geistkie

Originally Posted by geistkie

It isn 't wrong' date=' 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?

Unless you are looking at an approximation, if it has an error associated with it, it's wrong. [/quote']

You are confusing the issue. Neither you, Swansont, DH or J.C.McSwell has described with sufficient specificity what was "wrong". You take the phrase out of context then adopt an approach that prevents a rational response.

Steps: 1. Calculate the force from the shell theorem F = GmM/d^2 using convenient unit distances,. masses etc to simplify the calculations..

As the F above is an expression for force, there is no reason to claim that the "shell acts as if all the, mass was concentrated at the shell center". Why don't you just simplify the matter and specify w here in the development of the shell theorem did the concentration of mass occur? The statement says, "the force of the mass on a spherical shell of mass M located a distance d from m."

Step 2. Using the force calculated above segment the shell in equal ½ M amounts of point masses placed symmetrically with respect to the sphere location at the shell COM.. As the F expression represents only one mass entity it cannot be tested for mathematical or physical integrity, hence the need to segment the concentrated mass into two equal concentrated half shells.

Step 3. Calculate the force on m1 of the two shell halves separately.

Step 4. Calculate the total force on m1 from the two ½ shell halves.

Step 5. Calculate the cg of the segmented shell with respect to m1,

Step 6. Compare the two calculated m/1M3forces to the sphere, the segmented to ; which shows the segmented cg

 

Originally Posted by geistkie

Some say the sphere can be considered a point of concentrated mass. Well enough' date=' even though everyone realizes the physical impossibility of this condition being realized in nature. In any event I will get my concentrated mass of two 1/2 shells from the concentrated mass located at the center of the sphere.

 

Nobody says this is realized in nature. It is a mathematical realization that you get the same answer when you integrate over the volume, and thus is an allowable shortcut. A spherically symmetric mass behaves as if all the mass were concentrated at the center.

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

 

 

.

Sure we can. You haven't provided any justification for concentrating the mass at those points. You're just pulling numbers out of some dark' date=' foul region, and to almost nobody's surprise, you get the wrong answer. GIGO.[/quote']

Your concentrated mass is not even a pure mathematical abstraction as the force expression F = GmM/d^2 states only the force on m1 from the shell M located at d. However, you take the number as gospel So I simply 'liberate' the concentrated mass and cut it in two and then I place the two ½ M1 symmetrically around the COM of the shell.

Now calculating the forces of the two 1/2 shell segments can calculate the combined force and locate the system center of gravity which is off set from the COM of the sphere in the direction of m1 .

Mooeypoo, you must see that the sphere cannot be considered as a point mass for the reason that the shell has extension, it occupies space and the distribution of mass is asymmetric relative to m1, yet you insist that the forces generated is really other than it actual function – you are treating the shell as not did not obeying the universal gravitational law. You admit the shell isn't concentrated in fact, but it only acts like it is so concentrated which denies on one side of the mouth and agrees on the other side of the mouth.

BTW, the two point masses when arranged in space with extension is seen to behave normaly i.e. that the equal mass closest to m1 contributes more \force on a test particle than a mass farther away.

Posted by geistkie To accept the stated claim sustained through the years, from the 17th century to the present, one must accept the supremacy of the mathematical model over physical reality.

Originally Po

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;

 

I hate to break it to you' date=' but you have presented no physical evidence contradicting anything. All you have produced is a mathematical model, and are trying to supplant another mathematical model. However, the huge difference is that one is based on valid math, and the other is based on a supposition that is not true.[/quote']

You aren't giving me any news you are sharing an opinion.

Originally Posted by geistkie

But hasn't it been shown that concentrating the mass as some claim the shell theorem proves cannot be the case.

No' date=' because it only works for shells. Do the math.[/quote']

And you have a proof of the claim it only works for shells?

 

 

Originally Posted by geistkie

I have been sanctioned for improper answering of questions posed others as if I were distracting the flow of the questions or to avoid an embarrassing situation if proved in error here. Mr. Swansont' date=' Mr. DH, and to a lesser extent. J.C. McSwell, can you avoid irrelevant and unscientific resposes such that "you are wrong", "you don't understand the math" or, "the shell theorem only pertains to spheres?

"You're doing the math wrong" (or worse, "you aren't doing the math at all") and "the shell model only works for spherical systems" are not irrelevant nor are they unscientific.

