# Thread: Does global pull of a body vary with distance?

1. They say we can consider a massive body as if all mass is located at the centre, is it an approximation and that is true only at infinite distance ?
At a superficial inspection there are strong arguments to think that the pull on the surface must be somewhat different from the pull on the moon.
Is there any paper describing this?
Thanks

2.

3. Originally Posted by whizkid

At a superficial inspection there are strong arguments to think that the pull on the surface must be somewhat different from the pull on the moon.
What gives you this idea?

4. What tells you that if you are standing at the North pole the sum of the pull of the single masses must level out exactly at the average value?
Points nearer than the arctic are giving you almost no radial pull, points at the equator .35 , at South pole .25 , Australia , South america .30.
why should this babel of figures compensate each other exactly?
But even at a distance R from the surface the global pull cannot be the same on the Sun .

5. Originally Posted by whizkid
They say we can consider a massive body as if all mass is located at the centre.
Is it exacly so, or is it an approximation and that is true only at infinite distance ?
I understand this to be true under particular circumstances but is not fully general. A counterexample is to consider a spherical shell and a point inside that shell. The force at that point is zero rather than as if all the mass is at the centre.

6. Originally Posted by whizkid
What tells you that if you are standing at the North pole the sum of the pull of the single masses must level out exactly at the average value?
Points nearer than the arctic are giving you almost no radial pull, points at the equator .35 , at South pole .25 , Australia , South america .30.
1. The Earth is not a sphere, it is a geoid you should have learned that.
2. The Earth density is not constant, this affects the attraction force.
3. The Earth rotates, so there is centrifugal force being subtracted from the attractive force. The numbers that you are citing are adjusted for the effects of centrifugal force. You should have learned that as well, they teach it in high school.

But even at a distance R from the surface the global pull cannot be the same on the Sun
This is not a meaningful sentence, it is just nonsense.

7. As usual, it takes you some time to focus the problem.

8. Originally Posted by KJW
I understand this to be true under particular circumstances but is not fully general..
It is surely true at infinite distance, and of course we are talking of a perfect sphere.
It should be relatively simple to determine if the globall pull increases or diminishes when the object gets nearer to the surface.
My (un)educated guess is that at 10 m above the surface the pull is even smaller than 90%.

9. Originally Posted by whizkid
My (un)educated guess is that at 10 m abobe the surface the pull is even smaller than 90%.
Uneducated. More like ignorant. Don't let that stop u from spamming.

10. Originally Posted by xyzt
More like ignorant
Gratuitous, as usual.
Show your own science and wisdom, some time. If you have nothing positive, constructive to say, why do you answer my threads?
btw,
we are still waiting for your educated illuminated version of the shooting on the motorway.

11. Originally Posted by whizkid
Originally Posted by xyzt
More like ignorant
Gratuitous, as usual.
Not at all, I point out your mistakes, you should try to learn rather than continue to pile more mistakes on top of your previous mistakes.

If you have nothing positive, constructive to say, why do you answer my threads?
Your threads are full of glaring mistakes, I point them out in the hope that you'll learn from them. You should stop spamming, trying to look knowledgeable and start studying in earnest.

we are still waiting for your educated illuminated version of the shooting on the motorway.
It is a simple problem of kinematics. You should try solving it on your own sometimes.

12. Originally Posted by xyzt
You should try solving it on your own sometimes.
You are repetitive and ineffective. It is useless to denigrate mistakes if you do not give the correction.
You never give (intelligent) solutions. Janus gave one, I foretold it. If I , like all other members, could solve all problems by myself, nobody would post in this forum
if you do have a solution, put it forward.( first try to understand the problem I posed "does the pull vary with distance?")

If you have no positive contribution, please abstain!

13. "does the pull vary with distance?"
~ and the answer, yes.. is obvious because the greater distance you are from the mass.
The lesser will be it's space distortion factor ( gravity ). Try this as a mental exercise; 'If you were to tunnel into the planet would the gravity force measured be less as you descended' ?.. and yes.. because the all the mass is no longer below you..
Why are you asking ? I think you are aware of the mechanics of gravity.. You know that if we get far enough away the gravity force diminishes..
~ Some of the stars we observe are thousands of times larger than our sun. If the gravity did not diminish with distance we would be pulled into space.. The whole question is silly.. What is it you are asking ?

14. Originally Posted by astromark
The whole question is silly.. What is it you are asking ?
You should read more carefully, astromark,
of course gravity diminishes with the square of the distance, I think it is silly after so many posts to think that that is the question.
The issue here is if at different distances we can consider/calculate the pull as if it all the mass were concentrated in the centre.
Got it now?

15. @ Whizkid.
Are you trying to consider the effect of gravity as if you were on the outside of a thin shell globe?
It is the only way I can make sense of your posts.
KJW gives you a strong hint with the point interior to the sphere.
However the gravitational field inside a hollow sphere sums to zero at all internal points because the force of gravity cancels when you sum up the vectors. It does not matter where the location inside the hollow sphere is, the result you see is always a flat gravitaional field.

The case for what happens when outside of a thin shell is only slightly more complex.
You can find a treatment of this on Wikipedia.
Shell theorem - Wikipedia, the free encyclopedia

16. Originally Posted by whizkid
Originally Posted by xyzt
I point out your mistakes, y
You should try solving it on your own sometimes.
You are repetitive and ineffective. It is useless to denigrate mistakes if you do not give the correction.
I give corrections. The point is that you persist in making mistakes.

