance on mirror imaged dM pairs. The simple re arrn gement starts with adding a cos9neta) term to account for the dM of the mirror imaged dM. asnd to see thagt each pair of calculation places tnhe greater share of the two nbody force onto the closest dM such that when all dM are calculated the force center, i.e. the cg will be located in the nearest segment of the shell. If he refused to do it , fine, I would then provide at least an outline of the algorithm needed for paired force calculation. Don't you think you might have interferred in my grand plan and misjudged my post as a possible 'shifting of burden'? Even if i was totally lost and unable to prioduce a working model what does this prove about the necessary corrections required for the shell integral?.

It isn 't wrong, it has an error that is, nevertheless an error less than the error associated with mischaracterization of the shell results; also, the methodology takes advantage of the laws of physics that includes the fact that equal masses situated at different distances from a test mass m the closest mass contributes more force on m another equal mass located farther from m. This is obvious is it not? Some say the sphere can be considered a point of concentrated mass. Well enough, even though everyone realizes the physical impossibility of this condition being realized in nature. I will get my concentrated mass of two 1/2 shells from the concentrated mass located at the center of the sphere. For G = 1, m = M = 1 and the shell center located at d = 10. Therefore F = 1/100 = .010. No one can complain if I see what occurs when place .5M at 9 and .5M at 11 units distances from m. The closest 1/2 shell force is .5/81, the other force .5/121. Adding these forces gets, .0062 + ,0041 = .0103 = Fcf, and as Fs = ,01, Fcf/Fs = 1.03. Further, for the combined force, r^2 = 97.09, r = 9.85 which is what is expected. I have been screaming that the cg is off set along the axis toward m and here the %ERROR of the two methods has been identified.. What can be pointed to that focusses on the differences in the two systems? The .01 force either resulted from the mass effect of a shell centered at d relative to m, or the .01 force resulted from the placement of the concentrated mass of the shell at d. There is one argument that resolves the matter conclusively. If m is looking at a sphere centered at d from m, then it must be concluded that my calculations using the segmented concentrated mass in two locations is what is expected from the condition that the closest 1/2 shell contributes most of the total force. If we assume a condition claimed by shell theorem enthusists that there is no difference in the contributions of force from any two 1/2 shell halves, the calculations just completed show that indeed if a sphere is assumed the shifted cg in the direction of m is consistent with this arrangement. In the concentrated mass of the shell at the COM arrangement we have only one point to consider, and for this reason alone, can it be said that the concentrated mass, besides being mathematically concentrated, we are told that the sphere must not be subjected to the "equal mass force Fc, closest to m, is greater than equal mass Ff, farther from m and where Fc > Ff". Did Newton's genius extend to the realization of this phenomenon being a fantastic exception to the universal gravitational forces of attraction doctrine. To accept the stated claim sjustained through the years, from the 17th century to the present, one must accept the supremacy of the mathematical model over physical law regarding "closest equal mass contributes the greater force", and to accept instead, that for spheres only, the mass of the shell can be considered concentrated at the COM; This is not consistent with the front 1/2 of the sphere contributing more force than the farther 1/2 sphere. I had to think about it for a spell. All the mass on the stick is equivalent to shrinking the rings to collapse onto the stick, Therefore the vector linking a point on the stick to m is stronger as the mass is the same around the ring, but the distance is greater for each dM on the stick. Priojecting the force from the shell effectively results in losing some of the force due to off axis cancelation of vector forces perpendicular to the axis. Somewhat repetitive: Thje mass on the stick is same as the total mass of a ring circling the stick, but the distance to a point on the stick is less than the distance from m to the ring. Thnerefore by distance effects alone the stick should provide a greater force. When considering that the force from the ring was decreased by the cosine projection, no such loss is seen in the stick arrangement. - all forces are preserved. For these reasons I pick the stick to contribute more force than the shell. .