 

 

Originally Posted by geistkie

You thjee should consider the possibility that some of the motivation and intensity in which you maintain trust in the shell theorem, as you all learned in school from professors who learned in school from, . . , you know the drill. Please do not sanction yourself for believing in a matter that has never been wrung out as throughly as our scrutinizing processs was applied and from a very unlkely beginmning where imatter began with a clear and unambiguously polar opposite of opinion as could be produced in a problem with as wide a significant scope as is the applied shell theorem being so large ranged, it has been statistically highly improbable we have extended ourselves as we have these past few days.

 

Yeah, I know the drill all too well. Someone comes along and thinks they have shot down some scientific finding, only it's based on a flawed assumption that they are unwilling to shed. Then they insinuate that scientists have been brainwashed into thinking what they do.

 

You are actually geistkiesel, right, just re-registered with a slightly shorter name?

 

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?

 

Merged post follows:

 

Let's say I have two identical masses' date=' separated by some distance, and I am at a point on the line the bisects them. They are each a distance r away from me, at an angle of theta with respect to the line between them

 

The field contribution from each is Gm/r^2 (if my mass is M, then the force is GmM/r^2), but the sideways (x) contributions cancel, and only the y-component matters.

 

The y-component force is , making the total force

 

Now, what of I look at the force as if the mass were at their CoM. We will call this m'. The distance away is , which makes the force

 

 

OMG! m' is not 2m! Why? Because there's no reason to think it will be! It works for spherical symmetry, but not in general.[/quote']

 

Nice little expression proving that you haven't been paying attention to my calculations. Just because the calculations contradict your opinion I must be in error.

 

my calculations are with the two masses in line with the test mass. 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?

Unless you are looking at an approximation' date=' if it has an error associated with it, it's wrong. [//quote']

But you have no clue of how I make my calculations and why I make them.

 

 

Originally Posted by geistkie

Some say the sphere can be considered a point of concentrated mass. Well enough' date=' 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.

Nobody says this is realized in nature. It is a mathematical realization that you get the same answer when you integrate over the volume, and thus is an allowable shortcut. A spherically symmetric mass behaves as if all the mass were concentrated at the center.

 

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 m.

 

Sure we can. You haven't provided any justification for concentrating the mass at those points. You're just pulling numbers out of some dark, foul region, and to almost nobody's surprise, you get the wrong answer. GIGO.

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..

 

 

 

 

 

Originally Posted by geistkie

To accept the stated claim sustained through the years' date=' 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;

[quote=

I hate to break it to you, but you have presented no physical evidence contradicting anything. All you have produced is a mathematical model, and are trying to supplant another mathematical model. However, the huge difference is that one is based on valid math, and the other is based on a supposition that is not true.[/quote']

What is your supposition that is true? And what is my supposition that is false?

 

Originally Posted by geistkie

But hasn't it been shown that concentrating the mass as some claim the shell theorem proves cannot be the case.

 

No' date=' because it only works for shells. Do the math.[//quote'] I have, done the math, you just don't

understand it, and you never will.

 

 

Originally Posted by geistkie

I have been sanctioned for improper answering of questions posed others as if I were distracting the flow of the questions or to avoid an embarrassing situation if proved in error here. Mr. Swansont' date=' Mr. DH, and to a lesser extent. J.C. McSwell, can you avoid irrelevant and unscientific responses such that "you are wrong", "you don't understand the math" or, "the shell theorem only pertains to spheres?

 

"You're doing the math wrong" (or worse, "you aren't doing the math at all") and "the shell model only works for spherical systems" are not irrelevant nor are they unscientific.[/quote']

Then show me the proof that they are "scientific"

 

 

 

Originally Posted by geistkie

You three should consider the possibility that some of the motivation and intensity in which you maintain trust in the shell theorem' date=' as you all learned in school from professors who learned in school from, . . , you know the drill. Please do not sanction yourself for believing in a matter that has never been wrung out as thoroughly as our scrutinizing process was applied and from a very unlikely beginning where the matter began with a clear and unambiguously polar opposite of opinion as could be produced in a problem with as wide a significant scope as is the applied shell theorem being so large ranged, it has been statistically highly improbable we have extended ourselves as we have these past few days.

 

Yeah, I know the drill all too well. Someone comes along and thinks they have shot down some scientific finding, only it's based on a flawed assumption that they are unwilling to shed. Then they insinuate that scientists have been brainwashed into thinking what they do.

 

You are actually geistkiesel, right, just re-registered with a slightly shorter name?

[/quote'] Right Geistkiesel – I pushed the register button before editing the username.