You never give (intelligent) solutions. Janus gave one, I foretold it. If I , like all other members, could solve all problems by myself, nobody would post in this forum.
If you were less impertinent , you would get more help.

Now, please stop spamming/trolling. I do not need your jeers. I posed a simple question:
The point is that you do not "post simple questions". you embelish them with your crank, personal ideas.

17. Originally Posted by whizkid
The issue here is if at different distances we can consider/calculate the pull as if it all the mass were concentrated in the centre.
Got it now?
Yes, it does. There is a simple theorem that proves that. It requires knowing calculus.

18. Originally Posted by whizkid
The issue here is if at different distances we can consider/calculate the pull as if it all the mass were concentrated in the centre.
Whizkid, I think I understand your question.

You want to know if the standard gravitation pull equation that uses the inverse square of the distances between the centroids of the two bodies produces the same value as computing of the gravitational pull by integrating the gravitation between every particle in both bodies.

The answer is that the two methods produce slightly different values.

Simply consider the sum of the gravitational pull of two masses on another mass at an average distance, r. The standard gravitational equation will use 2/r², but the integration of the separate values will use 1/(r–dr)² and 1/(r+dr)². However, 2/r² does not equal 1/(r–dr)² + 1/(r+dr)² unless dr=0.

Let's use actual numbers. r=100 and dr=1. The standard equation will use 2/r², which is 2/100², which equals 2/10,000, which is 0.0002. The integration will use 1/99 + 1/101² which equals 0.0001020304... + 0.0000980296..., which is 0.00020006001... .

More simply 1/99 + 1/101² does not equal 1/100² + 1/100².

19. Originally Posted by jrmonroe

You want to know if the standard gravitation pull equation that uses the inverse square of the distances between the centroids of the two bodies produces the same value as computing of the gravitational pull by integrating the gravitation between every particle in both bodies.
The force exerted on a test probe by two spheres is simply the (vector) resultant of the two forces. You need to apply superposition of effects. This is not what "whizkid" appears to be doing/asking.

Simply consider the sum of the gravitational pull of two masses on another mass at an average distance, r. The standard gravitational equation will use 2/r²,
The above is true only if the test probe is located on the line connecting the centers of the two spheres, at equal distance (r) from their respective centers. It is not true in the general case, for an arbitrary position of the test probe. In addition, the simple composition of vectors tells you that the resultant is 0, not 2/r².

but the integration of the separate values will use 1/(r–dr)² and 1/(r+dr)². However, 2/r² does not equal 1/(r–dr)² + 1/(r+dr)² unless dr=0.

If the test probe is off center by dr, the resultant force is 1/(r–dr)² - 1/(r+dr)², not 1/(r–dr)² + 1/(r+dr)².
In the general case:where

are the masses of the two attractive bodies
is the mass of the probe
are the vectors of position of the test probe wrt the two attractive masses

20. It can also be solved using Gauss's flux theorem for gravity
Gauss's law for gravity - Wikipedia, the free encyclopedia

21. Originally Posted by dan hunter
It can also be solved using Gauss's flux theorem for gravity
Gauss's law for gravity - Wikipedia, the free encyclopedia
By his own admission, "whizkid" cannot follow the math in the derivation. Nor can he seem to agree with the theorem applications.

22. Originally Posted by whizkid
Originally Posted by astromark
The whole question is silly.. What is it you are asking ?
You should read more carefully, astromark,
of course gravity diminishes with the square of the distance, I think it is silly after so many posts to think that that is the question.
The issue here is if at different distances we can consider/calculate the pull as if it all the mass were concentrated in the centre.
Got it now?
Gauss theorem (if you can follow the math), teaches you that you are questioning an issue that has been long settled. For mainstream science, at least.

23. Originally Posted by jrmonroe
The answer is that the two methods produce slightly different values.
Let's use actual numbers. r=100 and dr=1. The standard equation will use 2/r², which is 2/100², which equals 2/10,000, which is 0.0002. The integration will use 1/99 + 1/101² which equals 0.0001020304... + 0.0000980296..., which is 0.00020006001... .
.
Hi jrmonroe, at last one intelligent post by an intelligent member.
I made that example in my post #7 and I got to the same conclusions and I observed that at a great distance differences are negligible expecially as the points on the side of the sphere make an angle near to 0° with the line joining the two centers of mass.

My question was if those different values differ a lot when distance gets shorter , expecially when the extreme points of the sphere make an angle in the excess of 45°. I made the example of a body 100 m above the NorthPole . If that difference can reach 10% of the value, as I ventured with my wild guess.

I suppose that many of you have the capacity of simulating on the computer program the interaction of a slice of a sphere , a semicircle, with a point A lying near one end of the diameter BA.

Thanks jr monroe

24. Originally Posted by whizkid
Originally Posted by jrmonroe
The answer is that the two methods produce slightly different values.
Let's use actual numbers. r=100 and dr=1. The standard equation will use 2/r², which is 2/100², which equals 2/10,000, which is 0.0002. The integration will use 1/99 + 1/101² which equals 0.0001020304... + 0.0000980296..., which is 0.00020006001... .
.
Hi jrmonroe, at last one intelligent post by an intelligent member.
I made that example in my post #7 and I got to the same conclusions and I observed that at a great distance differences are negligible expecially as the points on the side of the sphere make an angle near to 0° with the line joining the two centers of mass.

My question was if those different values differ a lot when distance gets shorter , expecially when the extreme points of the sphere make an angle in the excess of 45°. I made the example of a body 100 m above the NorthPole . If that difference can reach 10% of the value, as I ventured with my wild guess.

