Role of gravitational force F=GMm/d^2 b/t two objects?

I’m still wanting to make a real post but until then, is it just me or is it supposed to be r² not d²?

It can be either, as long as you know what it stands for, and I have seen it presented in many ways.
Some people use r for radius as that works nicely for an orbit and is consistent with F=m omega^2 r=m v^2/r.
Others use d for distance, sometimes to not confuse it with the radius of the object.

What is important is the meaning, not the variables used to represent them.

And if you want to get into the more technical level of how to represent them, as they represent physical quantities they should be in italic font, and then because they are both masses, both should be represented by a lowercase, italic m, with superscripts to distinguish between them.
e.g. F = G m₁ m₂ / r²

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In many ways, this is quite analogous to electrostatic interactions.

One word – not impressive! Although its very clear but I’m writing again. I would suggest reading the posts carefully again and again not just you – no rude at all as it is not about the win or loss. 

G is the proportionality constant. It depends upon the force b/w 2 masses and the square of o/c distance b/t them as F =Gm1m2/d^2 =mg where g = Gm1/d^2,

Considering “g” separately from force F as g = Gm1/d^2, means G is considered separately from force F which is impossible. Therefore I would say “G” is wrongly placed in g=GM/d^2.

Two objects are must for the presence of gravitational force “F” [F=GMm/d^2, where g=GM/d^2] and gravitational constant “G”. Gravitational force “F“ should generate gravitational acceleration “g” but “g” which is equal to GM/d^2 independent of second mass (falling mass) and gravitational force “F”. 

As said, gravitational constant “G” is wrongly placed in g=GM/d^2 as explained above.

If you say, “there is clearly a force” then there is clearly a second mass (falling one) as well. Then all objects do not fall at the same rate.

Still stuck on this I see.

Seems the only hope is for you to start plugging some numbers into the equations and see what happens.

I suggest this.

1.  Open up a spreadsheet as it’s the easiest way to see a bunch of different values side by side.  (Download Open Office if you don’t have one).

2.  First Row:  Mass m.  Write a bunch of  different masses for the falling object.  Perhaps  1kg, 10kg, 100kg, etc but it doesn’t matter.

3. Second row:  Force F.  Use F=GMm/d^2 with M as mass of Earth, d as radius of Earth (for gravity at sea level) and m taken from above row.

4.  Third row:  Acceleration a.  Use a = F/m using the numbers from  rows 1 and 2.

5.  Report back.  What do you see?

Quote from: E E K on September 02, 2020, 10:41:20 AM

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In many ways, this is quite analogous to electrostatic interactions.

One word – not impressive! Although its very clear but I’m writing again. I would suggest reading the posts carefully again and again not just you – no rude at all as it is not about the win or loss. 

G is the proportionality constant. It depends upon the force b/w 2 masses and the square of o/c distance b/t them as F =Gm1m2/d^2 =mg where g = Gm1/d^2,

Considering “g” separately from force F as g = Gm1/d^2, means G is considered separately from force F which is impossible. Therefore I would say “G” is wrongly placed in g=GM/d^2.

Two objects are must for the presence of gravitational force “F” [F=GMm/d^2, where g=GM/d^2] and gravitational constant “G”. Gravitational force “F“ should generate gravitational acceleration “g” but “g” which is equal to GM/d^2 independent of second mass (falling mass) and gravitational force “F”. 

As said, gravitational constant “G” is wrongly placed in g=GM/d^2 as explained above.

If you say, “there is clearly a force” then there is clearly a second mass (falling one) as well. Then all objects do not fall at the same rate.

Still stuck on this I see.

Seems the only hope is for you to start plugging some numbers into the equations and see what happens.

I suggest this.

1.  Open up a spreadsheet as it’s the easiest way to see a bunch of different values side by side.  (Download Open Office if you don’t have one).

2.  First Row:  Mass m.  Write a bunch of  different masses for the falling object.  Perhaps  1kg, 10kg, 100kg, etc but it doesn’t matter.

3. Second row:  Force F.  Use F=GMm/d^2 with M as mass of Earth, d as radius of Earth (for gravity at sea level) and m taken from above row.

4.  Third row:  Acceleration a.  Use a = F/m using the numbers from  rows 1 and 2.

5.  Report back.  What do you see?

As F=GMm/d^2 .......Eq #1,           And  g=GM/d^2....Eq #2

From Eq #1: G=Fd^2/Mm ...Eq #3      And  From Eq #2: G=gd^2/M .....Eq #4

Gravitational constant "G" depends upon M and m in Eq #3 while the same "G" depends only on M in Eq#4 so isn't "G" wrongly placed in Eq #2 and #4? You know "G" requires two masses M and m and F as well.

A single mass (if present alone in the universe) used in Eq #2 and #4 can't generate gravitational force "F" and determine gravitational "G". Both "G" and "F" require two masses M and m.

