Newton's formula F=GmM/d^2 has been used to great advantage so far
because it has proven to be valid and almost perfectly correct except
for the small discrepancy in the perihelion calculation of Mercury's
orbit, where Einstein's gravity formulas prove to be more precise.

So, Newton's formula is *almost* correct but not quite.

In this formula, the force is proportional to the product of the two
masses (m*M).

Suppose that body A has mass M=1000 and body B has mass m=1, so that
the force between the two bodies is proportional to 1000 (mM=1*1000).

If another unit mass 1 is added to body B, its mass doubles to m=2 and
the force acting between the two bodies also doubles, because it will
be proportional to 2000 (mM=2*1000).

But if the other unit mass is added to body A (instead of body B) the
mass of A will become equal to M=1001 (remaining almost unchanged) just
as the force between the two bodies remains practically unchanged and
will be proportional to 1001 (mM=1*1001).

Why does the force acting between the two bodies double if we add the
unit mass to body B and, substantially, does not change if we add it to
the mass of body A?

Luigi Fortunati

In article <vl0q35$28cau$1@dont-email.me> Luigi Fortunati wrote:
> Suppose that body A has mass M=1000 and body B has mass m=1 [[...]]
>
> If another unit mass 1 is added to body B, its mass doubles to m=2 and
> the force acting between the two bodies also doubles, [[...]]
>
> But if the other unit mass is added to body A (instead of body B) the
> mass of A will become equal to M=1001 (remaining almost unchanged) just
> as the force between the two bodies remains practically unchanged [[...]]
>
> Why does the force acting between the two bodies double if we add the
> unit mass to body B and, substantially, does not change if we add it to
> the mass of body A?

Why not? Why might we expect the effects of adding mass in one location
(A) to be the same as those of adding mass in a different location (B)?

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
on the west coast of Canada

In article <vl0q35$28cau$1@dont-email.me> Luigi Fortunati wrote:
> Suppose that body A has mass M=1000 and body B has mass m=1 [[...]]
>
> If another unit mass 1 is added to body B, its mass doubles to m=2 and
> the force acting between the two bodies also doubles, [[...]]
>
> But if the other unit mass is added to body A (instead of body B) the
> mass of A will become equal to M=1001 (remaining almost unchanged) just
> as the force between the two bodies remains practically unchanged [[...]]
>
> Why does the force acting between the two bodies double if we add the
> unit mass to body B and, substantially, does not change if we add it to
> the mass of body A?

Why not? Why might we expect the effects of adding mass in one location
(A) to be the same as those of adding mass in a different location (B)?

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
on the west coast of Canada

Jonathan Thornburg [remove -color to reply] il 01/01/2025 09:08:45 ha
scritto:
> In article <vl0q35$28cau$1@dont-email.me> Luigi Fortunati wrote:
>> Suppose that body A has mass M=1000 and body B has mass m=1 [[...]]
>>
>> If another unit mass 1 is added to body B, its mass doubles to m=2 and
>> the force acting between the two bodies also doubles, [[...]]
>>
>> But if the other unit mass is added to body A (instead of body B) the
>> mass of A will become equal to M=1001 (remaining almost unchanged) just
>> as the force between the two bodies remains practically unchanged [[...]]
>>
>> Why does the force acting between the two bodies double if we add the
>> unit mass to body B and, substantially, does not change if we add it to
>> the mass of body A?
>
> Why not? Why might we expect the effects of adding mass in one location
> (A) to be the same as those of adding mass in a different location (B)?

Yes, we *should* expect the same effects if we mean the same thing by
"effects."

I'm talking about masses (causes) and forces (effects): what effects
are you talking about?

