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