As F=GMm/d^2 .......Eq #1,           And  g=GM/d^2....Eq #2

From Eq #1: G=Fd^2/Mm ...Eq #3      And  From Eq #2: G=gd^2/M .....Eq #4

Gravitational constant "G" depends upon M and m in Eq #3 while the same "G" depends only on M in Eq#4 so isn't "G" wrongly placed in Eq #2 and #4? You know "G" requires two masses M and m and F as well.

No, the 2 equations are equivalent.

g is an acceleration.
Remember that F=ma.
With g, this is F=mg, which can also be written as g=F/m
So the F in eq #3 is actually mg.
Or conversely, in eq#4, g is actually F/m.

The 2 equations are equivalent, but written with different units.

A single mass (if present alone in the universe) used in Eq #2 and #4 can't generate gravitational force "F" and determine gravitational "G". Both "G" and "F" require two masses M and m.

The other way to think about it is g is the gravitational field.
A mass in that field will experience a force which is the product of the field and its mass (the mass of the object).

So that single object in the universe will generate a gravitational field, but not a force as it requires another object to have that force.

With a single object alone in the universe, there would be no way to determine g.

No, the 2 equations are equivalent.

g is an acceleration.
Remember that F=ma.
With g, this is F=mg, which can also be written as g=F/m
So the F in eq #3 is actually mg.
Or conversely, in eq#4, g is actually F/m.

The 2 equations are equivalent, but written with different units.

I understand what you say about F=ma. Here “F” and “a” depend upon “m” on which the force is applied but “g” in the Eq of “g=GM/d^2” doesn’t depend on the mass (second or falling mass) on which the gravitational force is applied.

The other way to think about it is g is the gravitational field.
A mass in that field will experience a force which is the product of the field and its mass (the mass of the object).

So that single object in the universe will generate a gravitational field

Similarly "G" in the Eq of g=GM/d^2 requires the presence of a second mass "m" for the determination of its constant value. No second mass "m" means no "G". No "G" means no g=GM/^2 of a single mass existed if lonely present in the universe.

As "G" requires gravitational force "F" and "F" requires the presence of two masses (M and m) therefore the presence of "G" in g=GM/^2 means that "g" in the foregoing Eq depend upon second mass "m" too. Thus "g" of any gravitating mass depends upon the mass of the falling object.

but not a force as it requires another object to have that force.

With a single object alone in the universe, there would be no way to determine g.

But Newton determines it to be “g=GM/d^2” for any object of mass "M" when he says “g” is independent of second or falling mass “m” after formulating the universal law of gravitation of F=GMm/d^2.  Here independent of second mass means "in the absence of a second mass".

It means gravitational field g=GM/d^2 is the innate property of any mass "M".

I understand what you say about F=ma. Here “F” and “a” depend upon “m” on which the force is applied but “g” in the Eq of “g=GM/d^2” doesn’t depend on the mass (second or falling mass) on which the gravitational force is applied.

The point is that the "a" is g. i.e. in the equation F=GMm/d^2, that IS F=m*g

Similarly "G" in the Eq of g=GM/d^2 requires the presence of a second mass "m" for the determination of its constant value. No second mass "m" means no "G". No "G" means no g=GM/^2 of a single mass existed if lonely present in the universe.

As "G" requires gravitational force "F" and "F" requires the presence of two masses (M and m) therefore the presence of "G" in g=GM/^2 means that "g" in the foregoing Eq depend upon second mass "m" too. Thus "g" of any gravitating mass depends upon the mass of the falling object.

Only for the determinisation of the value, not the values existence.
Also note that in practice you do not need to know the value of m.
All you need to do is measure its acceleration in free fall.
You can determine G by determining g, without knowing F or m.

By the need for a second object to determine it is not unique to gravity. It is fundamental for all forces.
For example, with the electrostatic force, you have F=kQq/d^2.
But the electrical field is just E=kQ/d^2.

A single object generates an electric field, but you need a second object to interact with the field to have a force.

Either way, the field (either E or g) exists from a single object alone in the universe.
But in order to be able to measure it you need a second object. In order to be able to quantify it in any way, you need a second object.

But Newton determines it to be “g=GM/d^2” for any object of mass "M" when he says “g” is independent of second or falling mass “m” after formulating the universal law of gravitation of F=GMm/d^2.  Here independent of second mass means "in the absence of a second mass".

It means gravitational field g=GM/d^2 is the innate property of any mass "M".

Here independent of the second mass means it does not depend on the second mass.

i.e. it doesn't matter if you have a second mass of 1 picogram or 1 ton, g is the same.

g is innate to the object of mass M, just like E is innate to the object of charge Q. (assuming a spherically symmetric distribution)

What is the role of gravitational force F=GMm/d^2 between objects A and B when

Object A falls on B @ rate of gb=GMb/d^2.
Here gb is innate to Mb and doesn’t depend on the magnitude of Ma even if it is zero and hence F as well.

Object B falls on A @ rate of ga=GMa/d^2.
Here ga is innate to Ma and doesn’t depend on the magnitude of Mb even if it is zero and hence F as well.

Both accelerations (ga and gb) produced don’t depend upon the F between them.

