Question About The Universal Law of Gravitation?

It depends on how you interpret it. Each mass always creates a gravitational force towards it on any other particle. It is just that inside the sphere, the forces created by all particles of the sphere cancel out.

No, it isn't strange.
Each part of the sphere does create a force, but they cancel.

Again - Gravitational forces are attractive. They never cancel out and you have agreed. Should they cancel each other, apple wouldn’t even fall towards the ground.

Isn’t below your statement.

As a consequence, for example, within a shell of uniform thickness and density there is no net gravitational acceleration anywhere within the hollow sphere.

No net gravitational acceleration means g = GM/d^2 = 0, GM/d^2 = 0  G=0

The final destination of the settlement of the particle that contains inside the hollow sphere should be the center of the hollow sphere if gravity were to exist.

Reasons: the “g” of larger mass has greater on one side of the cross-section than the “g” of smaller mass on the other side of the cross-section in the shell theorem.

Similarly, what would you do if the hollow sphere is not symmetrical, symmetrical, and unsymmetrical holes in it?

(That does not mean there is no gravity inside the sphere, the sphere does not 'cancel' other gravitational forces from other objects.)

but masses move toward the greater "g"

Again - Gravitational forces are attractive. They never cancel out and you have agreed.

No, I haven't, and this issue has NOTHING to do with gravity.

If 2 equal and opposite forces act on a single object, i.e. BOTH FORCES ACT ON ONE OBJECT, then they cancel each other out and result in no net force.

If 2 forces act on different objects, then they do not cancel.

This is not unique to gravity.
It applies to ALL forces.

In the case of an apple falling to the ground, we have the gravitational force acting on the apple and an equal and opposite force acting on Earth. These are NOT acting on the same object, so they cannot cancel.

But inside a hollow shell, you have the gravitational force from one part of the shell acting on an object inside, and a force from another part of the shell also acting on that SAME OBJECT inside. As it is acting on the same object, and due to the symmetry, the forces cancel.

No net gravitational acceleration means g = GM/d^2 = 0, GM/d^2 = 0  G=0

No, it doesn't.
As you are adding up a sphere, you also need the directionality, that is typically given by a unit vector in the d direction, which I will represent as v.
So what you actually have is g=GM  v/d^2

And you need to integrate this over the spherical shell, which can be done in a few ways.
And with that, what you end up with is the sum of v/d^2 = 0

So it isn't that G or M is 0, it is that the sum of the inverse of the square of the distance, multiplied by the direction to that distance, is 0.

This means that there is no net force.

The final destination of the settlement of the particle that contains inside the hollow sphere should be the center of the hollow sphere if gravity were to exist.
Reasons: the “g” of larger mass has greater on one side of the cross-section than the “g” of smaller mass on the other side of the cross-section in the shell theorem.

You can provide 2 completely different alleged outcomes, with 2 completely different reasons, which are both just as valid as each other.

Consider an object in a hollow sphere some distance from the centre of that sphere.
Draw a line from the centre to the object, and draw a plane perpendicular to that line.
Use that plane to divide the sphere into 2 parts.

Now we have a few options.
We could note, as you have done, that there is more mass on the centre side of the sphere, which means the attraction to it would be larger resulting in you moving towards the centre.

However we could also choose to focus on that centre side being further away, with the wall of the sphere closest being away from the centre, and claim that due to the lesser distance the force towards the wall of the sphere will be greater, resulting in you flying to the spherical shell.

Or we could recognise that both of these forces act and recognise that you need to add up the contribution from each part of the shell and see which wins.
If you do this, you end up with them perfectly balancing. The greater mass is the right distance away to result in no net force.

Similarly, what would you do if the hollow sphere is not symmetrical, symmetrical, and unsymmetrical holes in it?

A hole would act like a negative mass in an otherwise symmetric sphere, with an object inside the sphere appearing to be pushed away from it.

but masses move toward the greater "g"

There is no greater g.

@JackBlack

Over distance the force of gravity weakens and is inversely proportional to the square of the distance.

For a hollow sphere, net gravitational force inside the sphere would only be 0 if the distance between opposite sides of the inner surface of the sphere were negligible to the diminishment of the force over that distance.   

Hopefully, I'm wording that correctly. 

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It depends on how you interpret it. Each mass always creates a gravitational force towards it on any other particle. It is just that inside the sphere, the forces created by all particles of the sphere cancel out.