Determining absolute motion with momentum measurements Two inertial platforms move parallel and relative to each other – how do observers on each frame determine the absolute velocity of the other where the relative speed of the frames is V? Each frame is equipped with momentum balls (mbs) having mottled surfaces. Along the length of each frame and facing each other is a lengthy strip of a flat mottled surface suitable for bouncing the mbs... Frame R, moving right releases a few mbs perpendicular to the mottled surface on Frame L moving left relative to R. There are three cases. 1. L is at absolute zero velocity wrt to R; hence R contributes all the velocity to the relative motion. The mbs bouncing from the side of L will maintain a position consistent with the momentum induced by the R motion. Observers on R see the mbs moving in a straight line wrt the point the mbs left the R frame. The mbs will effectively bounce up the trajectory used when moving to L, less some friction loss incurred by the bounce. If the L frame were actually moving the mbs motion induced by the momentum component transferred from L can be measured and calibrated, hence each frame can determine the absolute motion of the other. 2. R is at absolute zero velocity. The mbs trajectory is seen the same as in 1. Above. When the mbs bounce will be directed to move in the direction of the induced momentum, towards the left. 3. Both frames have absolute motion moving at the same absolute speed, the momentum would be shared equally by the R and L frame where the loss in speed wrt the R speed is measured. Ditto for L. If Vr > VL then the changes in speed relative to each can be measured. The system requires a calibrated system capable of measuring angles and relative speeds – no big deal.

Are you ta;lking about a test mass inside a shell? I was discussing a test mass m external to the shell. Merged post follows: Consecutive posts merged If I am holding a chunk of a half shell in my hand and I attach a string at the shell point normally closest to a test mass [along the mM axis] and then I attach a string at some point on the shell surface that would be the midpoint of the mass drawn on a line from the shell center to m along the mM axis.I hold the 1/2 shell in the air by the first string. The open end of the 1/2 shell is parallel with the ground facing down. The string would point along the mM axis. Now I hold the 1/2 shell up by the other string. The 1/2 shell circular opening would be perpendicular to the horizon and the string would point to the mM axis directly below. From this I know the COM of the 1/2 shell which I will use as an approximation of the mass concentrated at the COM of the 1/2 shell. I do the same for the other 1/2 shell. Now I calculate the force on m from each 'point'. Then I find the total force of both points and then find the effective cg which is not in the center of mass of the two 1/2 shells. Where the Spheres and spherical shells can be modeled this way. Let us see if you are correct. If the1/2 shells were molded into spheres then I could use the COM of the sphere. But tnhis would have a different force effect on m due to the differences in the distribution of mass relative to m than the unmodified 1/2 shell. If I find the COM by calculatioin or 'string' I get the same location for the COM. If I use the COM as a point I do so knowing the force affect of the point would be different than the unmodified 1/2 shell. However, the errors can be corrected. I almost missed your statement "that would bring about an equivalent share of the force". I do not do this. I set the shell down and calulate the force a la the shell theorem for the in tact sphere. Then I segment the shell and calculate the two 1/2 shell points. where IOget the contradiction. First I calculate the the forces of each 1/2 shell where their COM was symetrically arranged around the COM of the shell - each of the two points are located equal distances from the COM of the whole shell. The total force on m from the segmented shell is greater than the force on whose mass is assumed concentrated at the shell center. For M = 10 located at 10 from m the force on m is 10/100 = .1. The force of 1/2 M llocated at 9 is 5/81. The force from 11 is 5/121 or .062 + .041 = 1.03 > .10. I do this for the shell segmented in 1/8ths or 1/16ths the calulations tend to have the errors go to zero. I don't follow why I would "have to locate the 1/2 shell that would bring about an equivalent share of the force" farther from the COM. About a matter of old business I would like to bring up what this post hints at. You agree that the closest 1/2 shell contributes the greater share of the total force. This is seen from inspectioon for one. Try calculating the force from each dM in mirror image pairs. The calculate the total force on m from each dM pair as well as calculating the effective cg of each pair and keep a running total of the cg point of each pair. As each pair has one dM in the closest 1/2 of the shell and its mirror image mate in the farthest 1/2 of the shell integral can spew out the total force frlom all dM on the shell and the exact location of the cg. This is one method to calculate the cg and forces. The other way is to do the calculations as I have been doling in this thread and in fact this post. Merged post follows: Consecutive posts merged DH We have been talking abiout different conditions as all my test masses are external to the shell. I think you are discuyssing confitions inside the shell. I have a proposition for you as your rank of 'expert' attests and I have no doubt you are probably over qualified for a simple but most important mission which should remove a bulk of our differences. Using the shell integral as you constructed in your demonstration to me and modify the algorithim slightly that rearranges the order of calculating the dM forces on the shell only and adds one other calculated parameter as follows: 1.When calculating the force attributable to one dM on a ring, place a mirror image ring in the opposite side of the shell and calculate the force from that mirror image dM. 2. Then calculate the total force of the pair and from this 3. calculate the position of the effective center of gravity of this pair on the mM axis, then, 4. Either keep a running total of the effective centers of gravity and make one final calculation or keep an up to date running center of gravity after calculating each force pair. If you do this all will be explained to you in a manner best suited to convince you of something without my saying another word.