Luigi Fortunati

In article <vl0q35$28cau$1@dont-email.me> Luigi Fortunati wrote:
> Suppose that body A has mass M=1000 and body B has mass m=1 [[...]]
>
> If another unit mass 1 is added to body B, its mass doubles to m=2 and
> the force acting between the two bodies also doubles, [[...]]
>
> But if the other unit mass is added to body A (instead of body B) the
> mass of A will become equal to M=1001 (remaining almost unchanged) just
> as the force between the two bodies remains practically unchanged [[...]]
>
> Why does the force acting between the two bodies double if we add the
> unit mass to body B and, substantially, does not change if we add it to
> the mass of body A?

In article <ltkbcoF9g4cU1@mid.dfncis.de>, I replied
| Why not? Why might we expect the effects of adding mass in one location
| (A) to be the same as those of adding mass in a different location (B)?

In article <vl3tv1$2sdba$1@dont-email.me>, Luigi replied
> Yes, we *should* expect the same effects if we mean the same thing by
> "effects."
>
> I'm talking about masses (causes) and forces (effects): what effects
> are you talking about?

Let's analyze a somewhat more general system: Suppose we have a pair
of masses A and B, and consider the effects of adding a mass C at either
position #1 or position #2.
[Luigi's original question had position #1 = position
of A, position #2 = position of B, mass A = 1000, mass
B = 1, and mass C = 1, but I find it useful to consider
the more generic case.]

A+B+C1 and A+B+C2 are *physically different* systems (going from one to
the other involves moving the mass C from position #1 to position #2).
So why should we expect any of the following Newtonian gravitational
effects to be the same between these two *physically different* systems:
* Newtonian gravitational potential U at some test point X
* Newtonian gravitational acceleration "little-g" at some test point X
(= - gradient of U)
* force between A+C1 and B versus force between A and B+C2

In fact, it's easy to see that all three of these "effects" differ... as
we should expect, because (again) we're comparing *physically different*
systems.

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
"[I'm] Sick of people calling everything in crypto a Ponzi scheme.
Some crypto projects are pump and dump schemes, while others are pyramid
schemes. Others are just standard issue fraud. Others are just middlemen
skimming off the top. Stop glossing over the diversity in the industry."
-- Pat Dennis, 2022-04-25

Jonathan Thornburg [remove -color to reply] il 03/01/2025 23:18:17 ha
scritto:
> Let's analyze a somewhat more general system: Suppose we have a pair
> of masses A and B, and consider the effects of adding a mass C at either
> position #1 or position #2.
> [Luigi's original question had position #1 = position
> of A, position #2 = position of B, mass A = 1000, mass
> B = 1, and mass C = 1, but I find it useful to consider
> the more generic case.]
>
> A+B+C1 and A+B+C2 are *physically different* systems (going from one to
> the other involves moving the mass C from position #1 to position #2).
> So why should we expect any of the following Newtonian gravitational
> effects to be the same between these two *physically different* systems:
> * Newtonian gravitational potential U at some test point X
> * Newtonian gravitational acceleration "little-g" at some test point X
> (= - gradient of U)
> * force between A+C1 and B versus force between A and B+C2
>
> In fact, it's easy to see that all three of these "effects" differ... as
> we should expect, because (again) we're comparing *physically different*
> systems.

It's true, you convinced me because your reasoning is completely
logical and shareable: we expect anything except that two physically
different systems give the same effects.

But then, why do two extraordinarily different systems like the Earth's
mass (6*10^24kg) generate the force of 90kg-weight on my body (mass
90kg) and my body generates the *same* opposing force of -90kg-weight
on the Earth?

Luigi Fortunati

Luigi Fortunati <fortunati.luigi@gmail.com> wrote
in <vlejo4$15i8k$1@dont-email.me>:
[...]
# But then, why do two extraordinarily different systems like the Earth's
# mass (6*10^24kg) generate the force of 90kg-weight on my body (mass
# 90kg) and my body generates the *same* opposing force of -90kg-weight
# on the Earth?

Asking "why" in physics usually means "is there a more elementary
explanantion?"

Do you accept

F = G*m1*m2/r^2 (1)

as an empirical observation?