No, it isn't strange.
Each part of the sphere does create a force, but they cancel.

Again - Gravitational forces are attractive. They never cancel out and you have agreed. Should they cancel each other, apple wouldn’t even fall towards the ground.

Again, slowly: Two forces on the same particle cancel out (if they are opposite direction and same magnitude). Two forces on different particles don't cancel out.

Isn’t below your statement.

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As a consequence, for example, within a shell of uniform thickness and density there is no net gravitational acceleration anywhere within the hollow sphere.

No net gravitational acceleration means g = GM/d^2 = 0, GM/d^2 = 0  G=0

No, this is just you applying an equation to a situation where you can not apply it, as has been repeatedly explained.

The final destination of the settlement of the particle that contains inside the hollow sphere should be the center of the hollow sphere if gravity were to exist.

Reasons: the “g” of larger mass has greater on one side of the cross-section than the “g” of smaller mass on the other side of the cross-section in the shell theorem.

That is just wrong, why do you think this would be the case?

Similarly, what would you do if the hollow sphere is not symmetrical, symmetrical, and unsymmetrical holes in it?

That is a different situation, then. Obviously.

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(That does not mean there is no gravity inside the sphere, the sphere does not 'cancel' other gravitational forces from other objects.)

but masses move toward the greater "g"

They do not. There is no greater "g". You are misunderstanding the basic concept of gravity and are phrasing your misunderstanding as a fact.

@JackBlack

Over distance the force of gravity weakens and is inversely proportional to the square of the distance.

For a hollow sphere, net gravitational force inside the sphere would only be 0 if the distance between opposite sides of the inner surface of the sphere were negligible to the diminishment of the force over that distance.   

Hopefully, I'm wording that correctly.

I'm not sure what you are trying to say.
If you are trying to say that there would be a net gravitational force if you were away from the centre of the hollow sphere, but still inside it, then you would be mistaken.

Being closer to one side of the sphere is balanced by there being more of the sphere to the other direction.

Quote from: NotSoSkeptical on March 04, 2023, 06:36:24 AM

@JackBlack

Over distance the force of gravity weakens and is inversely proportional to the square of the distance.

For a hollow sphere, net gravitational force inside the sphere would only be 0 if the distance between opposite sides of the inner surface of the sphere were negligible to the diminishment of the force over that distance.   

Hopefully, I'm wording that correctly.

I'm not sure what you are trying to say.
If you are trying to say that there would be a net gravitational force if you were away from the centre of the hollow sphere, but still inside it, then you would be mistaken.

Being closer to one side of the sphere is balanced by there being more of the sphere to the other direction.

Not from the same direction.  The more mass on the other side wouldn't all be pulling from the same vector.

There is no greater g.

I’m not sure if you have read the word unsymmetrical hollow sphere. The greater mass can be increased by increasing the thickness of the shell. Similarly, there is an infinite number of ways of unsymmetrical shapes of a hollow sphere that would change the math.

Moreover, GM = gd^2 if d = 0, G or M = 0

If 2 equal and opposite forces act on a single object, i.e. BOTH FORCES ACT ON ONE OBJECT, then they cancel each other out and result in no net force.

e.g. if you have an object O between A and B such that A applies a force of F onto O, and B applies a force of -F onto O, then the net force on O is F+(-F) = 0, i.e they cancel.

When the moon is b/t sun and the earth, shouldn’t the moon start orbiting the sun?

Quote from: JackBlack on March 04, 2023, 11:46:36 AM

Quote from: NotSoSkeptical on March 04, 2023, 06:36:24 AM

@JackBlack

Over distance the force of gravity weakens and is inversely proportional to the square of the distance.

For a hollow sphere, net gravitational force inside the sphere would only be 0 if the distance between opposite sides of the inner surface of the sphere were negligible to the diminishment of the force over that distance.   

Hopefully, I'm wording that correctly.

I'm not sure what you are trying to say.
If you are trying to say that there would be a net gravitational force if you were away from the centre of the hollow sphere, but still inside it, then you would be mistaken.

Being closer to one side of the sphere is balanced by there being more of the sphere to the other direction.

Not from the same direction.  The more mass on the other side wouldn't all be pulling from the same vector.

No, it wouldn't all be pulling from the same vector, but the sum of it all will cancel.

