Are all masses like black holes?

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Re: Are all masses like black holes?
« Reply #60 on: February 13, 2018, 03:36:07 AM »
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It would be different. Why do you ask?

Please explain how would you apply universal law gravitation in an aforesaid situation (an object inside a half-spherical shell)? - I wanna know
In the general case you need to sum all the individual forces between each small piece of one object with each small piece of the other  object.  The problem becomes a 3-D vector integration and a "bit above my pay-grade", which doesn't say much as my pay-grade is zero.

It is a little simpler if you just want the gravitational field at one point near a fairly regular object.
Often that object can be broken up into simpler parts, like rings etc, and these summed.

In general the only practical solution is to sum all individual contributions numerically, see:
Physics Stack Exchange, Gravitational Field from Irregular Object.
And to show the complexity of the problem, there is: Physics Forums, Gravitational field strength for irregular object.

Quote from: E E K
Edit: Would the praxis of shell theorem be amenable to hemispherical shell or quasi-cross-section, disorderly holes in the shell and asymmetrical shell?

I don't think so, but maybe someone more qualified than I can tell you more.

Someone who doesn't think satellites sit on tables maybe?

https://www.theflatearthsociety.org/forum/index.php?topic=73925.msg2024607#msg2024607

Said the guy who couldn't understand what the m stands for in F=ma

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Macarios

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Re: Are all masses like black holes?
« Reply #61 on: February 13, 2018, 04:11:50 AM »
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Newton discussed his laws for gravitational fields outside of objects.
- Any source

But Newton's universal law of gravitation states that any two masses attract each other with a force equal to a constant (constant of gravitation) multiplied by the product of the two masses and divided by the square of the distance between them - I don't see anything regarding gravitational fields outsides objects.

Ok, apply your own claim on explanation WHY homogenous infinite plane doesn't change attraction on object with object's distance from the plane.

Also, WHY is gravity force inside shell zero?
Show yourself HOW is that explained.
« Last Edit: February 13, 2018, 04:14:03 AM by Macarios »
I don't have to fight about anything.
These things are not about me.
When one points facts out, they speak for themselves.
The main goal in all that is simplicity.

Re: Are all masses like black holes?
« Reply #62 on: February 13, 2018, 04:05:51 PM »
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I don't think so, but maybe someone more qualified than I can tell you more.
- You seem to be an honest guy
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Ok, apply your own claim on explanation WHY homogenous infinite plane doesn't change attraction on object with object's distance from the plane.
- I don’t have any personal claim. I just follow the laws of physics. Can you rephrase your question I didn't understand
Quote
Also, WHY is gravity force inside shell zero?
- I never said gravity force inside shell is zero – Again G = 0, when gravity force or “g” = 0

Let an object located inside the spherical shell. The acceleration due to gravity “g” of an object is much greater (say 1000 times) than the “g” of a spherical shell. The radius of the object is much smaller than the radius of shprical shell. An object may be solid sphere, semi-solid spherical shell etc - What would you think about the aforesaid?

Also, how one can determine the acceleration due to gravity “g =GM/R^2” of a spherical shell, hemispherical shell, disorderly holes in the shell and asymmetrical shell etc  - Let the mass of each = 1 kg. and Radius = 1 meter - Just wondering

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markjo

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Re: Are all masses like black holes?
« Reply #63 on: February 13, 2018, 07:18:25 PM »
*sigh*
Science is what happens when preconception meets verification.
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Re: Are all masses like black holes?
« Reply #64 on: February 13, 2018, 08:53:47 PM »
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Drive an equation for the acceleration due to gravity at any depth within the earth’s surface
Is it derived by Newton?

I think you missed one of my posts. Anyhow, place an object at any depth within the earth’s surface

There is a mass of earth Ma above the object in horizontal plane

There is a mass of earth Mb below the object in horizontal plane

Ma pulls the object upward with g1  while
Mb pulls the object downward with g2

An object starts losing "g2" while gaining g1 as it goes down till g1 = g2 or both g1 and g2 turn into full “ge” at the center of earth.

I know you wouldn’t agree with this but IMPOV this is true as the effect of g1 is missing in derivation.

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rabinoz

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Re: Are all masses like black holes?
« Reply #65 on: February 13, 2018, 11:26:59 PM »
Also, how one can determine the acceleration due to gravity “g =GM/R^2” of a spherical shell,
The spherical shell is easy. Inside g = 0 and outside g =GM/R^2.


Quote from: E E K
hemispherical shell, disorderly holes in the shell and asymmetrical shell etc  - Let the mass of each = 1 kg. and Radius = 1 meter - Just wondering
Not at all easy, each has to be down by a complicated integration process. Often this is done numerically.

But any mass of only 1 kg has infinitesimal gravitation.
On the surface of a sphere of the size you give it would be about 2.67 x 1010 N/kg (or m/s2).