I suppose that many of you have the capacity of simulating on the computer program the interaction of a slice of a sphere , a semicircle, with a point A lying near one end of the diameter BA.

Thanks jr monroe
One ignorant reinforcing the errors of another ignorant.....Gauss must be turning in his grave.

25. Originally Posted by whizkid
They say we can consider a massive body as if all mass is located at the centre.
Is it exactly so, or is it an approximation and that is true only at infinite distance?
I addressed whizkid's first question on the assumption that the entirety of a body's mass is located at the center (and apparently of zero volume). And my answer is that it's an approximation and is only true/exact at an infinite distance. At infinite distance, dr/r 0, which is the assumption he mentioned, and thus, 1/(r–dr)² + 1/(r+dr)² → 2/r².

26. Originally Posted by jrmonroe
Originally Posted by whizkid
They say we can consider a massive body as if all mass is located at the centre.
Is it exactly so, or is it an approximation and that is true only at infinite distance?
I addressed whizkid's first question on the assumption that the entirety of a body's mass is located at the center (and apparently of zero volume). And my answer is that it's an approximation and is only true/exact at an infinite distance. At infinite distance, dr/r 0, which is the assumption he mentioned, and thus, 1/(r–dr)² + 1/(r+dr)² → 2/r².
Initially I thought that you just made an honest mistake but I see that you are persisting in posting crank stuff.

27. Originally Posted by whizkid
Originally Posted by astromark
The whole question is silly.. What is it you are asking ?
You should read more carefully, astromark,
of course gravity diminishes with the square of the distance, I think it is silly after so many posts to think that that is the question.
The issue here is if at different distances we can consider/calculate the pull as if it all the mass were concentrated in the centre.
Got it now?
~ Your rudeness and arrogance are ugly and grotesque.. You do not seem to understand simple physics.. I do.
Read what has been said by all of the contributors here.. we are trying to help.. Be less abusive and blunt.. and say what you mean..
I am not a clairvoyant.. If you do not want a discussion.. don't come here. Do you understand 'forum.' I think it is you that did not read or understand my response.. as it was intended.

28. Originally Posted by astromark
~ Your rudeness and arrogance are ugly and grotesque.. You do not seem to understand simple physics.. I do.
Read what has been said by all of the contributors here.. we are trying to help.. Be less abusive and blunt.. and say what you mean..
I am not a clairvoyant.. If you do not want a discussion.. don't come here. Do you understand 'forum.' I think it is you that did not read or understand my response.. as it was intended.
Astromark, you jumped in with "The whole question is silly" in your first reply to this thread, yet you had completely missed the point. whizkid was merely pointing that out.

29. I would never post crank stuff. Maybe I made a mistake, but I don't think so.

What's wrong with my math? Please address any error in my math. I compare two units of mass at distance r to the two units of mass, one at distance r–dr and the other r+dr. Is my math wrong? True, I don't show the entire equation for the gravitational pull because I left out the common parts.

Two units of mass at distance r contribute 2/r² to the gravitational pull equation.

Two units of mass, one at distance r–dr and the other r+dr contribute 1/(r–dr)² and 1/(r+dr)², respectfully.

As an example, with r=100 and dr=1, then

2/r² = 2/100² = 2/10,000 = 0.0002

And then

1/(r–dr)² = 1/(100–1)² = 1/(99)² = 1/9,801 = 0.00010203040506070809...

and

1/(r+dr)² = 1/(100+1)² = 1/(101)² = 1/10,201 = 0.000098029604940692089...

Summing results in 0.00020006001000140018...

This does not equal 0.0002 computed from the assumption model.

Thus, the equation assuming all mass at the center is not equal to the more accurate integration by parts. The all-mass-at-center assumption only holds true when r equals infinity because then dr/r=0. and, in my simple example, 2/r² equals 1/(r–dr)² + 1/(r+dr)² because dr/r=0 and dr = r×0 = 0, and so r–dr = r and r+dr = r.

This shows whizkid's point-of-view to be correct.

30. Originally Posted by jrmonroe
I would never post crank stuff. Maybe I made a mistake, but I don't think so.

What's wrong with my math? Please address any error in my math. I compare two units of mass at distance r to the two units of mass, one at distance r–dr and the other r+dr. Is my math wrong? True, I don't show the entire equation for the gravitational pull because I left out the common parts.

Two units of mass at distance r contribute 2/r² to the gravitational pull equation.
No, they don't, I already explained what's wrong with your math in post 18.

31. So, are you saying that gravitational forces are not additive?

Simply put, two masses at a distance r will exert a non-zero gravitational force on another mass.

32. Originally Posted by jrmonroe
So, are you saying that gravitational forces are not additive?
No, what I am saying is that you are adding the forces incorrectly. They add like vectors, you are erroneously adding them like scalars.

33. I'm adding like scalar quantities because I simplified the problem to one dimension. And my example does not have a probe.

34. Originally Posted by jrmonroe
I would never post crank stuff. Maybe I made a mistake, but I don't think so.

What's wrong with my math? Please address any error in my math.
OP question:
- if we consider gravitational masses in a perfect sphere A or we condider them all at the center of the sphere O, do we get the same pull on an object B at distance D from the center?... the answers so far have been no, there is a tiny +difference .
If you do not agree, disprove it

35. Just read the Wikipedia article on the shell theorem. Newton solved this problem hundreds of years ago. Why continue to debate the issue?
In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetrical body. This theorem has particular application to astronomy.
Isaac Newton proved the shell theorem[1] and said that:

1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre.
2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.