This is fairly well established.
For a force that follows the inverse square law, if it is being generated by a spherically symmetric shell, then outside the shell it can be treated as a single point at the centre of the shell with all the mass, and inside the shell the force is 0.

I’m not sure if you have read the word unsymmetrical hollow sphere.

As already said, if it isn't symmetric, then there will be a residual force.
i.e. the force inside an asymmetric hollow sphere will not necessarily be 0.

Moreover, GM = gd^2 if d = 0, G or M = 0

d is not 0.
If you want to consider a point at the centre of the spherical shell, you break the shell up into multiple separate components, and d is not 0 between the object and any part of the shell.

Again, for a point inside a spherically symmetric shell, the sum of (the unit vector towards a point on the shell divided by the square of the distance to that point) is 0.
No individual part is 0, but the sum is.

When the moon is b/t sun and the earth, shouldn’t the moon start orbiting the sun?

What do you mean start?

The Earth-Moon system is orbiting the sun.
So the moon is already orbiting the sun.

The orbital speed of the moon around Earth is ~ 1 km/s.
The orbital speed of the Earth-Moon system around the sun is ~30 km/s.
So even when the moon's orbit around Earth is going directly against the orbit of the Earth moon system around the sun, the moon is still going at roughly 29 km/s in its orbit around the sun.

This means you can trace the moon's path (or Earth's) and see it wobble side to side as it orbited the sun.

Here is the math:

Consider a uniform spherical shell of radius R and uniform area density s, with an object inside some distance l from the centre, and define the vertical axis to be passing through the object and the centre of the sphere.
Note that this vertical axis is an axis of symmetry, i.e. if the object experiences a force to the right due to some mass to the right of it, there will be a corresponding mass to the left cancelling out the horizontal component of the force. The only force which could remain is a vertical component (either up or down). So the only force we need to add up is that vertical component.

Consider a slice of the shell, defined as being some angle t away from the vertical axis (as measured from the top), and spanning an angle dt.
This slice is a circle with a radius of r = R*sin(t), a height of h=R*cos(t) above the centre of the sphere, and a width of R*dt.
The area of this slice is 2*pi*r*R*dt = 2*pi*R^2*sin(t)*dt.

Just to check we haven't done anything insane, if we integrate this from t=0 to 2*pi we get:
A=2*pi*R^2*[-cos(t)] from t=0 to 2*pi
A=2*pi*R^2*[-cos(2*pi) -(-cos(0))]
=2*pi*R^2*[-(-1)-(-1)]
=4*pi*R^2

Just like we would expect.
So this should be a valid way to analyse the spherical shell.

Now for the force due to gravity (or any other inverse square force).
The vertical component of the force will be equal for any point along this circle; and as said above, the horizontal force will cancel.
This means we can treat it as the entire mass of the ring is at one point along the ring, and calculate the vertical component of the force there. This is going to produce the same result as integrating the vertical component around the ring.

The mass of this ring is going to be M=s*2*pi*r*R*dt.
This ring, being a height of h above the centre of the circle, will be a height of h-l above our object.
This means the distance from the object to a point on the circle (d) will be given by:
d^2=r^2 + (h-l)^2 = r^2 + h^2 + l^2 - 2*h*l

The force of gravity is:
F=GMm/d^2

But this will have a horizontal component and a vertical component.
Conveniently, the forces add just like the distances, i.e. the ratios are the same.
so (h-l)/d = Fv/F
So the vertical force (Fv) is given by:
Fv = F*(h-l)/d
=GMm*(h-l)/d^3

This also has the advantage of making it signed.
If h-l is positive, this means the ring is above, so the force is directed upwards (in the +z direction).
If h-l is negative, this means the ring is below, so the force is directed downwards (in the -z direction).

And putting that all together we get this monstrous equation:
Fv = G*s*2*pi*r*R*dt*m*(h-l)/(r^2 + (h-l)^2)^(3/2)
=G*s*2*pi*R*sin(t)*R*dt*m*(R*cos(t)-l)/((R*sin(t))^2 + (R*cos(t)-l)^2)^(3/2)
=G*s*2*pi*R^2*m*sin(t)*dt*(R*cos(t)-l)/((R*sin(t))^2 + (R*cos(t)-l)^2)^(3/2)