What is the role of gravitational force F=GMm/d^2 between objects A and B when

Object A falls on B @ rate of gb=GMb/d^2.
Here gb is innate to Mb and doesn’t depend on the magnitude of Ma even if it is zero and hence F as well.

Object B falls on A @ rate of ga=GMa/d^2.
Here ga is innate to Ma and doesn’t depend on the magnitude of Mb even if it is zero and hence F as well.

Both accelerations (ga and gb) produced don’t depend upon the F between them.

First a note on when m is 0, if rest mass is 0, and you have a massless object then Newton's Laws break down. But note that even "massless" particles like light will be accelerated.

Ignoring that, note that object A falls at a rate of gb = GMb/d^2.
This means the force acting on the object (treating gravity as a force rather than curved spacetime) is F=Ma*gb=GMaMb/d^2

Object B falls on A at a rate of ga=GMa/d^2, and thus F=Mb*ga=GMaMb/d^2.

So we have the equal and opposite force.

I wouldn't say this doesn't depend on the force. Instead the acceleration of each object is independent of the mass of the object. If you increase the mass of A, that will not cause it to accelerate faster or slower, but it will increase the force and cause object B to accelerate faster or slower.

This was one of the strange properties of gravity, that the force was proportional to mass, so the acceleration was not.

But the simpler way to see the force is by allowing the objects to collide, or placing something to stop them moving.
You need to apply a force of F=GMaMb/d^2 to stop them accelerating towards each other.

The other way around

So we have the equal and opposite force.

Acceleration needs force for its generation - Right

How do accelerations produce when both equal and opposite forces of gravitating and falling cancel each other. There is no net force at all, which can accelerate either of the objects due to their cancellation. This net force is always zero no matter what the o/c distance is in between A and B -Right

Both ga and gb don’t require gravitational force “F” that exists in between them and their respective falling masses – Right. Because

We know there is a gravitational field of A, ga=GMa/d^2 innate to A, and a gravitational field of B, gb=GMb/d^2 innate to B = Right

As A falls on B @ gb=GMb/d^2 despite net “F=0”.
As B falls on A @ ga=GMa/d^2 despite net “F=0”

Therefore what is the role gravitational force “F” that exists in between A and B when both fall towards each other in the absence of net force?

The other way around

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So we have the equal and opposite force.

Acceleration needs force for its generation - Right

This is where you start going into philosophy.

From gravity as a force perspective, the g=GM/d^2 is not an acceleration, it is the strength of the gravitational field, which just happens to have the same units as acceleration.
You multiply this by the mass of the second object to get the force acting on it, and then divide it by that very same mass to determine the acceleration of the object. Due to the mass appearing twice, you can skip those steps and just use g as an acceleration, but that doesn't mean a force isn't involved.

From the gravity as curved spacetime perspective there is no actual acceleration.

This is distinct to the electrical field which instead has units of acceleration * mass / charge. In this case you multiply it by the charge of the object and divide it by the mass to determine the acceleration.

How do accelerations produce when both equal and opposite forces of gravitating and falling cancel each other. There is no net force at all, which can accelerate either of the objects due to their cancellation. This net force is always zero no matter what the o/c distance is in between A and B -Right

The net force on the AB system is 0. That means the centre of mass of the AB system will not move.
It does not mean there is no net force on either object.
The net force on A is the same as the net force on B which is GMaMb/d^2.

So there IS a net force acting on both objects.
It is only the AB system, where you consider A and B as a single object, that there is no net force.

This is just like if you had 2 objects tied together with a spring. You then pull them apart and let go. (ignoring gravity)
At the moment you let go, there is no net force on the system, so the centre of mass of the system remains stationary (remember, this is ignoring gravity).
But there is a net force on each object at the ends of the springs so the 2 objects accelerate towards each other.

Do not confuse the system (with no net force) with the object (which has a net force).

Lone point mass can’t have its gravitational field = g = GM/d^2 because “d” in the preceding Eq is not just a length but it is the o/c center distance between A and B. No second mass means, no “d” or d=0. And hence the value of g=GM/d^2 or gravitational strength = 0.

The net force on A is the same as the net force on B which is GMaMb/d^2.

But in opposite direction - Right. It is said

A attracts/accelerates B with a gravitational force Fa = Mbga
B attracts/accelerates A with a gravitational force Fb = Magb
Fa attracts Mb only
Fb attracts Ma only

The original equation of the gravitational force F=GMm/d^2 attracts two objects towards each other at the same time. This means that there is always exists ONE gravitational force F=GMm/d^2 between A and B which has to accelerate both A and B towards one another simultaneously.

Therefore the individual concept of the gravitational force of either Fa or Fb is wrong in principle.