36. Originally Posted by whizkid
if we consider gravitational masses at their real place in a perfect sphere A or we condider them all at the center of the sphere, do we get the same pull of an object B at distance D from the center? the answers so far have been no, there is a tiny +difference . If you do not agree, disprove it
The gravitational field external to a spherical mass is identical to the gravitational field that would be produced if all the mass were concentrated at the centre. The proof is in the shell theorem article.

37. Originally Posted by Harold14370
Just read the Wikipedia article on the shell theorem.

1. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.
Thanks Harold, has anyone ever found a result by actual integration? integrating a semicircle might suffice, wouldn't it?

Does that apply also to the case when the object is on the shell?

Then, in the case of a solid ball :

" A primitive function to the integrand is

and ...."

the function becomes just s
does that not influence the result?

38. Originally Posted by jrmonroe
I'm adding like scalar quantities because I simplified the problem to one dimension.
When you have only one dimension you need to subtract the scalars. You are adding them. A sign of being a crank is persisting even after you have been shown your errors.

And my example does not have a probe.
The mass acted by the two gravitational bodies is also called a "test probe".

39. Originally Posted by whizkid
Originally Posted by Harold14370
Just read the Wikipedia article on the shell theorem.

1. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell.
Thanks Harold, has anyone ever found a result by actual integration?
What do you think? That Gauss was just an ignorant, who did not know what he was doing? How arrogant (and ignorant) can you prove to be?

integrating a semicircle might suffice, wouldn't it?
No, it doesn't. You obviously do not understand the process, why don't you give up the wild ass guesses and LEARN, for a change?

Does that apply also to the case when the object is on the shell?

Then, in the case of a solid ball :

" A primitive function to the integrand is

and ...."

the function becomes just s
does that not influence the result?

I wrote that wiki entry. Now, stay away from the keyboard for a while and go take a calculus class. Stop posting until after you learn the subject.

40. Originally Posted by xyzt
I wrote that wiki entry.
Did you make that animation?

41. Originally Posted by whizkid
Originally Posted by xyzt
I wrote that wiki entry.
Did you make that animation?
No, I wrote the math. You would be smart if you started learning.

42. Okay Harold, I give up. I wasn't being a crank. You see my math. Maybe my error was oversimplifying the problem to a single dimension. My example of two bodies each with a mass of 1 doesn't come under Newton's proof, but was a lousy simplification of it.

Xyzt, I probably didn't make my example clear enough, but I think my math as-is is correct, although it's a lousy oversimplification of what whizkid needed, and I misdirected him.

A body with a unitless mass of 2 at a distance r does contribute a gravitational pull of 2/r². If the probe is measuring the gravitational force, then it is at the distance r from the body with a unitless mass of 2. In comparison, it's also at the same distance (r) from a point from which one body with a mass of 1 is dr closer to the probe and the second body with a mass of 1 is dr farther from the probe. Do you see my lousily oversimplified model?

But, yes, Gauss would roll over in his grave because I derived his so-called "Least Squares" Method without squaring or minimizing anything. In fact, squaring the errors eliminates their signs and forces the method to minimize their summation. With the errors left alone with their signs intact, their signs work as they should, and you can derive what also turns out to be the LSM. But that's another story (and thread).

Whizkid, stop banging your head against the wall and try this proof.

43. Originally Posted by xyzt
What gives you this idea?
Originally Posted by xyzt
1. The Earth is not a sphere, it is a geoid you should have learned that.
2. The Earth density is not constant, this affects the attraction force.
3. The Earth rotates, so there is centrifugal force being subtracted from the attractive force. The numbers that you are citing are adjusted for the effects of centrifugal force. You should have learned that as well, they teach it in high school.
Originally Posted by xyzt
Not at all, I point out your mistakes, you should try to learn rather than continue to pile more mistakes on top of your previous mistakes.
It is a simple problem of kinematics. You should try solving it on your own sometimes.
Originally Posted by xyzt
I wrote that wiki entry.
That's all the wisdom you produced, smartass.

You claim you wrote the wiki article and never quoted it
We had to wait 14 posts for dan hunter 34 for Harold and 41 for jrmonroe (and his great proof) to learn something.

You are just a rude bully, and a disgrace to this excellent forum.

44. Originally Posted by jrmonroe

Xyzt, I probably didn't make my example clear enough, but I think my math as-is is correct, although it's a lousy oversimplification of what whizkid needed, and I misdirected him.

A body with a unitless mass of 2 at a distance r does contribute a gravitational pull of 2/r². If the probe is measuring the gravitational force, then it is at the distance r from the body with a unitless mass of 2. In comparison, it's also at the same distance (r) from a point from which one body with a mass of 1 is dr closer to the probe and the second body with a mass of 1 is dr farther from the probe. Do you see my lousily oversimplified model?
Problem is that you wrote the force as . The correct expression is . This is the third time I point out your error to you.

45. Originally Posted by jrmonroe

Whizkid, stop banging your head against the wall and try this proof.
Thanks for the link, it is a clear article.

If I got it right the shell theorem says that if an object A is near the surface of a hollow sphere itgets the same pull it would get if the whole mass where concentrated at its center O.
I am not banging my head on the wall: before posting I made some rough calculation and that seems far from truth:
suppose we have a perfect hollow sphere of radius R = 10 and thickness 1, it has 1256.6 cubes, of mass 1,
suppose G = 100, if all mass were at O (radius = 0.0000...1), A would experience a force of 1256.6.
Are you saying that A gets the same pull from a sphere of radius = 10?