And if we simplify it by making l a factor of R, i.e. l=k*R, we can instead get:
Fv=G*s*2*pi*R^2*m*sin(t)*dt*(R*cos(t)-k*R)/((R*sin(t))^2 + (R*cos(t)-k*R)^2)^(3/2)
Fv=G*s*2*pi*R^3*m*sin(t)*dt*(cos(t)-k)/(R^2*(sin(t)^2 + (cos(t)-k)^2))^(3/2)
Fv=G*s*2*pi*R^3*m*sin(t)*dt*(cos(t)-k)/(R^3*(sin(t)^2 + (cos(t)-k)^2)^(3/2))
Fv=G*s*2*pi*m*sin(t)*dt*(cos(t)-k)/(sin(t)^2 + (cos(t)-k)^2)^(3/2)

And then we can a large part of this into a constant:
C=G*s*2*pi*m

To get:
Fv=C*sin(t)*dt*(cos(t)-k)/(sin(t)^2 + (cos(t)-k)^2)^(3/2)
=C*sin(t)*dt*(cos(t)-k)/(sin(t)^2 + cos(t)^2 + k^2 - 2*k*cos(t))^(3/2)
=C*sin(t)*dt*(cos(t)-k)/(1 + k^2 - 2*k*cos(t))^(3/2)
=C*sin(t)*dt*(cos(t)-k)/(1 + k^2 - 2*k*cos(t))^(3/2)


The math for this is hard, but if we integrate this w.r.t. t, we end up with:
C*(k*cos(t) - 1)/(k^2*sqrt(k^2 - 2*k*cos(t)+1))

(and it is fairly simple to go back the other way to confirm this integral is correct)

So if we integrate from 0 to pi, we get the total force Ft:
Ft = C*[(k*cos(pi) - 1)/(k^2*sqrt(k^2 - 2*k*cos(pi)+1)) - (k*cos(0) - 1)/(k^2*sqrt(k^2 - 2*k*cos(0)+1))]
Ft = C*[(k*(-1) - 1)/(k^2*sqrt(k^2 - 2*k*(-1)+1)) - (k*(1) - 1)/(k^2*sqrt(k^2 - 2*k*(1)+1))]
Ft = C*[-(k + 1)/(k^2*sqrt(k^2 + 2*k+1)) - (k - 1)/(k^2*sqrt(k^2 - 2*k+1))]
Ft = C*[-(k + 1)/(k^2*sqrt((k+1)^2)) - (k - 1)/(k^2*sqrt((k-1)^2))]

And now we reach a point where we need to be careful.
If we just process the algebra without thinking we end up with the answer for a point outside the sphere.
This is because we have something of the form sqrt(x^2).
By definition this is always positive.

So we look carefully at the 2 instances, remembering that we have defined k as a ratio indicating what portion of the radius we are away from the centre, and it is positive as we are above.
First we have sqrt((k+1)^2).
As k is >=0 and 1 is >0, this gives us (k+1)>0, so sqrt((k+1)^2)=(k+1)

Next, the important one, we have sqrt((k-1)^2).
If we have a point INSIDE the sphere, then k<1. This means (k-1)<0, so sqrt((k-1)^2)=-(k-1).
If instead we had a point OUTSIDE the sphere, k>1, so (k-1)>0, so sqrt((k-1)^2)=(k-1).

Just to show what happens for the point outside, we get:
Ft = C*[-(k + 1)/(k^2*(k+1)) - (k - 1)/(k^2*(k-1))]
Ft = C*[-1/k^2 - 1/k^2]
Ft = 2*C/k^2
Ft = 2*G*s*2*pi*m/k^2

And noting that k*R=l, we have k=l/R
Ft = 2*G*s*2*pi*m*R^2/l^2
Ft = G*s*4*pi*R^2*m/l^2
(And noting that 4*pi*R^2 gives the area of the shell, and the area density was s, so the product s*4*pi*R^2 gives the mass of the shell)
Ft = GMm/l^2

i.e. we get the law we know and love.

But if instead we consider the inside of the sphere, we need to make the substitution sqrt((k-1)^2)=-(k-1).
This instead gives:
Ft = C*[-(k + 1)/(k^2*sqrt((k+1)^2)) - (k - 1)/(k^2*sqrt((k-1)^2))]
Ft = C*[-(k + 1)/(k^2*(k+1)) - (k - 1)/(k^2*(-(k-1))]
Ft = C*[-1/k^2 + 1/k^2]
Ft = C*[0]
Ft = 0

So we see that inside the spherically symmetric shell we end up with a force of 0.