Actually, the derivation of g=GM/d^2 itself shown in the following link is wrong.
https://byjus.com/jee/acceleration-due-to-gravity/

Formula of Acceleration due to Gravity:
I don’t understand why are the equations of F=ma and F=GMm/d^2 are compared with each other.
Gravitational force of earth acting a mass m = ma=mg ….Eq #1
According to universal law of gravitation gravitational force between earth and the mass is F=GMm/d^2 ……Eq #2
Comparing Eq #1 and #2
We get mg= GMm/d^2
Hence g = GM/d^2

I don’t see the need of F=ma=mg in the presence of F=GMm/d^2 because there is only one force exits which is the gravitational force between the earth and mass m (M and m).

Lone point mass can’t have its gravitational field = g = GM/d^2 because “d” in the preceding Eq is not just a length but it is the o/c center distance between A and B.

It is the distance from the object.
It doesn't need a second object.

When you introduce a second object, the distance you care about is the distance between the 2 objects.
But the gravitational field is defined for any d.

This means that there is always exists ONE gravitational force F=GMm/d^2 between A and B which has to accelerate both A and B towards one another simultaneously.

No, it means there are 2 equal and opposite forces, just like you always find forces paired up.
There is the force A exerts on B which draws B towards A and there is the force B exerts on A which draws A towards B.
These forces are equal in magnitude and opposite in direction.

[/b]I don’t understand why are the equations of F=ma and F=GMm/d^2 are compared with each other.

Because the force is the same.
You could just write it like this:
F=ma=GMm/d^2=mg

They are 2 ways to write out the same force.

It is the distance from the object.

Distance from the object means when no second mass, d=0, and hence g = GM/d^2 = 0. For a person like you, it's not that difficult to work out.

No, it means there are 2 equal and opposite forces, just like you always find forces paired up.
There is the force A exerts on B which draws B towards A and there is the force B exerts on A which draws A towards B.
These forces are equal in magnitude and opposite in direction.

This means gravitational doesn’t exist anymore or always zero between the objects when these two forces cancel each other.
As asked, what is the role of gravitational force then b/t A and B when Fa and Fb cancel out?

Because the force is the same.
You could just write it like this:
F=ma=GMm/d^2=mg

They are 2 ways to write out the same force.

Nope they are two different forces
F=GMm/d^2 is the gravitational force b/t M and m while F=ma=mg means there is a force on m but m doesn’t exert force on anything.
If they are the same then why are equating it for “g”.

Similarly, if g = GM/d^2 doesn’t depend on falling mass "m" and force of gravity "F" then why does "M" settle more than "m" on the soft surface of the ground?

The equation of g = GM/d^2 is independent of gravitational force "F" and falling mass "m"

Both lM and m have the same g = GM/d^2

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It is the distance from the object.

Distance from the object means when no second mass, d=0, and hence g = GM/d^2 = 0. For a person like you, it's not that difficult to work out.

No, the field is defined everywhere.

For any point in space you pick, there is a distance between that point and the object, and thus a value of d.

Also, when d approaches 0, the field strength tends to infinity.

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No, it means there are 2 equal and opposite forces, just like you always find forces paired up.
There is the force A exerts on B which draws B towards A and there is the force B exerts on A which draws A towards B.
These forces are equal in magnitude and opposite in direction.

This means gravitational doesn’t exist anymore or always zero between the objects when these two forces cancel each other.
As asked, what is the role of gravitational force then b/t A and B when Fa and Fb cancel out?

No, it doesn't.

Again, there are 2 forces. One forces acts on A, the other acts on B.
As they are acting on different objects, they do not cancel.

It is only if you consider the entire AB system as a single object that the forces cancel, and all that means is that the centre of mass of the AB system will not accelerate. It tells you almost nothing about what is in the system and how they move relative to the system.

Just what do you think the net force acting on A is?
Not on the AB system, but on A?

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Because the force is the same.
You could just write it like this:
F=ma=GMm/d^2=mg

They are 2 ways to write out the same force.

Nope they are two different forces

It is the same force, no matter how much you don't want it to be.

You can determine the force as simply the product of the mass and the gravitational field, i.e. F=mg, or you can do it based upon the 2 masses and distance as F=GMm/d^2.
And regardless of which you pick, you can determine the acceleration of the object using F=ma.

This is just like the electrostatic force which you can do as F=qE, or F=kQq/d^2, and also have F=ma.

means there is a force on m but m doesn’t exert force on anything.

This is physically impossible. If there is a force on m, m MUST exert a force equal in magnitude and opposite in direction on something else.

Similarly, if g = GM/d^2 doesn’t depend on falling mass then why does a greater mass settle more than light mass on the soft surface of the ground?
Both lighter and greater masses have the same g = GM/d^2

Because g is the acceleration, not the force.
The force DOES depend on the mass.
And the force, and the related pressure, determines how much the ground is compressed.

And you are also slightly off, it isn't the greater mass, it is the greater mass per unit area.

I'm not good at convincing but here again one of the old posts of mine that may be offensive to someone.