See below where the error introduced by segmenting the sphere is less than if the shell sphere was not clorrected. So to your statement that the shells do not behave this way needs correcting. So I have assumed only that segmenting the shell results in a more accurate location of the shell center of gravity, than the misinterpreted results of the shell theorem.. First, I have never said that using half shells as I have been doing is error free. The very fact of the geometry of the half shells creates an asymmetric distribution of mass relative to m, but the error resulting in using the segmenting paradigm decreases as the segments double. ------------------------------------------------------------ Here is a form of Gauss' law I got from the Internet. Just as I remember from good ol' school days. the "∂V is any closed surface" Integral form The integral form of Gauss' law for gravity states: Int{[over ∂V] g.dA = -4piGM where ∂V is any closed surface, dA is a vector, whose magnitude is the area of an infinitesimal piece of the surface ∂V, and whose direction is the outward-pointing surface normal (see surface integral for more details), M is the total mass enclosed within the surface ∂V. The left-hand side of this equation is called the flux of the gravitational field. Note that it is always negative (or zero), and never positive.

What is wrong with it. If it converges to limit an error, theoretically the error converges to zero. What is worng with this? Let me remind you that monitors are literally breathing down my neck. I am making every effort to answer every post and question, but I have a problem with an unembellished "It is wrong" statement. Do you see my problem? Prove it to me that I am wrong and I will admit to it. I'd rather knoew the truth than to live a life of lies.

What I do is put the sphere at 10 say. Then arbitrarily I set 9 as the COM of the closest half shell and 11 as the COM of the farthest half shell. Or if I waant to segment the shel in quarters m1 at 8.5, m2 at 9.5 m3 at 10.5 and m4 at 11.5. This moves the cg found using two 1/2 masses toward m. The more segments the more accuracy and the cg creeping toward m slows with successive doubling of the segments. I agree. Here there is a slight glitch in logic. Replacing all dM on the rings with spheres halved by the plane r unit vector along the x axis. What do you suggest here? Should we use the shell theorem to treat each differential mass as concentrated at the COM of each sphere? or would this be circuitous? Previously you mentioned something to the effect that my treating shell halves as point masses and performing the calculations as I have been doing might not be "allowed". I am being general here and if I need correcting by all means go for it. However, when calculating the force on m for each dM there is no question of using dM as a point concentration of an infinitesimal chunk of mass. Now as the order of calculating the forces is insignificant then calculating the force due to the mirror image of the nearest shell dM on a mirror image ring in the farthest half sphere poses no perturbation to the total force as resulting from a the sum of single dM calculations at a time. I am being overly stating things but I want to make a point. When calculating the pair of dMs one dM in 1/2 of the shell, the other dM in the other 1/2 of the shell. Each dM in a shell has an exact counter part in the opposite shell half, and ergo all dM are counted in the pair system. From this it must be obvious that each dM in the closest shell contributes more force than the dM mate in the mirror image ring in the farthest shell half. Therefore the shell half closest nto m contributes a greater share of the force on m than does the farthest shell half. If calculating the force in pairs only and continuing with the shell integration the results will be identical to the unmodified integral model hence why bother? If when calculkating the force on m for two dMs, one force for the dM in the closest shell half, Fc, and one force force for the farthest dM mate, or Fc, the total force of these two dM pairs is Fp = Fc + Ff. Using F = GmdM/r^2 and setting dM = 2dM, Fc + Ff = Fp = Gm(2dM)/r^2, or r = sqrt(Gm(2dM))/Fp and the cg of each pair of forces is determined from which the final cg of the mM system can be calculated by some weighted average. This algorithim, more like numerical analysis determienws the cg exactly. To repeat, each calculated pair of forces places the cg of each pair in the nearest half shell to m, which means that the shell theorem must be read as the physics dictates, that is, the shell acts as if the mass of the shell was concentrated at the cg of the mM system and where the force F of the shell whose COM is located a distance d from m. I know the risk of saying "it is so clear", but it is so clear. I cannot understand the objection to changing, not the shell maths, even without the 'paired force' modification to the shell integral, observation justifies the correction proposed here. But then when the speed of gravitational forces are integrated into the paradigm I suspect an increase in discussions on the topic of, "conservation of angular mlomentum", that will begin to push "gravity" concepts as generally understood into the file of interesting historical attempts of physical descriptions of stellar phenomena.