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Macarios

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Re: Are all masses like black holes?
« Reply #66 on: February 13, 2018, 11:29:07 PM »
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Ok, apply your own claim on explanation WHY homogenous infinite plane doesn't change attraction on object with object's distance from the plane.
- I don’t have any personal claim. I just follow the laws of physics. Can you rephrase your question I didn't understand

If you follow the laws of physics then you will understand this:
Put small object in the center of big ball (or cube).
Left half of the ball will pull the small object to the left, right half will pull the object to the right.
Halves have same masses and inflict equal forces, each in own direction.
Forces of those halves will act on the same object, in opposite directions, and cancel each other.

Quote
Also, WHY is gravity force inside shell zero?
- I never said gravity force inside shell is zero – Again G = 0, when gravity force or “g” = 0

Let an object located inside the spherical shell. The acceleration due to gravity “g” of an object is much greater (say 1000 times) than the “g” of a spherical shell. The radius of the object is much smaller than the radius of spherical shell. An object may be solid sphere, semi-solid spherical shell etc - What would you think about the aforesaid?

Also, how one can determine the acceleration due to gravity “g =GM/R^2” of a spherical shell, hemispherical shell, disorderly holes in the shell and asymmetrical shell etc  - Let the mass of each = 1 kg. and Radius = 1 meter - Just wondering

You didn't say it, but physics did, and you say you follow laws of physics.
Inside homogenous hollow sphere acceleration g and gravitational force F is zero.

All sides of the hollow sphere pull in every direction and resultant is zero.
Imperfections or deformations of sphere will change balance and resultant will pull toward places with higher thickness or density.
Thinner or less dense parts will reduce that pull, but not completely.

The acceleration g (1000 times greater) you mention will work outside of the sphere, and be smaller and smaller as you go away from it.

Now, about that infinite plane.
It will pull small object towards it.
It will pull two objects towards two different spots, directly below each of them.
But unlike spheres, infinite plane gives constant g at all altitudes.
So, where is the center of mass of infinite plane?
How can you apply "F = GMm/d2" ?
Which d will you use?

As you can see, you don't use resultant as you would when objects are t some distance from each other.
You will use integral function that sums vectors of gravitational forces of individual parts of the plane.
Smaller the parts you choose, more of them you have, more accurate you calculate.

Laws of physics are tools.
Our mathematical explanations of nature, and ways to predict behavior, or to plan actions in order to get desired result.
As every other tool, you have to know how and where you can use it.
It is hard to tighten 12 mm nut with 9 mm wrench.
I don't have to fight about anything.
These things are not about me.
When one points facts out, they speak for themselves.
The main goal in all that is simplicity.

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markjo

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Re: Are all masses like black holes?
« Reply #67 on: February 14, 2018, 06:55:26 AM »
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Drive an equation for the acceleration due to gravity at any depth within the earth’s surface
Is it derived by Newton?
Yes.
https://www.math.ksu.edu/~dbski/writings/shell.pdf
Science is what happens when preconception meets verification.
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Besides, perhaps FET is a conspiracy too.
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Re: Are all masses like black holes?
« Reply #68 on: February 14, 2018, 07:02:19 PM »
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You didn't say it, but physics did, and you say you follow laws of physics.
Inside homogenous hollow sphere acceleration g and gravitational force F is zero.
Let's try again

Imagine a small solid sphere “A” located inside a very large hollow sphere “B” but their centers don’t coincide. The gravitational acceleration “g” of “A” is much greater than “g” of “B”.

There are two possibilities

•   Gravitational Force between two masses: Center of both “A” and “B” coincide due to universal law of gravitation

•   Gravitational Force on mass: Part of “B” located near “A” fall on the solid sphere due to less distance and greater accelerations due to gravities. The greater is the distance them the smaller is the “g”

The force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth but what makes the difference are accelerations due to gravities

You seem to sum all the individual forces and make the resultant zero but ignore their accelerations due to gravities.

I asked a question about gravity at the center of earth but you said: “it would make internal part of the mass zero” by limiting the value of R to zero in g=GM/R^2.

Gravity at the center of earth with a radius of R is not equal to zero. This is the center of gravity of earth or this is the center where gravity of whole mass of earth appears to act.

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rabinoz

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Re: Are all masses like black holes?
« Reply #69 on: February 14, 2018, 09:21:02 PM »
Quote
You didn't say it, but physics did, and you say you follow laws of physics.
Inside homogenous hollow sphere acceleration g and gravitational force F is zero.
Let's try again

Imagine a small solid sphere “A” located inside a very large hollow sphere “B” but their centers don’t coincide. The gravitational acceleration “g” of “A” is much greater than “g” of “B”.
I think that a lot of your confusion is trying to assign g's to every object. In my opinion the idea of assigning a gravitational acceleration to an object is useful only for massive  objects like the earth or the moon.