Fig #1
Mass of object A = Ma, Mass of object B = Mb = theoretical spherical earth, and Mass of object C = Mc
Mb >>>>>>>>>>>> Ma >>>>>>>> Mc. Similarly, ga, gb and gc are the acceleration due to gravities of Ma, Mb and Mc respectively such that gb >>>>>>>>>> ga >>>>> gc. The o/c distance b/t Ma and Mb = d = o/c distance b/t Mb and Mc. The position of both Ma and Mc are located on the opposite sides of the globe of Mb as shown. The point e and f are the antipodal points.   
                                                                                                                  Now imagine Ma and Mc fall simultaneously from the same height "d" on Mb as shown.

Galileo and Newton say that both Ma and Mc fall at the same rate  (Fig #2) and according to their statement both should strike the antipodal points f and e respectively at the same time. But since gb >>>>>>>>>> ga >>>>> gc, therefore, cognizance shows, Mb and Ma strike first before the striking of Mc and Mb, and for this reason, Mc and Ma don't fall at the same rate.

Explanation: The falling of Mb on Ma is greater than the falling of Mb on Mc due to the greater ga of Ma than gc of Mc therefore when Mb falls on Ma then the o/c distance b/t Mc and Mb increases during the fall of Mc on Mb and as a result, the falling rate of Mc decreases instead of at the same rate it was supposed to fall if the EARTH IS ROUND.

The same phenomenon can also be observed in Fig #3 as well where Mc is above Mb and Mb is above the Ma (earth).

Let Ma and Mc fall on Mb at the same time from the same height "d". Since ga >>>>>>>>>>>> gb >>>>>>>gc therefore here too Ma and Mb strike first and consequently the falling rate of Mc on Mb decreases.

Fig #2: When Ma and Mc are closer to each other then it's hard to notice the above contradictions to Gallilleo statements however theoretically Mb is still lean to Ma more than Mc as shown.   

So the falling mass matter to g=GM/d^2 - Right

Quote from: Unconvinced on September 05, 2020, 10:31:29 AM

Quote from: E E K on September 02, 2020, 10:41:20 AM

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In many ways, this is quite analogous to electrostatic interactions.

One word – not impressive! Although its very clear but I’m writing again. I would suggest reading the posts carefully again and again not just you – no rude at all as it is not about the win or loss. 

G is the proportionality constant. It depends upon the force b/w 2 masses and the square of o/c distance b/t them as F =Gm1m2/d^2 =mg where g = Gm1/d^2,

Considering “g” separately from force F as g = Gm1/d^2, means G is considered separately from force F which is impossible. Therefore I would say “G” is wrongly placed in g=GM/d^2.

Two objects are must for the presence of gravitational force “F” [F=GMm/d^2, where g=GM/d^2] and gravitational constant “G”. Gravitational force “F“ should generate gravitational acceleration “g” but “g” which is equal to GM/d^2 independent of second mass (falling mass) and gravitational force “F”. 

As said, gravitational constant “G” is wrongly placed in g=GM/d^2 as explained above.

If you say, “there is clearly a force” then there is clearly a second mass (falling one) as well. Then all objects do not fall at the same rate.

Still stuck on this I see.

Seems the only hope is for you to start plugging some numbers into the equations and see what happens.

I suggest this.

1.  Open up a spreadsheet as it’s the easiest way to see a bunch of different values side by side.  (Download Open Office if you don’t have one).

2.  First Row:  Mass m.  Write a bunch of  different masses for the falling object.  Perhaps  1kg, 10kg, 100kg, etc but it doesn’t matter.

3. Second row:  Force F.  Use F=GMm/d^2 with M as mass of Earth, d as radius of Earth (for gravity at sea level) and m taken from above row.

4.  Third row:  Acceleration a.  Use a = F/m using the numbers from  rows 1 and 2.

5.  Report back.  What do you see?

As F=GMm/d^2 .......Eq #1,           And  g=GM/d^2....Eq #2

From Eq #1: G=Fd^2/Mm ...Eq #3      And  From Eq #2: G=gd^2/M .....Eq #4

Gravitational constant "G" depends upon M and m in Eq #3 while the same "G" depends only on M in Eq#4 so isn't "G" wrongly placed in Eq #2 and #4? You know "G" requires two masses M and m and F as well.

A single mass (if present alone in the universe) used in Eq #2 and #4 can't generate gravitational force "F" and determine gravitational "G". Both "G" and "F" require two masses M and m.

G is a universal constant, it doesn’t depend on anything.  But you can calculate G if you have all the terms for either of the two equations.  There is no problem here.

Did you try my suggestion of actually putting some numbers into  the equations to see what happens with different values? It should make everything much clearer.

G is a universal constant, it doesn’t depend on anything.  But you can calculate G if you have all the terms for either of the two equations.  There is no problem here.

Did you try my suggestion of actually putting some numbers into  the equations to see what happens with different values? It should make everything much clearer.

Have you gone through my last reply where I showed that second mass m matter in the equation of acceleration due to gravity g=GM/d^2? This will answer your all questions.

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G is a universal constant, it doesn’t depend on anything.  But you can calculate G if you have all the terms for either of the two equations.  There is no problem here.

Did you try my suggestion of actually putting some numbers into  the equations to see what happens with different values? It should make everything much clearer.