Indeed, nobody has ever measured the effect of your body's gravitational
force on the Earth. The orders of magnitude for the respective
accelerations are too different. Verification of that formula is only
technically feasible for large pairs of masses, say Earth/Moon or
Sun/Jupiter by observing both bodies in orbit around their barycenter.
This requires each body being subject to equal but opposite forces.

The answer could be "because the masses in (1) appear without preference
for either." Or "because multiplication is commutative". Or "because
when (1) is written in vector notation, the force vectors have the same
magnitude, but opposing direction when the masses are exchanged."

Regards,

Jens
--
Jens Schweikhardt https://www.schweikhardt.net/
SIGSIG -- signature too long (core dumped)

Jens Schweikhardt il 06/01/2025 16:01:29 ha scritto:
> Luigi Fortunati <fortunati.luigi@gmail.com> wrote
> in <vlejo4$15i8k$1@dont-email.me>:
> [...]
> # But then, why do two extraordinarily different systems like the Earth's
> # mass (6*10^24kg) generate the force of 90kg-weight on my body (mass
> # 90kg) and my body generates the *same* opposing force of -90kg-weight
> # on the Earth?
>
> Asking "why" in physics usually means "is there a more elementary
> explanantion?"
>
> Do you accept
>
> F = G*m1*m2/r^2 (1)
>
> as an empirical observation?

The formula (1) has proven itself very well but has failed a bit on the
precession of the perihelion of Mercury's orbit and, *much* more so, on
the motion of the peripheral stars of our galaxy, which forces us to
imagine unobtainable dark matter and energy.

So it is reliable but not perfect.

> Indeed, nobody has ever measured the effect of your body's gravitational
> force on the Earth. The orders of magnitude for the respective
> accelerations are too different.

Exactly, we know perfectly well the gravitational force of the Earth on
me but we do not know at all my gravitational force on Earth.

And I find it very difficult to imagine and accept that the
gravitational force of the miserable mass of my body could exert a
force of 90kg-weight on the Earth or on any other body.

The real problem with formula (1) is that it concerns only one total
force that acts between the masses m1 and m2, without distinguishing
how much force m1 exerts on m2 and how much m2 exerts on m1.

This was remedied by accepting Newton's third law as valid.

> Verification of that formula is only
> technically feasible for large pairs of masses, say Earth/Moon or
> Sun/Jupiter by observing both bodies in orbit around their barycenter.

This is not true.

How can we verify a formula in which there are masses that no one has
ever measured? How much is the mass of the Earth, the Moon, Jupiter and
the Sun?

Do you know them? No.

You cannot know them exactly because no one has ever measured them
directly.

We can only assume that we have certain masses by deriving them
indirectly through formula (1) and, thus, we verify the validity of a
formula using values taken from the formula itself!

> This requires each body being subject to equal but opposite forces.

No, the opposite forces are equal only when the rotation occurs around
the midpoint and not when it occurs around a centre of rotation that is
closer to one mass than the other.

> The answer could be "because the masses in (1) appear without preference
> for either." Or "because multiplication is commutative". Or "because
> when (1) is written in vector notation, the force vectors have the same
> magnitude, but opposing direction when the masses are exchanged."

These are not proofs.

So far, we have accepted the equality between the action of m1 on m2
and that of m2 on m1, counting on the validity of Newton's third law,
which I have shown to be invalid.

I repeat: the claim that my miserable gravitational force can attract
the Earth with the same force (90kg-weight) with which the Earth
attracts me (as the equality between action and reaction claims) is
unacceptable!

Regards,
Luigi

In article <vlejo4$15i8k$1@dont-email.me>, Luigi Fortunati asked
> But then, why do two extraordinarily different systems like the Earth's
> mass (6*10^24kg) generate the force of 90kg-weight on my body (mass
> 90kg) and my body generates the *same* opposing force of -90kg-weight
> on the Earth?

and later, in article <vljh6k$28t5l$1@dont-email.me>, Luigi asked
> I repeat: the claim that my miserable gravitational force can attract
> the Earth with the same force (90kg-weight) with which the Earth
> attracts me (as the equality between action and reaction claims) is
> unacceptable!