Quote from: E E K
There are two possibilities

•   Gravitational Force between two masses: Center of both “A” and “B” coincide due to universal law of gravitation

•   Gravitational Force on mass: Part of “B” located near “A” fall on the solid sphere due to less distance and greater accelerations due to gravities. The greater is the distance them the smaller is the “g”
There is no gravitational field inside the  "very large hollow sphere B" due to B. So there is no force on A or B.

Quote from: E E K
The force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth but what makes the difference are accelerations due to gravities
No, the accelerations are due to the forces. One way to calculate "g" is to calculate the gravitational force on a test mass of 1 Kg.
In fact, while "g" is usually written in units of acceleration, m/s2, but equivalent and often better units are force per unit mass or N/kg.

Quote from: E E K
You seem to sum all the individual forces and make the resultant zero but ignore their accelerations due to gravities.
There are no separate "accelerations due to gravities". The acceleration of each object is simply dye to the one force acting on each object.

Quote from: E E K
I asked a question about gravity at the center of earth but you said: “it would make internal part of the mass zero” by limiting the value of R to zero in g=GM/R^2.
As I and others have said, there is no gravitational field inside a spherical shell, due to that shell. This is true for any thickness shell.
So at the centre of the earth, at a radius of zero, there can be no mass inside.

Quote from: E E K
Gravity at the center of earth with a radius of R is not equal to zero. This is the center of gravity of earth or this is the center where gravity of whole mass of earth appears to act.
No, I'm afraid that "Gravity at the center of earth with a radius of R is" equal to zero.

The whole mass of a sphere appears to act at the centre, only for points outside the sphere.

For points inside the sphere, at radius r from the centre, the gravitational force is only that due to the mass inside radius r .
And that is what everybody else here and all the freeness say.

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Macarios

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Re: Are all masses like black holes?
« Reply #70 on: February 15, 2018, 01:35:14 AM »
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You didn't say it, but physics did, and you say you follow laws of physics.
Inside homogenous hollow sphere acceleration g and gravitational force F is zero.
Let's try again

Imagine a small solid sphere “A” located inside a very large hollow sphere “B” but their centers don’t coincide. The gravitational acceleration “g” of “A” is much greater than “g” of “B”.

There are two possibilities

•   Gravitational Force between two masses: Center of both “A” and “B” coincide due to universal law of gravitation

•   Gravitational Force on mass: Part of “B” located near “A” fall on the solid sphere due to less distance and greater accelerations due to gravities. The greater is the distance them the smaller is the “g”

The force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth but what makes the difference are accelerations due to gravities

You seem to sum all the individual forces and make the resultant zero but ignore their accelerations due to gravities.

I asked a question about gravity at the center of earth but you said: “it would make internal part of the mass zero” by limiting the value of R to zero in g=GM/R^2.

Gravity at the center of earth with a radius of R is not equal to zero. This is the center of gravity of earth or this is the center where gravity of whole mass of earth appears to act.

Inside solid hollow homogenous sphere resultant g is zero, not because G is zero.
G is always 6.67408 x 10 -11, hence the name Gravitational CONSTANT. It is mathematical factor used in formulas. Our own tools.
Reason why g = 0 inside is because opposite sides of hollow sphere cancel each other's influence on the particle inside.
One pull one way, other pull the opposite.

Derived formula you are trying to use came from integral function of individual pull from parts of body.
It is just a resultant and you have to know where and when you can use it with satisfactory accuracy.
It works outside the sphere.

When the object is inside, one side of sphere is pullin in one direction and other side is pulling in other direction.
Newton gave proof of that.
Go to THIS ARTICLE, and scroll down to "Newton's Proofs".
There's a part named "Force on a Point Inside a Hollow Sphere".

Look at Fig. 2
Particle inside the sphere is pulled by mass distributed in sphere walls, not by empty spot S in the middle of the sphere.
Outside, the whole pull acts toward one side. Resultant would be toward S and we can use F = GMm / d2.
Inside the pull is toward everywhere. All around. The formula doesn't work here.

If particle P is closer to one side of wall it is by another side pulled weaker because of distance, but harder because from same angle is pulled by more mass.
There's more mass from K to L than from I to H.
Those two factors perfectly balance one another and resultant is equal to pull from closer part of the sphere with less mass.
Pull towards J is equal and opposite to pull towards M, and in all other directions situation is similar.
And all of that acts on the same particle P.
Total pull is zero.




------------------------------------------

Quote
The force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth but what makes the difference are accelerations due to gravities

Already told you:
Force by the Earth on apple is pulling apple, not Earth.
Force by the apple on Earth is puling Earth, not apple in that opposite direction.
Earth is pulling one object, apple is pulling another object.
Forces on two different objects don't cancel each other.