Have you gone through my last reply where I showed that second mass m matter in the equation of acceleration due to gravity g=GM/d^2? This will answer your all questions.

I'll take that as a no then.

I have no questions on the law of universal gravitation.  I'm trying to help answer your questions.

Why massively change to a different topic?

Do you at least accept that there is a net force on each object?

The position of both Ma and Mc are located on the opposite sides of the globe of Mb as shown. The point e and f are the antipodal points.
Galileo and Newton say that both Ma and Mc fall at the same rate

No, they don't.
They discuss 2 objects dropped from the same location, not antipodal points.
When you have 2 objects dropped from antipodal points, you now have 3 objects in a straight line.

Each object is attracted to the other 2.

But in practice, on Earth that difference is negligible.

So the falling mass matter to g=GM/d^2 - Right

No, because what you actually have now when you consider object A falling is:
g=G*(Mb/dab^2 + Mc/dac^2).
And something similar for object B falling and object C falling, noting you also need to note the direction, where the object in the middle is pulled in 2 opposite directions so the overall acceleration is smaller.

You are also equating object B moving towards object A as making object A fall faster, rather than just it reducing the distance object A need to fall to collide with object B.

As another simple example, you have 4 trains.
Train A is travelling at 100 km/hr east towards a train moving at 50 km/hr to the east initially 150 km away.
Train B is travelling at 100 km/hr west towards a train moving at 50 km/hr to the east initially 150 km away.

Which is travelling faster? A or B?

The answer is that they are both travelling at 100 km/hr. It doesn't matter than with train A the other train is moving away from it, making it take 3 hours to collide while train B is heading towards the train making it take only 1 hour to collide.

Earth moving towards the object doesn't mean the object is accelerating faster.

Have you gone through my last reply where I showed that second mass m matter in the equation of acceleration due to gravity g=GM/d^2? This will answer your all questions.

You didn't.
You poorly showed the third mass matters.

They discuss 2 objects dropped from the same location, not antipodal points.

Reference, please!!!!!

Small objects do fall at the same rate even if dropped from the antipodal points simultaneously because the acceleration of the earth is ignored.
A thought experiment can be done with the help of two identical towers if located on the opposite side of the globe. Here

Falling masses are too small as compared to the mass of the earth
They are closed to each other
Falling height is also small

As explained things changed when a mass of one or both the falling masses increase. 
It’s hard to notice in the Galileo experiment but the mass of the earth is still leaned to greater mass more than lighter mass. This changes its falling rate theoretically. There should be no if and but in the theory. Although Newton didn’t notice this is what Newton says in his theory, not me.

Question: Why do cooler /denser molecules of air take the position of warmer molecules on or above the surface of the earth when “g” is independent of temperature and pressure?

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They discuss 2 objects dropped from the same location, not antipodal points.

Reference, please!!!!!

One example would be this:
https://en.wikipedia.org/wiki/Galileo%27s_Leaning_Tower_of_Pisa_experiment

But that isn't how it works. You claim a contradiction, you claim they are wrong, so the burden is on you to show that what they are claiming actually applies to this.

Small objects do fall at the same rate even if dropped from the antipodal points simultaneously because the acceleration of the earth is ignored.

Under the assumption that the only significant gravitational field is that from Earth. Then all other masses are ignored and the motion of Earth is ignored.

This changes its falling rate theoretically.

No, it doesn't. It means Earth falls towards it.
That reduces the distance. It doesn't make the object fall faster.

Question: Why do cooler /denser molecules of air take the position of warmer molecules on or above the surface of the earth when “g” is independent of temperature and pressure?

Because pressure is dependent upon density.
For a fluid with a uniform density, the pressure gradient across it can be given by dP=rho*g*h.
So if you have a column of dense air next to a column of low density air, the pressure will be greater at the bottom of the dense air. This pushes the lower density air up.

They did the experiment at one place but still, the exact locations of the falling masses are different, not the same. Similarly, the idea was to prove “All objects fall at the same rate on earth if fall from the same height simultaneously”. It means above said fall from any point on the contour line of “g” above the earth will be the same. Don't forget we are assuming the earth is a theoretical sphere of uniform density.

Moreover, the value of G=6.67408 × 10-11 m3 kg-1 s-2 is so small. If it can be used in g=GM/d^2 then why do we ignore the acceleration of earth if we increase the size of one of the two falling masses to a greater size or close to the earth?

Because pressure is dependent upon density.

Great!! Gravity depends upon the density (gamma = mass per unit volume) of the object. The mass of the denser molecules has greater acceleration due to gravity while the mass of the warmer molecules has lesser acceleration due to gravity. So it means denser objects fall faster than lighter objects. You yourself choose period.

Technically pressure (gamma x height =F/A) is not even a force. It's a measurement of how much force is being applied per unit area. As said, there are two masses of molecules. The mass of the denser molecules can exert F = PA on earth from the original height during fall. I mean it doesn’t have to come down and takes the place of the mass of the warmer molecules. F=PA of denser mass and warmer masses of the molecule can just simply be added. 