It doesn't matter where you are inside the shell, the force will be 0.
You can be quite close to one wall, or near the middle, the force will still be 0.

The only point that isn't covered by this analysis is a point which is part of the shell.

When the moon is in b/t earth and sun in a line

The gravitational force of attraction b/t the sun and moon > The gravitational force of attraction b/t the sun and earth

Since the net gravitational force of attraction b/t the sun and moon is greater therefore technically the moon should have had its own separate orbit around the sun instead of orbiting around the earth.

Similarly, Galileo says that all objects fall at the same rate but isn't the moon and the earth a perfect example that they don’t fall at the same rate on the sun in the absence of air resistance?

Moreover if “g = GM/d^2” is the “g” of the hollow sphere then we can find the “g” of the hollow sphere at any distance “d” from its center.  As “d” can be < the radius of the hollow sphere therefore we can still calculate “g” of the hollow sphere within the shell.

Concentric circles within a shell represent the contour of “g” but these concentric ring changes when the geometry of the shell change as explained above.

The mass has a gravitational field on all its sides. So within a shell, an object would be attracted toward the sides, not the center.

so above are two different opposite scenarios i.e attraction toward the center and attraction toward the inner side of the shell.

Since the net gravitational force of attraction b/t the sun and moon is greater therefore technically the moon should have had its own separate orbit around the sun instead of orbiting around the earth.

Again, it DOES orbit the sun.
Earth and the moon together orbit the sun.

It wont just magically ignore the gravitational attraction to Earth.
So that means Earth will perturb the orbit of the moon, causing the moon to circle it as well. To be in its own orbit it needs to be further separated than the L1 or L2 point.

Similarly, Galileo says that all objects fall at the same rate but isn't the moon and the earth a perfect example

No.
Galileo was referring to a limited set of circumstances, dropping 2 objects on Earth.
The moon and Earth are not examples for this, as you can't drop Earth on Earth.

Moreover if “g = GM/d^2” is the “g” of the hollow sphere then we can find the “g” of the hollow sphere at any distance “d” from its center.  As “d” can be < the radius of the hollow sphere therefore we can still calculate “g” of the hollow sphere within the shell.

Again, see the math above.
You can calculate "g" for the hollow sphere inside the sphere, and it is 0.
Not because gravity magically doesn't exist inside, but because the spherical symmetry results in no net force due to the gravity of the spherical shell.
Again, the big difference is the treatment of the term sqrt((k-1)^2). When you are outside the shell, it is (k-1), but when you are inside it is -(k-1).

so above are two different opposite scenarios i.e attraction toward the center and attraction toward the inner side of the shell.

Which serve to demonstrate your very basic and primitive analysis is wrong.

It shows that there are competing effects, and you need to work out how they compare.
Does one work out to be the winner, with attraction towards it, or do they work out cancelling.

And note that the same is not true when outside the shell, as that way, all the force is towards the shell.

When the moon is in b/t earth and sun in a line

The gravitational force of attraction b/t the sun and moon > The gravitational force of attraction b/t the sun and earth

Since the net gravitational force of attraction b/t the sun and moon is greater therefore technically the moon should have had its own separate orbit around the sun instead of orbiting around the earth.

Distance.

When the moon is in b/t earth and sun in a line

The gravitational force of attraction b/t the sun and moon > The gravitational force of attraction b/t the sun and earth

Since the net gravitational force of attraction b/t the sun and moon is greater therefore technically the moon should have had its own separate orbit around the sun instead of orbiting around the earth.

Similarly, Galileo says that all objects fall at the same rate but isn't the moon and the earth a perfect example that they don’t fall at the same rate on the sun in the absence of air resistance?

Moreover if “g = GM/d^2” is the “g” of the hollow sphere then we can find the “g” of the hollow sphere at any distance “d” from its center.  As “d” can be < the radius of the hollow sphere therefore we can still calculate “g” of the hollow sphere within the shell.

Concentric circles within a shell represent the contour of “g” but these concentric ring changes when the geometry of the shell change as explained above.

The mass has a gravitational field on all its sides. So within a shell, an object would be attracted toward the sides, not the center.

so above are two different opposite scenarios i.e attraction toward the center and attraction toward the inner side of the shell.

Yes, in a homogeneous gravitational field without any other forces present, all things fall at the same rate. Ignoring tidal forces for now, the earth and the moon fall towards the sun at the same rate, caused by the gravitational field of the sun.
However, the earth and the moon also exert a gravitational pull on each other, which causes the moon to orbit the earth.