It only means that one object pulling another will not remain indifferent by pull by that other object back.
Similar thing you also have in magnetic fields.
One magnet pulling another will not stay in place.
They will both "jump" to each other.
Only, heavier magnet will jump less.

That is why, if you pull me with some rope, you have to support yourself with your leg to stay where you are.
If you don't, you'll also get pulled to central point between us.
« Last Edit: February 15, 2018, 02:53:10 AM by Macarios »
I don't have to fight about anything.
These things are not about me.
When one points facts out, they speak for themselves.
The main goal in all that is simplicity.

Re: Are all masses like black holes?
« Reply #71 on: February 15, 2018, 06:47:50 PM »
Quote
Inside solid hollow homogenous sphere resultant g is zero, not because G is zero.


Case #1: Resultant "g" is zero at the center of solid hollow homogenous sphere (HS) when there is no mass M at all inside HS

Case #2: Presence of M inside HS, however, M can be more in numbers too

There are two gravitational fields inside HS

1-   Due to  HS
2-   Due to M

Now there are three conditions

The “g” of HS < M, The “g” of HS = M, The “g” of HS > M

Also, the size, mass, and shape of M can be varied and its posture & location inside HS as well

All above is self-explanatory however I can explain in there is any question

Quote
Total pull is zero.
Most physicists agreed that gravities do not cancel each other even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero.

Mass closer to one side of the wall of HS: it depends upon the “g” of M and "g" of the mass of HS closer to the M. We know “g” increases with a decrease in distance.

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rabinoz

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Re: Are all masses like black holes?
« Reply #72 on: February 15, 2018, 07:41:20 PM »
Quote
Inside solid hollow homogenous sphere resultant g is zero, not because G is zero.


Case #1: Resultant "g" is zero at the center of solid hollow homogenous sphere (HS) when there is no mass M at all inside HS

Case #2: Presence of M inside HS, however, M can be more in numbers too

There are two gravitational fields inside HS

1-   Due to  HS
2-   Due to M
No, as has been presented numerous times and with proof.
"Inside solid hollow homogenous sphere resultant g" due to the HS is zero everywhere.

Quote from: E E K
Now there are three conditions

The “g” of HS < M, The “g” of HS = M, The “g” of HS > M
Since the “g” of HS inside the HS is everywhere zero, obviously the only case is “g” of HS < M.

Quote from: E E K
Also, the size, mass, and shape of M can be varied and its posture & location inside HS as well

All above is self-explanatory however I can explain in there is any question
There is no question at all the total pull is zero.

Quote from: E E K
Quote
Total pull is zero.
Most physicists agreed that gravities do not cancel each other even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero.

Mass closer to one side of the wall of HS: it depends upon the “g” of M and "g" of the mass of HS closer to the M. We know “g” increases with a decrease in distance.
Where did you drag this up from, "Most physicists agreed that gravities do not cancel each other even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero."
No physicist would have made a statement like that.

Gravitation has both magnitude and direction (a vector) so, what you call "gravities" most certainly can cancel.
That is how there is no gravity inside a hollow sphere, due to that sphere.

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Macarios

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Re: Are all masses like black holes?
« Reply #73 on: February 15, 2018, 10:37:48 PM »
Quote
Inside solid hollow homogenous sphere resultant g is zero, not because G is zero.


Case #1: Resultant "g" is zero at the center of solid hollow homogenous sphere (HS) when there is no mass M at all inside HS

Case #2: Presence of M inside HS, however, M can be more in numbers too

There are two gravitational fields inside HS

1-   Due to  HS
2-   Due to M

Now there are three conditions

The “g” of HS < M, The “g” of HS = M, The “g” of HS > M

Also, the size, mass, and shape of M can be varied and its posture & location inside HS as well

All above is self-explanatory however I can explain in there is any question

Quote
Total pull is zero.
Most physicists agreed that gravities do not cancel each other even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero.

Mass closer to one side of the wall of HS: it depends upon the “g” of M and "g" of the mass of HS closer to the M. We know “g” increases with a decrease in distance.

Mass on one side is closer, but there is more mass on another side.

If you don't understand English, try to pretend that you understand Hindi.
It might seem strange, but I'm sure you will understand this guy very well.

I'm not joking.
I don't speak Hindi, but I see what is he talking about.
Give it a try:

I don't have to fight about anything.
These things are not about me.
When one points facts out, they speak for themselves.
The main goal in all that is simplicity.