Now, tell me one more thing, if we patiently calculate the pull given by each individual cube of unit mass and sum them up, do we get a reasonable approximation? how bad can we get? 5%, 10% or more?

46. Originally Posted by whizkid
Originally Posted by jrmonroe

Whizkid, stop banging your head against the wall and try this proof.
Thanks for the link, it is a clear article.
I hope you can spare some time and have the patience to explain simply what escapes me.

If I got it right the shell theorem says that if an object A is near the surface of a hollow homogeneous sphere gets the same pull it would get if the whole mass where concentrated at its center O.
Not hollow. It is a ball. This means full, like the Earth , for example. But the shell theorem works for any value of the inner radius.

If so, I am not banging my head on the wall, but before posting I made some rough calculation and that seems very far from truth.
Once more:

suppose we have a perfect hollow sphere of radius R = 10 and thickness 1, it has 1256.6 cubes of volume 1 and mass 1 perfectly distributed.
suppose now G = 100, if all mass were at O, A would experience a force of 1256.6.
Are you saying that A gets the same pull from a sphere of radius= 10?
That is what the shell theorem teaches you.

Now tell me one more thing, if we patiently calculate the pull given by each individual cube of unit mass and sum them up, do we get a reasonable approximation? how bad can we get? 5%, 10% or more?
If you do it right, you get an exact match. No "approximation". This is what the shell theorem teaches you. You will need to learn calculus (integration) in order to understand how it works.

47. A solid sphere can be thought of as an infinite number of nested hollow balls. So it works for a solid sphere as well.

48. Xyzt, okay, different scenario.

Let's say a probe is at the origin of a single dimension and there's objects (of unitless masses of 1) at coordinates a and b, both positive numbers. Do they contribute to the gravitational pull on the probe as 1/a² and 1/b², and so, their combined pull is 1/a² + 1/b²? Because that's what I'm trying to say. In other words, a = r – dr and b = r + dr. I just don't understand the subtraction. I think now you'll see why I've been so insistent.

Whizkid, xyzt can answer these questions better than I can.

Harold, so if I went down into the Earth to a radius of r from the center, I could pretty much ignore the "shell" above radius r, which gives me zero grav pull, and just compute the grav pull from the "sphere" of radius r beneath me?

49. Xyzt knows his stuff, whizkid just can't bear being corrected...

50. Originally Posted by PhDemon
Xyzt knows his stuff, whizkid just can't bear being corrected...
He doesn't know alphabetical order.

51. Think about it 3 dimensions of space (often denominated as x,y and z) and one of time (t)...

52. Originally Posted by jrmonroe
Xyzt, okay, different scenario.

Let's say a probe is at the origin of a single dimension and there's objects (of unitless masses of 1) at coordinates a and b, both positive numbers. Do they contribute to the gravitational pull on the probe as 1/a² and 1/b², and so, their combined pull is 1/a² + 1/b²?
Only IF they are at the same side of the test probe and IF they occupied the same exact position. Which is physically impossible. As I explained to you, repeatedly (this is going to be the fourth time), if the test probe is between the two attractive masses, then the resultant is 1/a² - 1/b². Not 1/a² + 1/b².

Because that's what I'm trying to say. In other words, a = r – dr and b = r + dr. I just don't understand the subtraction. I think now you'll see why I've been so insistent.
You get r-dr and r+dr in the case the test probe is between the attractive masses. In this case you also get the resultant force as . Do you finally understand your mistakes?

Whizkid
, xyzt can answer these questions better than I can.
True.

Harold
, so if I went down into the Earth to a radius of r from the center, I could pretty much ignore the "shell" above radius r, which gives me zero grav pull, and just compute the grav pull from the "sphere" of radius r beneath me?
Also true.

53. Originally Posted by PhDemon
Think about it 3 dimensions of space (often denominated as x,y and z) and one of time (t)...
yep :-)

54. Originally Posted by PhDemon
Xyzt knows his stuff, whizkid just can't bear being corrected...
So, he makes new mistakes, in order to cover up his earlier mistakes....

55. I stopped trying to help him after he forayed into QM, he kept trying to apply classical physics after being repeatedly told it didn't work in the case he was considering and said I was "arrogant" for pointing out his misconceptions
:shrug: he seems to be carrying on in the same vein here. Even Strange, one of the most patient posters on the forum gave up trying to educate him as he has the belief anything he can't understand (which seems to include anything involving calculus) can't be correct...

56. Originally Posted by PhDemon
I stopped trying to help him after he forayed into QM, he kept trying to apply classical physics after being repeatedly told it didn't work in the case he was considering and said I was "arrogant" for pointing out his misconceptions
:shrug: he seems to be carrying on in the same vein here. Even Strange, one of the most patient posters on the forum gave up trying to educate him as he has the belief anything he can't understand (which seems to include anything involving calculus) can't be correct...
This is a syndrome of a type of mental illness, I can't remember how it is called, when one insists on debating subjects that are way out of the person's competence.

57. Originally Posted by jrmonroe
Xyzt, okay, different scenario.

Let's say a probe is at the origin of a single dimension and there's objects (of unitless masses of 1) at coordinates a and b, both positive numbers. Do they contribute to the gravitational pull on the probe as 1/a² and 1/b², and so, their combined pull is 1/a² + 1/b²? Because that's what I'm trying to say. In other words, a = r – dr and b = r + dr. I just don't understand the subtraction. I think now you'll see why I've been so insistent.