You know I am sure you mean well but you argue in generalities and non-specifics which prevents a rational response from me. I make the simple request that you leave me alone.

Merged post follows:

Let's say I have two identical masses' date=' separated by some distance, and I am at a point on the line the bisects them. They are each a distance r away from me, at an angle of theta with respect to the line between them

 

The field contribution from each is Gm/r^2 (if my mass is M, then the force is GmM/r^2), but the sideways (x) contributions cancel, and only the y-component matters.

 

The y-component force is , making the total force

 

Now, what of I look at the force as if the mass were at their CoM. We will call this m'. The distance away is , which makes the force

 

 

OMG! m' is not 2m! Why? Because there's no reason to think it will be! It works for spherical symmetry, but not in general.[/quote']

The y component you should use is ® cos(theta). making the total force

2GmMr(cos(theta))/ r^2.

You are substituting m' for the effective mass located at the midpoint of the two equal masses m. You want to equate the forces to determine m'

2GmM[cos(theta}]/r^2 = Gm /r^2 (cos(theta))^2

 

m' = 2m[cos^2(theta)]

Your OMG drama is petty –and you have done nothing to this thread except distract from any useful discussion.

 

I will finish the responses to the unaswered posts at a later time.

 

[/indent]

Why can't you do it? This is shifting the burden of proof.

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?.

Then you can't use it! If it's wrong, then you can't get the right answer by using it!

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.

The shell model gives NO error when the COM. using it. If you want to show it is in error' date=' you can't used flawed math to do so. [/quote']

 

But hasn't it been shown that concentrating the mass as some claim the shell theorem proves cannot be the case. When I segmented the point mass the two segments indicated agreement with the 'closest mass greatest force' paradigm, which means the shell theorem did not concentrate the mass mathematically at d, the center of the shell, the results of the shell theorem places the shell centered at m and the the accuracy in placing the cg is directly proportional to the number of segments chosen for calculation. There has to be some clever looping algorithm to place the cg as accurately as desired.

 

I have been sanctioned for improper answering of questions posed others as if I were distracting the flow of the questions or to avoid an embarrassing situation if proved in error here. Mr. Swansont, Mr. DH, and to a lesser extent. J.C. McSwell, can you avoid irrelevant and unscientific resposes such that "you are wrong", "you don't understand the math" or, "the shell theorem only pertains to spheres?

 

You thjee should consider the possibility that some of the motivation and intensity in which you maintain trust in the shell theorem, as you all learned in school from professors who learned in school from, . . , you know the drill. Please do not sanction yourself for believing in a matter that has never been wrung out as throughly as our scrutinizing processs was applied and from a very unlkely beginmning where imatter began with a clear and unambiguously polar opposite of opinion as could be produced in a problem with as wide a significant scope as is the applied shell theorem being so large ranged, it has been statistically highly improbable we have extended ourselves as we have these past few days.

 

We may not be listed as the 'number one whizz bang gravitationa subjects research group', but from my perspective we got a huge job completed, we the job done. For those insisting on maintaining the status quo of shell dogma, good luck and whatever personal differences may have been hinted at, no harm no foul.

It is any closed surface' date=' but there's a dot product, so you are only looking at the the component normal to the surface. If you want to get the whole (i.e. correct) answer, you must have the flux lines going through it normal to the surface. Otherwise you have to do a rather messy integral.[/quote']

 

You say the dot product gives the normal component only and that the correct answer requires the flux lines going through it be normal.

OK. I'll give this one tlo you. [/indent]

---

Merged post follows:

Consecutive posts merged

Geiestke.

 

Imagine a sphere lying with it's COM directly along the x axis from a test mass

 

Now replace the sphere with a magic stick.

 

1. The magic stick has the same COM as the sphere. It's COM is positioned in the same place as the COM of the sphere was.

 

2. The axis of the stick lies along the x axis

 

3. The magic stick has the same length as the sphere's diameter

 

4. The magic stick has the same mass as the sphere

 

5. The magic stick has the same mass distribution as the sphere in the x direction but it is magically concentrated along the x axis

 

Now the question is: Is the force of gravity between the test mass and the magic sphere greater, lesser, or the same as it was between the test mass and the sphere? Do you see the difference?

 

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.

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.

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:

Geistke. The half spheres cannot be modeled as mass points at their respective centers of mass for this purpose (the inverse square law). You agree with that, correct? Yet you are using that assumption to prove that it cannot be true for the sphere itself.