Re: Are all masses like black holes?
« Reply #74 on: February 16, 2018, 11:20:16 PM »
Case #1: Resultant "g" is zero at the center of solid hollow homogenous sphere (HS) in the absence of mass both inside and outside

The whole mass of HS is concentrated at its center as mentioned many times earlier but if we examine the pull of individual particle of HS then the total pull between any two particles located opposite to each other on the inner wall of HS is zero and hence the resultant “g” zero at the center of particle only IMPOV

Quote
Mass on one side is closer, but there is more mass on another side.
There are three masses and three centers of gravities “cg”
Closer mass is gravitating mass if g of HS > g of particle
More mass is gravitating mass if g of HS > g of particle
Particle – Falling mass

So total pull may be zero if

The value of “g” of closer mass at a height h from its “cg” to center of particle =  The value of “g” of more mass at a height h1 from its “cg” to center of particle

But the value of “g” of particle at height h or at the ”cg” of closer mass is greater than the value of “g” of particle at height h1 or at the “cg” of more mass

No need to explain the following as its quite easy to understand

Case #2: Presence of M inside HS, however, M can be more in numbers too
There are two gravitational fields inside HS
1-   Due to  HS
2-   Due to M
Now there are three conditions
The “g” of HS < M, The “g” of HS = M, The “g” of HS > M
Also, the size, mass, and shape of M can be varied and its posture & location inside HS as well

Most physicists agreed that gravities do not cancel each other even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero.
Mass closer to one side of the wall of HS: it depends upon the “g” of M and "g" of the mass of HS closer to the M. We know “g” increases with a decrease in distance.
« Last Edit: February 17, 2018, 06:30:16 AM by E E K »

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rabinoz

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Re: Are all masses like black holes?
« Reply #75 on: February 17, 2018, 02:17:20 AM »
Case #1: Resultant "g" is zero at the center of solid hollow homogenous sphere (HS) in the absence of mass both inside and outside
It's no point you saying the same ting over and over again. Whether you accept it or not,
"Resultant g is zero" everywhere inside a "hollow homogenous sphere (HS) in the absence of mass both inside and outside".

There are numerous references to this well know property of a spherical shell, here is one:
Quote
Gravity Inside a Spherical Shell
For application of the law of gravity inside a uniform spherical shell of mass M, a point is chosen on the axis of a circular strip of mass. The problem is envisioned as dividing an infinitesemally thin spherical shell of density σ per unit area into circular strips of infinitesemal width. The choice of such a point involves no loss of generality because for any point inside the shell, the mass elements could be chosen so that the point is on their symmetry axis.
<< I'll let you look up the calculations if you feel qualified. >>
From: Gravity Force Inside a Spherical Shell
You could also look at: Wikipedia, Shell theorem or Quora.com Why is gravity inside a spherical shell considered to be zero?

Quote from: E E K
Most physicists agreed that gravities do not cancel each other
No! That is completely untrue. Gravitation has direction and so the gravitation from two object's can cancel.

Quote from: E E K
even if it does then again you are making “g” zero at the center of particle P inside the HS and hence G = 0 at the center of particles. Similarly, the weight of the imaginary earth on our real earth is also = 0; Two equal and opposite pulls so total pull is zero.
Mass closer to one side of the wall of HS: it depends upon the “g” of M and "g" of the mass of HS closer to the M. We know “g” increases with a decrease in distance.
The bit you are ignoring is that the closer part is smaller in area than the farther part by exactly the correct ratio for their gravitational forces to be equal and opposite.

Look, go and read the references that hopefully explain it better than I.

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Macarios

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Re: Are all masses like black holes?
« Reply #76 on: February 17, 2018, 04:29:59 AM »
Your link to Quora points to pretty clear image of distribution of forces:

I don't have to fight about anything.
These things are not about me.
When one points facts out, they speak for themselves.
The main goal in all that is simplicity.

Re: Are all masses like black holes?
« Reply #77 on: February 17, 2018, 10:30:54 PM »
Quote
Your link to Quora points to pretty clear image of distribution of forces:
- This is one way where is particle pull?

Following is the edit that I had done in the previous post. here it is if missed

Particle mass is Gravitating mass
Closer mass of HS is Falling mass
More mass of HS is Falling mass

But the value of “g” of particle at height h or at the ”cg” of closer mass is greater than the value of “g” of particle at height h1 or at the “cg” of more mass

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Macarios

  • 1190
Re: Are all masses like black holes?
« Reply #78 on: February 18, 2018, 02:45:41 AM »
Quote
Your link to Quora points to pretty clear image of distribution of forces:
- This is one way where is particle pull?

Following is the edit that I had done in the previous post. here it is if missed

Particle mass is Gravitating mass
Closer mass of HS is Falling mass
More mass of HS is Falling mass

But the value of “g” of particle at height h or at the ”cg” of closer mass is greater than the value of “g” of particle at height h1 or at the “cg” of more mass

That makes you closer to understanding why resultant "g" is zero inside homogenous hollow sphere.
Center of mass of one side is closer to inner particle, but mass of that side is smaller.
Center of mass of other side is farther from inner particle, but mass of that part is bigger.
Pull between the particle and one side in one direction becomes equal to pull between the particle and other side in opposite direction.
I don't have to fight about anything.
These things are not about me.
When one points facts out, they speak for themselves.
The main goal in all that is simplicity.