Whizkid, xyzt can answer these questions better than I can.

Harold, so if I went down into the Earth to a radius of r from the center, I could pretty much ignore the "shell" above radius r, which gives me zero grav pull, and just compute the grav pull from the "sphere" of radius r beneath me?
Harold could answer better, but yes.
You would not feel any gravitional attraction to the shell you were inside of.
You would still feel the weight of the ground above you because it would still be responding to the gravity beneath you, but you would not be feeeling any gravitaional attraction to it.

Bizarre, huh.
It took me a few times to get through this too. The idea that it did not matter how close you were to the inside wall of the shell was totally counterintuitive to my mind.

58. Xyzt, I understand you; you are correct. In my first example, the "probe" (for lack of a better word) was r–dr from the closer mass and r+dr from the farther mass, and both were on the same side of the probe.

So xyzt, let me present you a situation, and you provide the math — I will refrain from doing so.

Two objects, each with a unitless mass of 1 exist along the same straight line from a probe, one behind the other.
One object is at a unitless distance of 10, the other is at 9.
What m/r² value does each mass alone contribute to the grav pull on the probe?
What do they contribute combined?

59. Originally Posted by jrmonroe
Xyzt, I understand you; you are correct. In my first example, the "probe" (for lack of a better word) was r–dr from the closer mass and r+dr from the farther mass, and both were on the same side of the probe.
Well, if you do this, your exercise is devoid of any meaning in demonstrating anything of any interest. Why are you even choosing the distances to be ?

60. Originally Posted by jrmonroe

[color=red]Thus, the equation assuming all mass at the center is not equal to the more accurate integration by parts.[/coloe] The all-mass-at-center assumption only holds true when r equals infinity because then dr/r=0. and, in my simple example, 2/r² equals 1/(r–dr)² + 1/(r+dr)² because dr/r=0 and dr = r×0 = 0, and so r–dr = r and r+dr = r.
Ahh, I see what you are trying to prove. It falls into the category "not even wrong".
The redlined sentence is patently false, the shel theorem produces an EXACT answer. So, you are just reinforcing "whizkid"'s earlier error.

This shows whizkid's point-of-view to be correct.
Nope, it is just an example of GiGo. You put garbage in, you got garbage out. You are just reinforcing whizkid's error.

Your "counter-example" is false. You do not understand how the integration is executed in the shell theorem. Do you know integral calculus?