 

The half spheres, not being spheres or spherical shells, cannot be modeled this way.

Look in particular at the closer half sphere' date=' the half that is responsible for the majority share of the force. If it had to be replaced by a point mass (of equal mass) that would bring about an equivalent share of the force, this point mass would have to be positioned [b']further

[/b]

from the test mass than the center of mass of the half sphere is positioned. I don't think you realize this point. It is not so much that your logic is wrong, as it is the assumptions you are making.

Merged post follows:

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.

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.

No, it's not. You have assumed that a half-shell behaves this way, ironically, to show that shells do not behave this way.

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..

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']

 

Not quite. I have shown that using the the half shell can lead to a contradiction of the shell theorem. By this I mean that even though the 1/2 shell considered concentrated at the COM of the half shells has an error, the error is less than that of the uncorrected shell. The shell has an intrinsic error, due, of course, to the asymmetric disribution of the 1/2 shell mass relative to m.

 

If using 1/4 segments, or 1/8th segments the errors diminish rapidly as the result of the 2, 4 or 8 mass centers are calculated and where the cg slowly creeps in the direction of m as the segmentation lof the shell increases. If such a calculation was performed in the shell integral such that mirror image pairs of forces are calulated the dM closest to m contributes a greater share of force on m than the shell farther from m. All mirrore image pairs places the center of gravity in the half closest to m.

 

This last statement does not distort the shell into segmented parts and the calculations are performed on a differential level the only differfence with this model is the order of calculating the dM centered forces acting on m. This eliminates all error in calculation of the cg. Your objection vanishes when all dM in the shell are in the form of differential spheres pointed to by x terminating at the differential sphere COM defined by the plane r where r is a unit vector pointing down the x axis, the plane segmenting the differential sphere in two equal spheres where x is always perpendicular to the plane for all dM.

 

A number of persons have told me in this thread that using of 1/2 shell masses assumed concentrated at their centers of mass that the shell theorem does not work for this configuration. I never said it did. Even though the segmenting the shell into equal halves, quarters, eights etc creates an error, the error is minimized with succeeding increases in the segmentation - just like the differential the error tends to zero as the segmentation increases.

 

Taking your assessment that the half shell model is not covered by the shell theorem, then I assume you are saying is do nothing - the caclulation results in errors therefore leave the shell alone. I am sure it is because of my poor powers of communication that neither yourself, nor DH (JC MCSwell is starting to grasp the implications of my correction thesis) , see that the segmenting of the shell leads to a test of the validity of the shell theorem.Validity should be replaced by accuracy. How so?

 

Say a shell is centered at distance d = 10 with a mass M = 10. Calculating this force with m =1 and G = 1. F = 10/100 = .1. When the shell is segmented in two parts one half centered at 9 and the other at 11, the force on the closest segment is 5/81 the farthest segment is 5/121. The total force of the half shells is .1030, which has an error which would decrease if quarter, eights or 16th segments were used and where each succeeding doublin g of the segments increases accuracy considerably. I never said this was 'shell theorem' compatible, but then I've seen no proof that my system is just plain "wrong" to quote JC McSwell.

 

The ratio of the the segmented system to the shell sphere calculated force is .1030/.1 = 1.03 a 3 % error which has the effect of significantly affecting many thousands of astronomical calculations. I had an inspiration that dark matter was just swept from the astronomical literature with geistkie's correction. Maybe so. I can handle it.

 

What I have just done above is to use a cal culation assuming the mass of the shell was concentrated at the shell COM. Then by segmenting the shell I show the errors in the assumption that the mass of the shell is located at the COM of the shell. The more the number of segments, not unlike the rings distributed in the shell integral, the more accurate the determination of the cg of the shell.

 

This just dawned on me. For sure the segmenting of the shell as I have been doing introduces errors, but these errors are less than that if the shell sphere was not corrected for and even for the half shell configuration the error is lessened from using the uncorrected force using the shell theorem constraints.

---

Merged post follows:

Consecutive posts merged

I think geistkie is talking about a test mass outside the sphere, but the underlying problem is still there, and so is the need to prove the assumption mathematically.

---

Merged post follows:

Consecutive posts merged

 

 

The "sphere only" restriction is in the derivation of gauss's law, where you do the math — it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.

 

At there point there is no acceptable response other than a derivation supporting this claim or an admission that it's wrong

 

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.

 

Originally Posted by geistkie

 

 

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 want to segment the shell 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.