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rabinoz

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  • Real Earth Believer
Re: Are all masses like black holes?
« Reply #79 on: February 18, 2018, 02:54:46 AM »
Quote
Your link to Quora points to pretty clear image of distribution of forces:
- This is one way where is particle pull?

Following is the edit that I had done in the previous post. here it is if missed

Particle mass is Gravitating mass
Closer mass of HS is Falling mass
More mass of HS is Falling mass

But the value of “g” of particle at height h or at the ”cg” of closer mass is greater than the value of “g” of particle at height h1 or at the “cg” of more mass
Here is another attempt at explaining that there is no "gravity" inside a spherical shel, due to thst shell:
Quote from: Physics Stack Exchange
Gravitational field intensity inside a hollow sphere
It is quite easy to derive the gravitational field intensity at a point within a hollow sphere. However, the result is quite surprising. The field intensity at any point within a hollow sphere is zero.

Answer from Hypnosifl
One intuitive way I've seen to think about the math is that if you are at any position inside the hollow spherical shell, you can imagine two cones whose tips are at your position, and which both lie along the same axis, widening in opposite direction. Imagine, too, that they both subtend the same solid angle, but the solid angle is chosen to be infinitesimal. Then you can consider the little chunks of matter where each cone intersects the shell, as in the diagram on this page:

One intuitive way I've seen to think about the math is that if you are at any position inside the hollow spherical shell, you can imagine two cones whose tips are at your position, and which both lie along the same axis, widening in opposite direction. Imagine, too, that they both subtend the same solid angle, but the solid angle is chosen to be infinitesimal. Then you can consider the little chunks of matter where each cone intersects the shell, as in the diagram on this page:

You still need to do a bit of geometric math, but you can show that the area of each red bit is proportional to the square of the distance from you (the blue point) to it--and hence the mass of each bit is also proportional to the square of the distance, since we assume the shell has uniform density. But gravity obeys an inverse-square law, so each of those two bits should exert the same gravitational pull on you, but in opposite directions, meaning the two bits exert zero net force on you. And you can vary the axis along which the two cones are drawn so that every point on the surface of the shell ends up being part of a pair like this, which leads to the conclusion that the entire spherical shell exerts zero net force on you.

Physics Stack Exchange, Gravitational field intensity inside a hollow sphere

But,  if you can't,  or won't do the maths yourself and won't believe the references from those that can, I don't know how to explain it any further.

Re: Are all masses like black holes?
« Reply #80 on: February 19, 2018, 11:58:47 AM »
Hope you may convince in my last try

NEWTON LAW OF GRAVITATION: The gravitational force BETWEEN two masses = F GMm/d^2

SPLIT ANALYSIS of Earth and Apple:

When an Earth is a gravitating mass and an Apple is a falling mass
Acceleration due to gravity of earth; ge = GM/R^2 (on the the surface of earth); where R = radius of earth.
So an apple fall on earth at the rate of ge

When an Apple is a gravitating mass and an Earth is a falling mass
Acceleration due to gravity of apple; ga = Gm/r^2 (on the surface of apple); where r = radius of apple
So an earth falls on an apple at the rate of ga

Force of Earth on Apple = Force of Apple on Earth
An apple falls due to ge on Earth. The earth also moves upwards towards apple due to ga of apple but by such a minuscule amount to be noticed or measured.

Shell Theorem

The presence of any mass M inside homogeneous Hallow Sphere HS. Two gravitational accelerations “g” are involved in this problem

1-   Acceleration due to gravity “g” of HS
2-   Acceleration due to gravity “g” of M

Three possible conditions

The “g” of HS < M, The “g” of HS = M, The “g” of HS > M

Also, the size, mass, and shape of M can be varied and its location inside HS as well

According to shell theorem: The entire spherical shell exerts zero net force on M

This may be true as I said earlier but this is not the end of story. There is mammoth difference between

“Gravitational Force ON a mass and Gravitational Force BETWEEN two masses”

Newton’s gravitational force is between two masses. Gravitational force on a mass is only considered during split analyses like “Earth on apple” and “Apple on earth”. Their combined effect appears in F = GMm/d^2. As gravitational force is a force that attracts any objects with mass therefore HS attracts M but M also attracts HS.

The miscalculation in shell theorem is that HS attracts M but M doesn’t attract HS.

SPLIT ANALYSIS of HS amd M

When HS is a gravitating mass and M is a falling mass
Mass on one side is closer, but there is more mass on another side.