No, it doesn't. It means Earth falls towards it.
That reduces the distance. It doesn't make the object fall faster.

Reduction in the distance means changes in the rate of falling - it starts falling with higher types of motion

If it can be used in g=GM/d^2 then why do we ignore the acceleration of earth if we increase the size of one of the two falling masses to a greater size or close to the earth?

We don't.
For example, to accurate determine the path of the Earth and moon around the sun, we consider both of them.

We ignore the acceleration of Earth when it is insignificant.

The mass of the denser molecules has greater acceleration due to gravity while the mass of the warmer molecules has lesser acceleration due to gravity. So it means denser objects fall faster than lighter objects.

No, it doesn't.
It means the acceleration of Earth towards it is slightly more, and that amount is negligible.

It doesn't mean it accelerates towards Earth more.
So no, denser objects do not fall faster than lighter objects, at least not in a vacuum.

Technically pressure (gamma x height =F/A) is not even a force.

And I never said it was.
But it being a pressure is important.
Unlike a force which simply pushes down, the pressure in the air pushes in ALL directions.

In order to keep the dense air on top it would need to be perfectly balanced. As soon as there is any slight variation (which will occur), the denser, higher pressure will push the lower pressure out of the way. This is because it also pushes sideways and up.

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No, it doesn't. It means Earth falls towards it.
That reduces the distance. It doesn't make the object fall faster.

Reduction in the distance means changes in the rate of falling - it starts falling with higher types of motion

No, it doesn't. It means Earth is moving as well.
You continually asserting the same nonsense will not make it true.

Air molecules push outwards to the sides equally, but in my understanding, their weights due to gravity indeed push down harder than they push upwards. For the same volume of air, cold air has more molecules than hot air. This means cold air weighs (W=F=mg) more than hot air. Hot air can have a lower density but the same pressure as cold air. Therefore cold air sinks due to gravity, not because of pressure. The hot air then floats up or goes sideways until it meets air of the same density in the atmosphere.

No, it doesn't. It means Earth is moving as well.
You continually asserting the same nonsense will not make it true.

In other words to the offensive/nonsense: Light mass enters quickly than it is supposed to be into the gravitational field of the earth. The quick increase in its velocity means there is a quicker change in its velocity as well than usual.

Air molecules push outwards to the sides equally, but in my understanding, their weights due to gravity indeed push down harder than they push upwards.

For a gas, there is a pressure gradient.
The pressure at the top is less than the pressure at the bottom.
But if you take an infinitely thin slice of air, it will push up and down equally.
If you take a large column, then the bottom will be pushing down more than the top is pushing up due to the pressure gradient in it.
This also means the bottom has the sides pushing out more than the top.

Hot air can have a lower density but the same pressure as cold air. Therefore cold air sinks due to gravity, not because of pressure.

It is not simply the pressure, but the pressure gradient.
If you have a region of hot and cold air quite close to each other, such that one is denser than the other, and there is any region where they are side by side, then with the pressure being equal at the top, the denser air will have a greater pressure at the bottom which will push the less dense air out of the way.

In other words to the offensive/nonsense: Light mass enters quickly than it is supposed to be into the gravitational field of the earth.

No, that is the nonsense.
Earth moving towards the object doesn't mean the object is accelerating faster.

Lets take the example of footing settlement.

Contact pressure: Response of soil against applied pressure of footing. It depends upon

1-   Stiffness of footing (rigid or flexible)
2-   Stiffness of soil (loose or hard)
3-   Type of load  (point or uniform)

Settlement of the rigid footing is uniform but the contact pressure distribution depends on soil type. For sandy soil: Max @ center while min @ edges. For Clayly soil, it's the opposite. So

Isn’t the aforementioned applied load, which is used as a force per unit area gravitational?

No, that is the nonsense.
Earth moving towards the object doesn't mean the object is accelerating faster.

Similarly,

Let g1=9.8 m/s/s,g2=9.7m/s/s..................g5=9.4m/s/s and g6=9.3 m/s/s are zones of gravitational field above earth.

Case #1: Erath moves upward while mass m is stationary: Mass m is at rest at a height where the value of g of earth is g6. The mass “m” seems to enter into g5 zone of the earth slowly as soon as the earth starts accelerating towards “m” @ the rate of “g” of “m”. After some time it crosses the zone of g5 and then g4,g3,g2, and finally, the earth hits the mass “m”.

Case #2; Erath is stationary while mass m is moving: Mass m is at rest at a height where the value of g of earth is g6. The mass “m” starts entering into g5 zone of earth slowly as soon as it starts accelerating towards “earth” @ the rate of “g6” of “earth” initially. After some time it crosses the zone of g5 and then g4,g3,g2 and finally hit the earth @ the rate of  g1.

Case #3: Now imagine both start accelerating towards each other simultaneously. They experience the same gravitational force but the acceleration of m towards earth is greater than the acceleration of earth towards “m”. Since mass “m” enters the gravitational field of earth faster than before therefore there will be an increase in the velocity of “m” and acceleration as well.