Again, it DOES orbit the sun.
Earth and the moon together orbit the sun.

I know.

But the net gravitational force of attraction = gravitational force b/t the sun and moon – gravitational force earth and moon, say 7-2 = 5 N

So automatically gravitational force of the earth and the moon is eliminated. The moon should drop independently on the sun just like you cancel out gravity forces in the shell.

No.
Galileo was referring to a limited set of circumstances, dropping 2 objects on Earth.
The moon and Earth are not examples for this, as you can't drop Earth on Earth.

Why moon and earth are not objects? Don’t forget they theorize the solar system from the Galileo statement.

What if two objects are dropped simultaneously from the same height but the location of heights are on the opposite side of the globe?

Again, see the math above.
You can calculate "g" for the hollow sphere inside the sphere, and it is 0.

The math is correct. M of the hollow sphere can be calculated from the difference of diameters of the outer and inner of the shell. The center of gravity of the shell would still be the center of the shell.

It's not necessary that “g” would always be zero. Again the thickness of half of the shell can be changed.

An object would fall if released from the ceiling of a deep well if its top is closed.

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Again, it DOES orbit the sun.
Earth and the moon together orbit the sun.

I know.

But the net gravitational force of attraction = gravitational force b/t the sun and moon – gravitational force earth and moon, say 7-2 = 5 N

Okay, staying with your example: Say is a gravitational acceleration of 7-2 = 5 m/s^2 the moon. But there is also a gravitational acceleration of 7m/s^2 on the earth. So earth and moon both accelerate ('fall') towards the sun. However, there is also a 2 m/s^2 difference between the acceleration on the earth and the one on the moon, which is the cause of their orbital motion.

So automatically gravitational force of the earth and the moon is eliminated. The moon should drop independently on the sun just like you cancel out gravity forces in the shell.

No. This is again a baseless assertion without any evidence to back it up.

The math is correct. M of the hollow sphere can be calculated from the difference of diameters of the outer and inner of the shell. The center of gravity of the shell would still be the center of the shell.

It's not necessary that “g” would always be zero. Again the thickness of half of the shell can be changed.

Tell me you don't understand Newton's shell theorem (or the equations of gravity for extended bodies) without telling me you don't understand Newton's shell theorem.

An object would fall if released from the ceiling of a deep well if its top is closed.

Your point being?

But the net gravitational force of attraction = gravitational force b/t the sun and moon – gravitational force earth and moon, say 7-2 = 5 N

So automatically gravitational force of the earth and the moon is eliminated. The moon should drop independently on the sun just like you cancel out gravity forces in the shell.

No, it shouldn't.

In order for it to just orbit the sun, and ignore Earth, the force from Earth would need to be 0, while the force from the sun would need to be 7 (just using your numbers).

Instead, you have the moon having a force of 5. This is too low for it to orbit the sun.
This means instead of following a circular orbit, it will be moving out, away from the sun.
This causes it to fall behind in its orbit (as its orbital path is now larger, yet it is travelling at the same velocity).

This causes the Earth-Moon-Sun system to fall out of alignment. So now instead of Earth just pulling the moon away from the sun, it is also trying to pull it further along its orbit. This causes it to start speeding up.
These combined effects cause the moon to go to the outside of Earth, and speed up past it.
During this it reaches a point where we again have alignment, but again, the moon is not just being affected by the sun, it is also being pulled in the same direction by Earth, causing it to accelerate towards the sun more than it should for its orbit, meaning it will move closer.

This causes the moon to periodically fall behind Earth, move to the outside, overtake Earth, move to the inside, and repeat.
In effect, it is circling Earth, i.e. orbiting Earth.

You need to go all the way to the L1 point for it to be perfectly balanced for them to remain in a line, or conversely all the way out to the L2 point.
If you are closer to Earth than either of these points, the effect of Earth's gravity is too significant, and it will result in it orbiting Earth.
If you are further away, the effect of Earth is too small, and it will fall away from Earth and orbit the sun.

Why moon and earth are not objects? Don’t forget they theorize the solar system from the Galileo statement.

They are objects, but you can't drop Earth on Earth.

While gravity which lead to the gravitational model of the solar system may have been influenced by Galileo's statement, it isn't that statement directly applying.

It requires a uniform gravitational field, without the objects in question having a significant gravitational field.