There are three masses and three centers of gravities “cg”

A represents the “cg” of Closer mass
B represents the “cg” More mass
C represents the “cg” of M

The distance between A and C is h
The distance between B and C is h1

Obviously, h1 > h. The entire spherical shell exerts zero net force on M – OK for the sake of arguments BUT the entire M also exerts a force on Closer mass as well as More mass of the HS. Here

When M is a gravitating mass and HS is a falling mass
Closer mass of HS falls on M with “g” of M
More mass of HS also falls on M with “g” of M
As h < h1 therefore the value of "g" of M at A > the value of "g" of M at B. Since net pull of M on two different parts of HS is not zero, therefore, Closer mass falls on M with little bit resistance from more mass of HS

Conclusion: Accelerations due to gravities of Closer mass and More mass of HS may be cancelled at the center of M BUT the value of “g” of M at A is greater than the value of “g” of M at B due to the difference in heights therefore movement happens because of unbalance pull of M on two parts of the HS
« Last Edit: February 19, 2018, 12:05:03 PM by E E K »

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Macarios

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Re: Are all masses like black holes?
« Reply #81 on: February 19, 2018, 12:53:30 PM »
If feather and hammer fall together, they hit Earth together, because they have combined pull and Earth intercepts them together.
If they fall in succession, one and then another, hammer would in deed fall a tiny bit faster, but that tiny bit is many times tinier than our measuring abilities.
Mathematically it can be shown, but there is no way (not even slightest chance) to measure such small difference in reality.
I don't have to fight about anything.
These things are not about me.
When one points facts out, they speak for themselves.
The main goal in all that is simplicity.

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Shifter

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Re: Are all masses like black holes?
« Reply #82 on: February 19, 2018, 02:12:50 PM »
If feather and hammer fall together, they hit Earth together, because they have combined pull and Earth intercepts them together.
If they fall in succession, one and then another, hammer would in deed fall a tiny bit faster, but that tiny bit is many times tinier than our measuring abilities.
Mathematically it can be shown, but there is no way (not even slightest chance) to measure such small difference in reality.

Wrong
That's like saying someone with an open parachute and someone without a parachute will land at the same time.

Aerodynamics and the much lighter weight of the feather could result in the feather being picked up by the wind and not landing somewhere else a long time after.

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Re: Are all masses like black holes?
« Reply #83 on: February 19, 2018, 03:52:51 PM »
Quote
If feather and hammer fall together, they hit Earth together, because they have combined pull and Earth intercepts them together.
If they fall in succession, one and then another, hammer would in deed fall a tiny bit faster, but that tiny bit is many times tinier than our measuring abilities.
Mathematically it can be shown, but there is no way (not even slightest chance) to measure such small difference in reality.
- i think you missed one of my previous posts regarding hammer and feather

Try feather and hammer separately from the same altitude

case#1: Drop feather from the same height h on earth
Split Analysis of Feather and Earth

Feather falls on earth due to ge of earth
Earth falls on feather due gf of feather - Earth can be seen from feather

Case #2 Drop hammer from the same height h on earth
Split Analysis of Hammer and Earth

Hammer falls on earth due to ge of earth
Earth falls on Hammer due to gh of Hammer - Earth can be seen from Hammer

Since ge > gh > gf therefore earth falls on hammer at greater accelaration than falls on feather - quite simple

Earth moves towards hammer if both hammer and feather drop at the same time from the same altitude on antipodes 
« Last Edit: February 19, 2018, 04:06:02 PM by E E K »

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rabinoz

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Re: Are all masses like black holes?
« Reply #84 on: February 19, 2018, 04:39:16 PM »
Since ge > gh > gf therefore earth falls on hammer at greater accelaration than falls on feather - quite simple

Earth moves towards hammer if both hammer and feather drop at the same time from the same altitude on antipodes
Would you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?

All you talk of separate ge , gh and gf is a totally ridiculous waste of time because the typical masses would be:
  • Earth: 5.972 Χ 1024 kg,
  • Hammer: 2 kg and
  • Pigeon Tail Feather: 50 mg.
If you have any concept of relative values you  might get the message.

PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.

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Shifter

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Re: Are all masses like black holes?
« Reply #85 on: February 19, 2018, 05:46:14 PM »
Since ge > gh > gf therefore earth falls on hammer at greater accelaration than falls on feather - quite simple

Earth moves towards hammer if both hammer and feather drop at the same time from the same altitude on antipodes
Would you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?

All you talk of separate ge , gh and gf is a totally ridiculous waste of time because the typical masses would be:
  • Earth: 5.972 Χ 1024 kg,
  • Hammer: 2 kg and
  • Pigeon Tail Feather: 50 mg.
If you have any concept of relative values you  might get the message.

PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.