61. Originally Posted by pzkpfw
Originally Posted by astromark
~ Your rudeness and arrogance are ugly and grotesque.. You do not seem to understand simple physics.. I do.
Read what has been said by all of the contributors here.. we are trying to help.. Be less abusive and blunt.. and say what you mean..
I am not a clairvoyant.. If you do not want a discussion.. don't come here. Do you understand 'forum.' I think it is you that did not read or understand my response.. as it was intended.
Astromark, you jumped in with "The whole question is silly" in your first reply to this thread, yet you had completely missed the point. whizkid was merely pointing that out.
~~~ Oops and YES, yes, yes... I should not come into a running conversation when I should have been sleeping.. " I am sorry..and I did get that upside down..I was wrong.." consider it withdrawn..
~ but there is something about the manor of that contributor that prompted my error.. Oh well.. It leaves me sorry and wrong..

62. Yes, I was trying to compare the grav pull of a body of mass 2 at distance r to two bodies of mass 1 at distances r–dr and r+dr., where dr<<r, and all on the same straight line as the probe.

So, maybe you can explain the grav pull in my system of two bodies of mass 1 at distances r±dr. I'm all ears.

Yes, I know integral calculus, but I was unaware of the shell theorem until someone mentioned it here.

Aha, I see from my journal entries that I last studied astrophysics in 1686, and Newton published his Shell Theorem in 1687. Well, that explains it. —jk

63. Originally Posted by jrmonroe
Yes, I was trying to compare the grav pull of a body of mass 2 at distance r to two bodies of mass 1 at distances r–dr and r+dr., where dr<<r, and all on the same straight line as the probe.

So, maybe you can explain the grav pull in my system of two bodies of mass 1 at distances r±dr. I'm all ears.

Yes, I know integral calculus, but I was unaware of the shell theorem until someone mentioned it here.

Aha, I see from my journal entries that I last studied astrophysics in 1686, and Newton published his Shell Theorem in 1687. Well, that explains it. —jk
The problem with your "counter-example" is that it doesn't really prove your point. the Shell Theorem is exact, contrary to your attempt at proving that it is just an approximation.
I trace your failure to two things:

1. You used a unidimensional example but the integration in the Shel Theorem is a volume integral (three dimensional)

2. You used scalar addition of forces (due to the error in judgement at point 1). But the Shell Theorem relies on vector addition of forces.

The two errors taken together render your attempt as a failure.

64. Originally Posted by PhDemon
I stopped trying to help him.
Originally Posted by xyzt
This is a syndrome of a type of mental illness,.
You keep insulting, you have a few supporters, it seems.
It would be wiser to follow PhD's example, just give me a break.

65. Originally Posted by whizkid
Originally Posted by PhDemon
I stopped trying to help him.
Originally Posted by xyzt
This is a syndrome of a type of mental illness,.
You keep insulting, you have a few supporters, it seems.
It would be wiser to follow PhD's example, just give me a break.
If you stop posting crank stuff, I'll stop answering. As long as you continue posting crank stuff, I'll debunk it. This is supposed to be a science forum, there is no room for your fringe ideas. It is that simple.

66. Originally Posted by Harold14370
A solid sphere can be thought of as an infinite number of nested hollow balls. So it works for a solid sphere as well.
Hi Harold, thanks for your attention. So, the theorem does not refer to a hollow shell, but can we agree that if we consider A on a hollow shell, the pull P_10 is about half the pull A would get if all mass were at O (P_0):
P_10/P_0 = 0.5 ?

Originally Posted by xyzt
Originally Posted by whizkid
.. if we patiently calculate the pull given by each individual cube of unit mass and sum them up, do we get a reasonable approximation?
If you do it right, you get an exact match. No "approximation". .
I have patiently summed up all the individual cubes of the shell, and I got P_10 < 1

67. Originally Posted by whizkid
Originally Posted by Harold14370
A solid sphere can be thought of as an infinite number of nested hollow balls. So it works for a solid sphere as well.
- Do you agree that if we consider A on a hollow shell, the pull P_10 is about half the pull A would get if all mass were at O P_O P_10/P_O = 0.5?
- If the theorem refers to a solid ball, does it take into account that for each of the 10 shell the distance from A is different?
- As you go deeper into the shells the ratio P_x/P_O increases and reaches 1 only at the center, where distance is near infinite.
Now if you sum up all the pulls from .5 to 1, how can you ever get an average of 1 as a final result?
easy, you need to learn how to do integrals. Until you learn, you will continue to post nonsense.

68. Yes xyzt, I failed.

I oversimplified. I performed the summation of the least number of parts along a single dimension in an attempt to give the simplest explanation to someone who seemed to need an easily digestible answer.

I envisioned a sphere cut in two perpendicular to the axis that ran from the sphere's center to the probe.

I invented my own simplistic shell theorem, not knowing that Newton's Shell Theorem already existed.

69. Originally Posted by jrmonroe
Yes xyzt, I failed.
Nice to admit to error, this is a good step.

I oversimplified. I performed the summation of the least number of parts along a single dimension in an attempt to give the simplest explanation to someone who seemed to need an easily digestible answer.

I envisioned a sphere cut in two perpendicular to the axis that ran from the sphere's center to the probe.

I invented my own simplistic shell theorem, not knowing that Newton's Shell Theorem already existed.
I understand all that. The problem is that your conclusion was wrong, the Shell Theorem gives an exact answer, not an approximate one, as in whizkid's fringe misconception (he still clings to it). So, you inadvertently supported a crank. Now that you know about the Shell Theorem do you still think that it works only for infinite distance?

70. Originally Posted by whizkid
So, the theorem does not refer to a hollow shell, but
- can we agree that if we consider A on a hollow shell (post #44, r = 10) , the pull P_10 is about half the pull A would get if all mass were at O (P_O):
P
_10/P_O = 0.5?
(I have patiently summed up all the individual cubes of the shell
No. The shell theorem says it is the same as if all mass is concentrated at the center, not half. Show us how you summed up the individual cubes., and perhaps we can see where you are going wrong.

71. Originally Posted by Harold14370
Show us how you summed up the individual cubes., and perhaps we can see where you are going wrong.
Thanks Harold,
as per post #44, A is standing at the Northpole, and southpole B is 20 units away. All points P on a circle give same pull to A, (which is 100/x^2 *cos phi), but circles near the equator E have more mass.
The angle phi is at A (PAB) and AP = x = cos phi*20, so the net pull by P is (100/20 = ) 5/x

In the Southern hemisphere pull vary from .3535 (at E) to .25 at B,
http://www.wolframalpha.com/input/?i...+from+0+to+20+
taking the gratest value far all (.35*628 = 222)
In the Northern hemisphere pulls vary from .35 at E to 4 at A. If the average of this hemisphere is less than 1.645 (400pi-222)/200pi = 1.646, then the global pull is less than 1

Is there anything wrong in this procedure?
Thanks again

72. As per post 44, A is standing at the Northpole and the southpole B is 20 units away. All points P on a circle give the same pull to A, which is 100/x^2 *cos phi