Conclusion: Although it's incorrect but for simplicity we can say case #3 = case #1 + case #2. We can add the velocities involved in case #1 and case #2.

Lets take the example of footing settlement.

Contact pressure: Response of soil against applied pressure of footing. It depends upon

1-   Stiffness of footing (rigid or flexible)
2-   Stiffness of soil (loose or hard)
3-   Type of load  (point or uniform)

And the actual load.
This actual load depends on the mass of the object.
It isn't helping you support your prior claims.

Let g1=9.8 m/s/s,g2=9.7m/s/s..................g5=9.4m/s/s and g6=9.3 m/s/s are zones of gravitational field above earth.

Case #1: Erath moves upward while mass m is stationary: Mass m is at rest at a height where the value of g of earth is g6. The mass “m” seems to enter into g5 zone of the earth slowly as soon as the earth starts accelerating towards “m” @ the rate of “g” of “m”. After some time it crosses the zone of g5 and then g4,g3,g2, and finally, the earth hits the mass “m”.

Case #2; Erath is stationary while mass m is moving: Mass m is at rest at a height where the value of g of earth is g6. The mass “m” starts entering into g5 zone of earth slowly as soon as it starts accelerating towards “earth” @ the rate of “g6” of “earth” initially. After some time it crosses the zone of g5 and then g4,g3,g2 and finally hit the earth @ the rate of  g1.

Case #3: Now imagine both start accelerating towards each other simultaneously. They experience the same gravitational force but the acceleration of m towards earth is greater than the acceleration of earth towards “m”. Since mass “m” enters the gravitational field of earth faster than before therefore there will be an increase in the velocity of “m” and acceleration as well.

Conclusion: Although it's incorrect but for simplicity we can say case #3 = case #1 + case #2. We can add the velocities involved in case #1 and case #2.

That is the point though, it is incorrect.
Earth accelerating towards the object changes the distance and thus changes the rate of acceleration.
It is not the object itself accelerating towards Earth faster, it is Earth moving towards the object.
And for most things, that motion is negligible.

That is the point though, it is incorrect.
Earth accelerating towards the object changes the distance and thus changes the rate of acceleration.
It is not the object itself accelerating towards Earth faster, it is Earth moving towards the object.
And for most things, that motion is negligible.

The motion is negligible for small masses but it can't be ignored when the size of the falling mass is increased. Also, we are discussing theory.
Case #3 = Case #1 +case#2
We know g = GM/d^2 is the gravitational strength of any point mass. As soon as the on-center distance “d” b/t the two objects starts reducing from both ends simultaneously, the “g “ of earth and the “g” of mass start increasing at the same times than it were considered to be like in case #1 or case #2. Both masses are gravitating and falling at the same time therefore they enter into the gravitational field of one another very quickly not like in case#1 or Case#2.

And the actual load.
This actual load depends on the mass of the object.
It isn't helping you support your prior claims.

This is what I’m trying to explain that greater the density of mass is, the greater will be the settlement in ground but you say it doesn’t support my claim of “ g depends on the falling mass” or “all objects don’t fall at the same rate”

As “g” neither depends on falling mass nor gravitational force therefore if all objects would fall at the same rate then those objects should have shown at least equivalency/ similarity in settlements. For example:
 
1-   Two identical spheres of different masses resting on soft ground should have shown uniform settlement if fall at the same rate towards the center of earth.
2-   If the size and density of sphere A is greater than sphere B. As A has more surface area therefore B should penetrate more than A in the ground if both fall at the same “g”.

The motion is negligible for small masses but it can't be ignored when the size of the falling mass is increased. Also, we are discussing theory.
Case #3 = Case #1 +case#2
We know g = GM/d^2 is the gravitational strength of any point mass. As soon as the on-center distance “d” b/t the two objects starts reducing from both ends simultaneously, the “g “ of earth and the “g” of mass start increasing at the same times than it were considered to be like in case #1 or case #2. Both masses are gravitating and falling at the same time therefore they enter into the gravitational field of one another very quickly not like in case#1 or Case#2.

But that doesn't mean that the mass is making it accelerate faster.
The change in distance to the other object, by the other object moving is what does.
It doesn't show any fault with the theory.

As “g” neither depends on falling mass nor gravitational force therefore if all objects would fall at the same rate then those objects should have shown at least equivalency/ similarity in settlements. For example:

No, it shouldn't.
Falling is different to sitting on a surface.

Just like if you throw 2 different objects at a wall at the same speed.
A lightweight object can bounce off with nothing happening to the wall while a much denser and heavier object can break straight through the wall.

Even with both moving at the same velocity, their interaction with the wall is different.

The same applies with falling and settling.
Even if 2 objects fall at the same rate, that doesn't mean they will cause the same indentation in the ground, even if gently placed on the ground.

Once more, the rate of acceleration is g.
The force, is m*g.

The settling is based upon the force, not the acceleration.