Where is it falling? Obviously must be gravitationally affected by something which in turn means the object with heavier mass is pulling on that object more than the object with a lighter mass. Also the object with a lighter mass will be gravitationally affected by the falling object that is heavier. An infinitesimal amount sure but enough that God himself would consider it a variable in the experiment. The idea of experiments is to eliminate the variables
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Twerp

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Re: Are all masses like black holes?
« Reply #86 on: February 19, 2018, 06:00:21 PM »
Since ge > gh > gf therefore earth falls on hammer at greater accelaration than falls on feather - quite simple

Earth moves towards hammer if both hammer and feather drop at the same time from the same altitude on antipodes
Would you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?

All you talk of separate ge , gh and gf is a totally ridiculous waste of time because the typical masses would be:
  • Earth: 5.972 Χ 1024 kg,
  • Hammer: 2 kg and
  • Pigeon Tail Feather: 50 mg.
If you have any concept of relative values you  might get the message.

PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.

And if they were being dropped from the same location at the same time, their combined mass would act on the earth so they would still hit the earth simultaneously. EEK thinks he has discovered something noteworthy but it is in fact extraordinarily insignificant.

Re: Are all masses like black holes?
« Reply #87 on: February 19, 2018, 07:13:33 PM »
Quote
Whatever you claim Newton's Shell Theorem is correct
Quote
EEK thinks he has discovered something
- Neither claim nor discovery but its just a truth

Quote
PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.
What do you think if the "g" of M >>>>>>> the "g" of HS? For Example, Earth (ideally sphere) inside a homogeneous HS. The "g" of earth >>>>>>>> the "g" HS

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rabinoz

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Re: Are all masses like black holes?
« Reply #88 on: February 19, 2018, 08:52:09 PM »
Since ge > gh > gf therefore earth falls on hammer at greater accelaration than falls on feather - quite simple

Earth moves towards hammer if both hammer and feather drop at the same time from the same altitude on antipodes
Would you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?

All you talk of separate ge , gh and gf is a totally ridiculous waste of time because the typical masses would be:
  • Earth: 5.972 Χ 1024 kg,
  • Hammer: 2 kg and
  • Pigeon Tail Feather: 50 mg.
If you have any concept of relative values you  might get the message.

PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.

Where is it falling? Obviously must be gravitationally affected by something which in turn means the object with heavier mass is pulling on that object more than the object with a lighter mass. Also the object with a lighter mass will be gravitationally affected by the falling object that is heavier. An infinitesimal amount sure but enough that God himself would consider it a variable in the experiment. The idea of experiments is to eliminate the variables
I did ask E E K, "Would you now calculate how long it would take a hammer of mass 1 kg and a feather of mass 0.05 g to fall 10 m in a perfect vacuum?"

But no, "The idea of experiments is" not "to eliminate the variables", but to find the magnitude of their effects.
Maybe that does eliminate the variable, but it might also simply mean that effect of the variable is too small for that experimental method to resolve.

Exactly that happened to the early proposals for the heliocentric solar system.
Aristarchus of Samos proposed it, but it was rejected because it was reasoned, correctly, that the annual motion of the sun should cause stellar parallax. As no stellar parallax could be observed, heliocentrism was rejected.
Again, Copernicus proposed heliocentrism, but it was rejected, at least by Tycho Brahe because even with his far more accurate measurements he still observed no stellar parallax.

But this stellar parallax was finally observed and was far smaller than Tycho Brahe could have ever measured without a telescope. The largest stellar parallax of only of 0.772-arcsec is for Proxima Centauri.

One reason for doing such calculations is to get some idea of what precision is required in the experiment.
Of course neither Aristarchus of Samos nor Tycho Brahe could have made these calculations, but for these fall times, we can.

Possibly a simple manual drop an a stop-watch would be adequate.
Then maybe a sophisticated electronically controlled drop mechanism coupled to a high precision electronic timer might be needed.

 ;D :D So, maybe you could calculate those times and let's know what equipment might be needed to measure the times. :D ;D

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rabinoz

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Re: Are all masses like black holes?
« Reply #89 on: February 19, 2018, 09:10:13 PM »
Quote
Whatever you claim Newton's Shell Theorem is correct
Quote
EEK thinks he has discovered something
- Neither claim nor discovery but its just a truth

Quote
PS: Whatever you claim Newton's Shell Theorem is correct and the gravity inside a spherical shell, due to that shell is zero.
What do you think if the "g" of M >>>>>>> the "g" of HS? For Example, Earth (ideally sphere) inside a homogeneous HS. The "g" of earth >>>>>>>> the "g" HS
If the earth, of mass say Me, were inside and concentric with a huge HS of any mass, say mHS, then:
  • Inside HS, g is due only to the earth. HS contributes nothing.
  • Outside HSg is due to the mass of the earth, Me, plus the mass of HS, mHS, all centred at the common centre.
Your ">>>>>>>>" is unnecessary, provided the geometries are perfect the result applies to any masses.

I hope this is close enough to what you were asking.