If you look at it from a broad POV, you see the Earth and moon both accelerating towards the sun at the same rate.
If you zoom in, you see that Earth and the moon have significant masses exerting a significant gravitational influence on each other.

What if two objects are dropped simultaneously from the same height but the location of heights are on the opposite side of the globe?

Again, not what the statement is covering.
But, the acceleration of Earth is negligible, so the acceleration of the objects towards Earth would be basically the same.

The math is correct. M of the hollow sphere can be calculated from the difference of diameters of the outer and inner of the shell. The centre of gravity of the shell would still be the center of the shell.

The "centre of gravity", which is really centre of mass of the shell would be at the centre. But that doesn't mean an object inside the shell is attracted to the centre.

It's not necessary that “g” would always be zero. Again the thickness of half of the shell can be changed.

It doesn't matter how thick the shell is. As long as it is spherically symmetric, it wont exert a net gravitational force on the inside.

An object would fall if released from the ceiling of a deep well if its top is closed.

Because that isn't falling into a hollow spherically symmetric shell.
That is falling into a mostly solid object, with a hole in it.
Even if it was a mostly hollow spherically symmetric shell with a hole in the surface, it still doesn't match.

Like I said before, you can treat the hole as a negative mass overlapping a spherically symmetric shell (so in the region of the hole, the positive mass of the shell and the negative mass of the hole gives 0 mass in total).
This means you can treat the majority of it as a series of spherical shells.
The shells you are outside of will attract you towards them.
The shells you are inside will do nothing.
Then in addition you have the negative mass of the holes below you repelling you (in reality, it is the shell with the hole attracting you slightly less), and the holes above you also repel you, but push you into the shell.

Even once you are entirely inside the hollow shell, you still have the negative mass holes pushing you, accelerating you towards the opposite wall.
And in addition, then negative mass of the holes

Here are two scenarios inside a shell
 
1-   The equation of g= GM/d^2

If d needs the adjustment then d=h-r, where h is the distance originating from the center of mass and r is the radius of the shell. If h = r then either g is undefined or infinity but if h is less than r, we can still have g inside a shell

2-   Since flat earth can also have “g” on all sides therefore inside wall of the shell can also have “g” which can attract things toward it. The change in the thickness of the partial (half, quarter, etc.) walls of the shell can also change the symmetry of the spherical shell.  So canceling gravity forces on a particle inside the shell is not always the case as the net gravitational force will be towards the wall which has greater mass due to its greater thickness.

So which one of the above is true?

Similarly, both unequal masses of the pan balance fall at the same rate toward the ground but the pan balance is lopsided due to the net gravitational force.

I don’t understand why the same is not true in the aforementioned unsymmetrically shell and when the moon is in between the earth and sun when they are in line.

Would the two unequal masses of a pan balance have them fall at the rate if dropped the wholw pan balance from a height h?  initially, both the said masses are at h. I think we should try it.

Here are two scenarios inside a shell
 
1-   The equation of g= GM/d^2

If d needs the adjustment then d=h-r, where h is the distance originating from the center of mass and r is the radius of the shell. If h = r then either g is undefined or infinity but if h is less than r, we can still have g inside a shell

This equation does not need adjustment, other than putting in a directionality.
You just need to apply it to every small bit of matter on the shell, rather than treating the shell as a point at its centre.

2-   Since flat earth can also have “g” on all sides therefore inside wall of the shell can also have “g” which can attract things toward it. The change in the thickness of the partial (half, quarter, etc.) walls of the shell can also change the symmetry of the spherical shell.  So canceling gravity forces on a particle inside the shell is not always the case as the net gravitational force will be towards the wall which has greater mass due to its greater thickness.

Something applying to a flat surface will not necessarily apply to a spherically symmetric shell.
Sure you can think of part of the wall of the spherical shell as having a value of g attracting things towards it, but you need to consider the contribution from every part.
And if it is symmetric, that will cancel.

Yes, you can also have an unsymmetrical one, but then you need to do more complex math.

I don’t understand why the same is not true in the aforementioned unsymmetrically shell and when the moon is in between the earth and sun when they are in line.

The unsymmetrical shell will have a net force, but you can't treat the shell as a point, as it isn't spherically symmetric and you aren't outside it.

I don't understand what your issue is for the moon.
The Earth moon system orbits the sun, with the moon and Earth orbiting their barycentre.