Using the mass points is of course incorrect, but the result is more accurate if using as many pairs of points in the shell integral and calculalting the cg of each pair (mirror image pairs). Do not misread me. I am not opposed to concentrating the mass at some point as long as it is the correct point. I calulated the force on m from each "point" of the half spheres COM. I computed the resulting force of these two points and then determined the location of the cg I arbitrarily put the sphere at 10 and for R = 2, the COM for the closest sphere is located at 9. The farthest sphere at 11. Equal widths of shell strips have equal mass, ergo for R = 2 the COM of the two half shells are located at 9 and 11. If I use 4 equal segments the mini cgs are located at 8.5, 9.5, 10.5 and 11.5 for a force total of .0105 the cg is at 9.75 or for 8 points the cg is 9.74 with a total force of .01052 So the force gets bigger the more segments used as the cg creeps toward m which is consistent with my "off set from the COM in the direction of m". First, I calculated the force on the sphere of mass M located a distance 10 from m using, for m = M = 1 the force is .01 the lowest of all calculated forces and the cg farthest from m. So using the conmcentrated mass a of the shell at the COM gives the lowest of all forces, which is consiustent with the "closest mass contributes the greater share of the forces." If the force from the shell were measured at .01 the calculations would apply as stated. F8/Fs = .0105/.01 = 1.052 or a 5% error from the ell calculatioin as a whole sphere, the magnitude of which is dependent on the scale of the variables used. Merged post follows: Consecutive posts merged Is Gauss' Law valid only for spheres? I finally got the fact that you were discussing the conditions inside the sphere which I discussed above. Merged post follows: Consecutive posts mergedJ.C.McSwell I don't know if you picked up another 'asymmetry' in the mass distribution but each dM in the shell development has an intrinsic error due to the closest/farthest problem we have been discussing. I eliminiated the gross complexities by assuming each dM was a differential sphere with a unit vector r defined along the x axis and where the resiultant plane bisects the differential sphere and where the plane is always perpendicular to x, duh, that's what thr r does right?. There is of course, the gross closest/farthest conditions we have been discussing.

H;491354]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. Each ring contains differential masses dM acting on m a distance d from the center of the shell and where the net vector force is projected along the mM axis. I merely substituted a differential sphere for the various configuations offerred in various models explaining the shell theorem. When I suggest that the differential sphere must be treated like an asymmetric distribution of mass relative to m I mean that from the inverse distance squared condition of the gravity law, masses of equal mass closer to a test mass contribute more force on m than an equal mass located farther away. Forces of attraction being confined to mass-mass interactions are not geometrically conditioned such that only spheres can be treated with shell logic and physics - spheres [shells] do not such that the closest mass to m of 1/2 a sphere contributes the same force as all the mass in the farthest half shell. This is saying that a cubic mass M equal to a shell volume acts like the closest half cube to m does contribute a greater share of force to m than the farthest half cube, or that a cylindircal shell of equal M and equal volume [with or without the cylinder hole covered] acts like the cube but the sphere is geometrically such that the shell acts as if the mass of the shell was concentrated at the COM of the shell. If the shell is so unique in the law of gravity such that for all other shapes the general rule of equal masses closest to a test mass contribute a greater share of the total force on m than equal masses further from m holds rigidly, but that for spherical shells and solid sheres the contribution of force on m ree equal from all segments of the shell - this is saying that distance of physical mass from a test mass is irrelevant and that that the pure mathematical abstracted condition of mass [as opposed to actual concentration of mass] concentrated at the shell COM is physically demonstrated Who made that claim? If anyone here made it, its wrong.

