Galactic Rotation, Outboard Mass Included?

[Doc Al]

How about this: Do the exercise in post #31. It should take you about 1 minute.

If you think that your result (that the field everywhere within a thin ring is zero) is "basic physics", then you should have no trouble whatsoever in finding a standard reference stating that. After all, rings of charge are discussed all the time in electrostatics. And such a simple result should be well known, just like the result for a spherical shell is well known.

I'll wait.

 

[sankalpmittal]

Nonsense. You can add Newton's 3rd law to the list of things you don't understand.

If you can apply then Newton 3rd law everywhere then Newton could have not derived his 3rd law . To derive it he took two balls of same mass and applied equal force on them and so they covered same displacement (after collision).

(mass x)Ball 1 ---------------------><-------------------Ball 2 ( mass x)
Same force of 1 kg ms^-2 on two balls .

<-----------------------Ball1 Ball2------------------------------------>
Same distance y covered after the collision of two balls in opposite direction .

"Newton's third law is only applicable when the force applied by one body would be equal to the resistive force of another body . "

No they don't. The area covered is proportional to the square of the distance from the cone's origin. For a spherical shell, that's true. (After correcting the error you made regarding 'equal areas'.)No they don't. In the case of a ring, the mass subtended is proportional to the distance from the cone's origin. (Not the square of the distance.) But you failed to treat the boundary! You just assumed that it works just like the sphere.Why don't you actually do the calculation? Start by drawing a diagram, showing your cones.

Yes , This turns out to be surprisingly simple! We imagine the shell to be very thin, with a mass density kg per square meter of surface. Begin by drawing a two-way cone radiating out from the point P, so that it includes two small areas of the shell on opposite sides: these two areas will exert gravitational attraction on a mass at P in opposite directions. It turns out that they exactly cancel.



This is because the ratio of the areas A1 and A2 at distances r1 and r2 are given by : since the cones have the same angle, if one cone has twice the height of the other, its base will have twice the diameter, and therefore four times the area. Since the masses of the bits of the shell are proportional to the areas, the ratio of the masses of the cone bases is also . But the gravitational attraction at P from these masses goes as , and that r2 term cancels the one in the areas, so the two opposite areas have equal and opposite gravitational forces at P.



In fact, the gravitational pull from every small part of the shell is balanced by a part on the opposite side—you just have to construct a lot of cones going through P to see this. (There is one slightly tricky point—the line from P to the sphere’s surface will in general cut the surface at an angle. However, it will cut the opposite bit of sphere at the same angle, because any line passing through a sphere hits the two surfaces at the same angle, so the effects balance, and the base areas of the two opposite small cones are still in the ratio of the squares of the distances r1, r2.) Source : http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm

 

[Doc Al]

If you can apply then Newton 3rd law everywhere then Newton could have not derived his 3rd law .

Please explain this sentence. Give me an example of two balls colliding where Newton's 3rd law is not satisfied.

"Newton's third law is only applicable when the force applied by one body would be equal to the resistive force of another body . "

You may as well try to explain this sentence also.

After reading the following from your earlier post, it's hard for me to take you seriously:

If Newtons third law works everywhere then no work can be done by anybody .

And what does Newton's 3rd law have to do with the topic of this thread anyway?

At least the stuff you wrote about the shell theorem was correct. Of course I discussed that several pages ago--I even gave the same reference. Note added: Of course your 'cutting and pasting' from that website misses the point entirely, which is that the same reasoning does not apply to the thin ring.

 

[sankalpmittal]

Please explain this sentence. Give me an example of two balls colliding where Newton's 3rd law is not satisfied.

You may as well try to explain this sentence also.

After reading the following from your earlier post, it's hard for me to take you seriously:And what does Newton's 3rd law have to do with the topic of this thread anyway?

At least the stuff you wrote about the shell theorem was correct. Of course I discussed that several pages ago--I even gave the same reference.

Dude , How can you push a chair ? Answer me .

Because action>reaction .

Another example :
If someone slaps you hard then who will get more hurt ??
Of course you not the person because when person slaps his hands possesses acceleration or dv/dt . Thus the force of slap become more than the resistive force of your cheeks .

Rather difficult for you to convince the person . :)

 

[Doc Al]

Dude , How can you push a chair ? Answer me .

Because action>reaction .

Nope. "Action" always equals "reaction". You push a chair, the chair pushes back on you.

To move a chair, the net force must be non-zero. That's Newton's 2nd law. Nothing to do with the 3rd law.

Another example :
If someone slaps you hard then who will get more hurt ??
Of course you not the person because when person slaps his hands possesses acceleration or dv/dt . Thus the force of slap become more than the resistive force of your cheeks .

More nonsense. The force that your hand exerts on the cheek is equal and opposite to the force that the cheek exerts on your hand.

Same thing if you bang your thumb with a hammer--the force that hammer and thumb exert on each other is equal and opposite. The effect of the same force is different for each object--the hammer is undamaged, while your thumb is crushed. Nonetheless, Newton's 3rd law still applies, like always.

 

[sankalpmittal]

Nope. "Action" always equals "reaction". You push a chair, the chair pushes back on you.

To move a chair, the net force must be non-zero. That's Newton's 2nd law. Nothing to do with the 3rd law.


More nonsense. The force that your hand exerts on the cheek is equal and opposite to the force that the cheek exerts on your hand.

Same thing if you bang your thumb with a hammer--the force that hammer and thumb exert on each other is equal and opposite. The effect of the same force is different for each object--the hammer is undamaged, while your thumb is crushed. Nonetheless, Newton's 3rd law still applies, like always.

Thanks for your reply but please do illustrate an example mathematically .

 

[Doc Al]

Thanks for your reply but please do illustrate an example mathematically .

Illustrate what mathematically? That action equals reaction? F_a/b = -F_b/a?

 

[sankalpmittal]

Illustrate what mathematically? That action equals reaction? F_a/b = -F_b/a?

That i know
F₁=-F₂

Well ok ok ok !

No need !