I guess you mean that the pull is (100/x^2) cos phi, where x is the straight line distance from A to the mass, and phi is the angle PAB. I agree with that.
, but circles near the equator E have more mass than the others.
The angle phi is at A (PAB) and x = cos phi*20 , so the net pull by P is (100/20 = )/x
If x is the straight line distance from A to P, then why do you say x is 20 cos phi? I'm not seeing that.

73. Originally Posted by jrmonroe
Yes xyzt, I failed.
.
Not really

74. Originally Posted by whizkid
Originally Posted by Harold14370
Show us how you summed up the individual cubes., and perhaps we can see where you are going wrong.
Thanks harold, it is very kind of you to listen to my questions.

As per post 44,
A is standing at the Northpole and the southpole B is 20 units away. All points P on a circle give the same pull to A, which is 100/x^2 *cos phi, but circles near the equator E have more mass than the others.
The angle phi is at A (PAB) and AP = x = cos phi*20, so the net pull by P is (100/20 = ) 5/x

In the Southern hemisphere pull vary from .35 (at E) to .25 at B:
y= 5/x from 10sqrt2 to 20 - Wolfram|Alpha.
It is useless to do thousands of sums to get the exact average, let's concede (to prevent any objection) that all points here (1/2 of the total mass: 2pi*r^2) give the maximum pull (.35*628, P = 220)

In the Northern hemisphere pulls vary from .35 (at E) to 1 (when phi= 76°) and go from 1 to 4.2 (phi > 87°, for a fractional number of points at distance 1 from A):
y= 5/x from 0 to 10sqrt2 - Wolfram|Alpha

Granting that all points give pull 1 (P = 628)
the global pull is 1+.35/2 = .6748 P_O , P (= 628+220) = 848 / P_O (1256)
(but doing individual sums we get about [probably less than] 1/2 P_O (.3+.6/2))

Is there anything wrong in this procedure?
Thanks again
This has been explained to you:

1. The Earth is not spherical, it is a geoid (a type of ellipsoid). Therefore the attraction force varies with position on the Earth surface.

2. The force listed is the RESULTANT of the attractive force AND the centrifugal force. Therefore the attraction force varies with position on the Earth surface.

So, far from invalidating the Shell Theorem, your continued droning simply proves your ignorance and, what is worse, your refusal to learn.

75. Xyzt, I think he is just trying to get an intuitive idea of why the shell theorem works by approximating it with discrete masses. I don't see anything wrong with that. We all have different ways of learning. Yes, he could just go through the proofs using calculus which are published in Wiki and elsewhere. But if he wants to do this, what's the harm?

Whizkid, I'll look at this later. Got to go back to work.

76. Originally Posted by Harold14370
Xyzt, I think he is just trying to get an intuitive idea of why the shell theorem works by approximating it with discrete masses. I don't see anything wrong with that. We all have different ways of learning. Yes, he could just go through the proofs using calculus which are published in Wiki and elsewhere. But if he wants to do this, what's the harm?

Whizkid, I'll look at this later. Got to go back to work.
Based on my experience with him (the same experience POhDemon had), whizkid has an agenda, he's not trying to learn, he's trying to disprove mainstream science. Thesame agenda surfaced in his discussions on QM (with PhDemon) and on GR (with me and Janus). This, combined with his very obnoxious, petulant demeanor, makes his posts highly objectionable. Note how he insists on the variation of attractive force despite the fact that I explained to him that the numbers he's been rattling reflect a non spherical, rotating Earth.

77. Originally Posted by xyzt
Originally Posted by Harold14370
Xyzt, I think he is just trying to get an intuitive idea of why the shell theorem works by approximating it with discrete masses. I don't see anything wrong with that. We all have different ways of learning. Yes, he could just go through the proofs using calculus which are published in Wiki and elsewhere. But if he wants to do this, what's the harm?

Whizkid, I'll look at this later. Got to go back to work.
Based on my experience with him (the same experience POhDemon had), whizkid has an agenda, hes not trying to learn, he's trying to disprove mainstream science. This, combined with his very obnoxious, petulant demeanor, makes his posts highly objectionable. Note how he insists on the variation of attractive force despite the fact that I explained to him that the numbers he's been rattling reflect a non spherical, rotating Earth.
I think that mainstream science will withstand the onslaughts of whizkid.

78. Originally Posted by Harold14370
I think that mainstream science will withstand the onslaughts of whizkid.]
Yes, you are right :-)
But his demeanor needs serious work. he needs to drop the attitude and start learning.

79. Originally Posted by xyzt
Now that you know about the Shell Theorem do you still think that it works only for infinite distance?
It works for any distance.

Originally Posted by whizkid
It is useless to do thousands of sums to get the exact average
Whizkid, if you mean that integration is thousands of sums, then your statement here is not right. Integration transcends sums and takes the solution to a completely different level of understanding and provides an exact solution. If you want, I think I can give you an example that shows this.

I agree with Harold that we all have different ways of learning.

80. Originally Posted by jrmonroe
Originally Posted by xyzt
Now that you know about the Shell Theorem do you still think that it works only for infinite distance?
It works for any distance.
Excellent, we are good.

81. Originally Posted by whizkid
x = cos phi*20, so the net pull by P is (100/20 = ) 5/x
Now you're going to have to explain this part to me because the gravitational pull is inversely proportional to x-squared, not x.

82. Originally Posted by jrmonroe
Whizkid, if you mean that integration is thousands of sums, then your statement here is not right. Integration transcends sums and takes the solution to a completely different level of understanding and provides an exact solution. If you want, I think I can give you an example that shows this.
.
Integration may go beyond the physical limits

83. Originally Posted by Harold14370
Originally Posted by xyzt
Originally Posted by Harold14370
Xyzt, I think he is just trying to get an intuitive idea of why the shell theorem works by approximating it with discrete masses. I don't see anything wrong with that. We all have different ways of learning. Yes, he could just go through the proofs using calculus which are published in Wiki and elsewhere. But if he wants to do this, what's the harm?

Whizkid, I'll look at this later. Got to go back to work.
Based on my experience with him (the same experience POhDemon had), whizkid has an agenda, hes not trying to learn, he's trying to disprove mainstream science. This, combined with his very obnoxious, petulant demeanor, makes his posts highly objectionable. Note how he insists on the variation of attractive force despite the fact that I explained to him that the numbers he's been rattling reflect a non spherical, rotating Earth.
I think that mainstream science will withstand the onslaughts of whizkid.
Excellent, I will let you deal with the incredible mess he's creating :-)

84. Originally Posted by whizkid
What tells you that if you are standing at the North pole the sum of the pull of the single masses must level out exactly at the average value?
Basic physics.

Points nearer than the arctic are giving you almost no radial pull,
This is obviously false.

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