m will always be subject to the strongest force bearing down on it. If the force vectors are all aligned along the mM axis then using the point as a concentration of the half sphere mass on the mM axis is understood. The force vectors will always point along the mM axis. The only question left unanswered is should the half shell act as if all the mass were concentrated at its center? In thinking about this I stumbled a tad. Actually the half sphere doesn't act as if the mass were concentrated at the COM of the half sphere. However, using the half sphere as indicated, the cg of the mM system is more accurately located. Slicing the sphere up into 1/4 and 1/8 etc merely increases the accuracy of locating the 'net' CG. for the sphere. To get the exact location this calculatioon using mirror image pairs that are then used to calulate the effective cg of the pair can be determined exactly. You ask what allows using a point source of mass concentrated at the COM of the shell half? The same allowance for using point sources in performing the calculations in the shell integral. As stated elsewhere if the shell integral is evaluated from d-R to d and if the differential masses, say differential spheres of mass dM as used in the integral [depending on the shell model used] the limits stated should result in a force greater that when compared with the integral evaluated from d to d + R., Remember, the dM on all rings all show a different distribution of mass to m, whichj was solved by using the differential spheres, where the half closest to m contributes a greater share of force on m than the half shell farther away. Now don't argue shell theorem regarding the differential mass of spheres used as an essential element in forming the basis for the shell theorem. Merged post follows: Consecutive posts mergedOriginally Posted by D H 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. I have a serious problem here. If the force on m from the closest half shell were +1 and the force of the farthest half shell was -1, then the total sums to zero. Is this what you meant? I have a serious problem here. If the force on m from the closest half shell were +1 and the force of the farthest half shell was -1, then the total sums to zero. Is this what you meant? Gauss' theorem does not distingjuish between spheres or arbitrary shaped masses, it is concerned with the flux in/out of a a physical body. Merged post follows: Consecutive posts merged

I am merely stating the obvious that is, that the closest half shell contributes the greater share of the total force and this from the law of gravity which is conditiomned on the inverse distance squared. To say that the shell theorem on works for spheres is an arbitrary statement - that the sphere is symmetrical is true but not relative to m, where the mass distribution on the shell is asymmetrical - to assert the 'sphere only' rule requires that the law of gravitation has a separate law for spheres in thagt the closest halh shell on a sphere would contribnute equal force on m as does the farthest half shell, which is nit a tenable conclusion. Where is it proved about the ;sphere only' law.? yes I am taking the weighted average of the locations of the centers of gravity. the shell theorem conclusion is that the 'shell really acts as though the mass M of the shell were concentrated at the COM of the shell'. All of this comes from F = GmM/d^2 which states only the force on a test mass m whose COM is located a distance d from m. I have seen the words that the shell theorem only works for spherical symmetry, yet the theorem itself makes no such claims, nor is there any such conclusion inferred from the maths. SO, if spherical symmetry is the only configuration that the theorem pertains then a sphere stretched out along the x axis with (the mM axis) would provide a different resultwith both masses equal and symmetrically spread on both halves - the halves mirror each other -- would not give a similar result as the shell theorem. This is doubtful.

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 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.

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. 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.. 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. 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.

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'.

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 I don't get this. 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

I actually got this earlier, the understanding you refer to that is, the first time approximately 40 years ago. Why do you need to know this? I thought you were an expert on Newton's Shell theorem. 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]

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. 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 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.

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.