 

What makes you arbitrarily do this? It is wrong.

 

Segmenting further into quarters with m1 at 8.5' date=' m2 at 9.5 m3 at 10.5 and m4 at 11.5. is more wrong

 

The more segments the more wrong it gets even if it converges to limit you error. [/quote']

If the error is limited or corrected within some experimental error what is wrong? Why do I do this? I do this because the laws of physics hold that masses closest to a test mass contribute a greater share of the total force on m than equal masses located farther away. You are saying or so it appears to me that the mass in a shell cannot be segmented in a way that selectively isolates the location of the force centers seen by m looking along the mM axis.

 

Is this it?

 

What about making the caculations in the shell integral in pairs as I suggested? Does this corrupt the integral, that is to merely vary the order in which various forces attributable to various dMs in the shell is a 'wrong'? Certainly not. I am calculating the total force between paired dMs and calculating the cg of these two dMs. When all pairs are calulated and the totla florce of the shell is determined, the cg has also been determined, which is not the case ion the shell integral. I mean by this that in developing nthe shell integral, even as modleed by DH in an earlier post, does not provide for a claculation iof the shell cg , center of gravity [FORCE] as I emphacise.

 

 

 

What makes you arbitrarily do this? It is wrong.

 

Segmenting further into quarters with m1 at 8.5' date=' m2 at 9.5 m3 at 10.5 and m4 at 11.5. is [b']more wrong

[/b]

 

The more segments the

more wrong it gets

even if it converges to limit you error.

 

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.

Not sure if If I follow what you are saying exactly, but by continuously using the COMs along the x axis in your calculations you end up with higher forces, because you are generally picking a point closer to the test mass than the average distance from the test mass of all the points it represents.

If, say, your segments were an infinitesimally thin set of discs perpendicular to the x axis, and you used the COM of each disc, each point chosen to represent the disc would be the closest point on each disc to the test mass and lead to an overestimation of the force, since you continuously underestimate the effective distance.

For some shapes this effective distance is further than the COM, for others it is closer, but for the sphere it is exactly the right distance.

Geistke. The half spheres cannot be modeled as mass points at their respective centers of mass for this purpose (the inverse square law). You agree with that, correct? Yet you are using that assumption to prove that it cannot be true for the sphere itself.

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

The half spheres' date=' not being spheres or spherical shells, cannot be modeled this way.[/quote']

 

I calculated the force from the sphere M =1 = m located at 10. The Fs = .01' Next the two shell halves were calculated and using R = 1 and the COM of each sphere located at 9 and 11 for a total force of .0103 and a calculated cg of 9.85 F12/Fs = .0103/.01 = 1.03 a three % discrepancy. The location difference for the cg was .15 out of 10 or .015 a 1.5 % discrepancy.

 

Spheres and spherical shells can be modeled this way.

 

Look in particular at the closer half sphere, the half that is responsible for the majority share of the force. If it had to be replaced by a point mass (of equal mass) that would bring about an equivalent share of the force, this point mass would have to be positioned further from the test mass than the center of mass of the half sphere is positioned. I don't think you realize this point. It is not so much that your logic is wrong, as it is the assumptions you are making.

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:

Note: I was talking about points interior to the spherical shell. It is a direct consequence of Gauss' law of gravitation and the additivity of Lebesgue integrals for measurable subsets of a space.

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:

J.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.

Split a sphere in half. In fact' date=' 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.[/quote']

 

Here, like the paragraph above, an object in some location within a shell will be closer to some mass and farther away from other segments - I recognize the accepted shell theorem that states what was claimed re zero forces on the object will result inside the shell. The statement, however, that, "

these two parts on the test point will be equal in magnitude and opposite in sign

" is hard to accept for two reasons. First, the shell theorem evaluated within the sahell produces a 'zero' result, but this says nothing about the fiorces on the test mass being equal and opposite in sign - the evaluated integral for the force is simply zero.

 

Take a point on the shell axis a short distance from the midpoint of the sphere. Your statement says that all the force in the shell segment having slightly less than half the mass will nevertheless contribute a force equal in absolute value as that of the shell segment having slightly larger a quantity of mass. Further, the statement says that as the object recedes from the near midpoint along the axis the forces remain balanced in absolute value and differ only in sign. Again, this rule of physics indicates an extreme departure from the F = - GmM/x^2 for objects subject to gravitational forces where the mass is oriented in a thin spherical shell.

 

Besides the incredulity of the claim, I have never been exposed to a mathematical proof of that claim. I have, of course familiarity with then generally understood claim of the shell theorem regarding the mass ncentrated at the shell COM.

 

Do you have a reference or can you construct a math model indicating compliance with your claim?

Think of it this way. If a sin(theta) term does creep in' date=' that means half adn each according tlo gravity law are imposing attractive force on the objective 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.[/quote']

 

This thread is clear - the challenge to the statement that "the shell acts as if all the mass was concentrated at the shell COM" is simple. The statement should read that "the shell acts as if the force on m was from a sphere [shell] with COM located at d." Whether I submit to computing the force exerted by a ring of mass on a test mass m has absolutely nothing to do with the thread. All I have to do is get a shell model from the Internet and show it to you. What would be proved?

 

cAN YOU PROIV

 

DH would you please indicate to me exactly where the shell theorem jutifies itself in the conclusion people assigned to the result? The resulting exzpression after the integration is F = - GmM/d^2 says nothing about concentrated mass. The vector projection of GmdM/x^2, projects a force onto the mM axis, it does not project mass into the center of the shell. The expression is simply a term for the total force calculated from the shell integral..

 

Can you rpove your "equal and opposite theory"?

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

 

 

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?

Merged post follows:

Originally 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?

Your entire argument against the shell theorem is based on this faulty assumption.

I think geistkie is talking about a test mass outside the sphere' date=' but the underlying problem is still there, and so is the need to prove the assumption mathematically.

 

[/quote'] YUes I am talking about a test mass external to the shell.

The shell theorem says the shell acts as if all the mass was concengtrated at the sphere COM. I say the 'shell theorem' does not say the foregoing. The expression F =[ GmM/d^2 says the force acting on m from a shell [sphere] located a distance d from m. The development of the shell integral does not include the calculation of the cg, it calculates the force from a sphershell located at d.

 

Merged post follows:

 

 

Originally Posted by geistkie

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.?

The "sphere only" restriction is in the derivation of gauss's law, where you do the math — it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.

 

At there point there is no acceptable response other than a derivation supporting this claim or an admission that it's wrong

---

Merged post follows:

Consecutive posts merged

Originally 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?

Your entire argument against the shell theorem is based on this faulty assumption.

I think geistkie is talking about a test mass outside the sphere' date=' but the underlying problem is still there, and so is the need to prove the assumption mathematically.

 

[/quote'] YUes I am talking about a test mass external to the shell.

The shell theorem says the shell acts as if all the mass was concengtrated at the sphere COM. I say the 'shell theorem' does not say the foregoing. The expression F =[ GmM/d^2 says the force acting on m from a shell [sphere] located a distance d from m. The development of the shell integral does not include the calculation of the cg, it calculates the force from a sphershell located at d.

 

Merged post follows:

 

 

Originally Posted by geistkie

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.?

The "sphere only" restriction is in the derivation of gauss's law, where you do the math — it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.

 

At there point there is no acceptable response other than a derivation supporting this claim or an admission that it's wrong

---

Merged post follows:

Consecutive posts merged

Where is it proved about the ;sphere only' law.?

The "sphere only" restriction is in the derivation of gauss's law, where you do the math — it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.

 

At there point there is no acceptable response other than a derivation supporting this claim or an admission that it's wrong

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:

Otiginally by Geistkie -"Where is it proved about the 'sphere only' law.?"

The "sphere only" restriction is in the derivation of gauss's law' date=' where you do the math —

 

it's a surface integral, i.e. done over a closed surface. If you think you can replace a half-shell with a point mass at the COM, prove it mathematically.[/quote']

 

The half-shell was used as a point mass to make a point thaqt there is a contradiction if the sphere is used as a point mass in locating the cg at other than the shell COM.

use of half shell, or 1/4 or 1/8 etc shells are used to demonstrate the increase in accuracy in locating the shell cg that would resolve on 100% accuracy is calculated using the shell integral calculating mirror image pairs together with a runniung acciunt of the cg location, just like the force is a running account of forces until the integral completess.

No, it's not. You have assumed that a half-shell behaves this way, ironically, to show that shells do not behave this way.

 

 

 

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.

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.

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:

No, it's not. You have assumed that a half-shell behaves this way, ironically, to show that shells do not behave this way.-

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.

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.

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 x

r

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 x

r

.

 

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.

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.

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' 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?

Read the rules of this site. You need to answer this question if you have any hopes of keeping this thread alive.

[/indent]

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.