To generalize Luigi's questions, if we have a pair of masses M1 and M2,
fixed in position (with respect to a Newtonian inertial reference frame
(IRF), to keep things simple) some distance apart, with M1 not equal to
M2, is there any good reason to think that the gravitational force of M1
acting on M2 is equal in magnitude and opposite in direction to the
gravitational force of M2 acting on M1?

There are actually a couple of useful lines of reasoning, each of which
suggests that the answer is "yes":

To start with, notice that Newton's law of gravitation specifically
states that the answer is "yes". So we're basically asking whether
Newton's law of gravitation is in fact an accurate description of
reality (in the domain where we expect it to work, i.e.,
non-relativistic non-quantum systems).

(1) We can directly measure the gravitational force between masses in
a laboratory, and do indeed find them to agree with Newton's laws.
See
https://en.wikipedia.org/wiki/Cavendish_experiment
for an introduction to the "classic" experiments on this (dating
back to the late 1700s), and
https://en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_experiment
for an introduction to experiments verifying one somewhat subtle
aspect of the Newtonian gravitational interaction.

(2) Think about what would happen if the force of M1 on M2 were NOT
equal-in-magnitude-and-opposite-in-direction to the force of M2
on M1: the only way the sum of two vectors can be zero is if the
two vectors are equal-in-magnitude-and-opposite-in-direction, so
if the force of M1 on M2 were NOT
equal-in-magnitude-and-opposite-in-direction to the force of M2 on
M1, then the (vector) *sum* of these two forces, i.e., the total
gravitational force on M1+M2, would be nonzero.

Let's now imagine that M1 and M2 are held apart by a light stick
so as be at a fixed distance from each other, forming a "dumbbell"
(still at reset in a Newtonian IRF, and let's say floating out in
space far from any other masses). Then (if the force of M1 on M2
were NOT equal-in-magnitude-and-opposite-in-direction to the force
of M2 on M1), that nonzero "total gravitational force on M1+M2"
would accelerate the dumbbell with respect to the Newtonian IRF,
violating the law of conservation of momentum.

Moreover, that acceleration would result in the dumbbell having
kinetic energy, so we've also violated the law of conservation of
energy. For example, if we put our dumbbell sideways on a turntable
(e.g., if we're looking down on the turntable, put the dumbell
oriented horizontally, attached to the turntable's 12-o-clock
position), this acceleration would start the turntable rotating.
So, if we put an electric generator on the turntable's axis, we
would have a "free" source of energy, i.e., a perpetual motion
machine.

So, to summarize, we've shown that if the force of M1 on M2 were
NOT equal-in-magnitude-and-opposite-in-direction to the force of
M2 on M1, then you could violate the laws of conservation of momentum
and conservation of momentum, and build a perpetual motion machne.

For a variety of good reasons that I won't go into here, we think
that the laws of physics forbid violations of the laws of conservation
of momentum or energy (and hence forbid the existence of perpetual
motion machines),
so this argument (strongly) suggests that in fact the force of M1 on
M2 *IS* equal-in-magnitude-and-opposite-in-direction to the force of
M2 on M1.

[In the context of general relativity "conservation of
momentum" and "conservation of energy" are rather tricky
concepts, because there's no good way to add up energy/momentum
"here" and energy/momentum "there" to get a total energy/momentum.
And in general relativistic cosmology things get trickier still.
I'm going to ignore all of these subtleties here, and stick to
Newtonian gravity/mechanics.]

(3) In my gedanken system of M1 and M2 being joined by a light stick,
the stick isn't actually necessary. You could actually have M1 and
M2 in orbit about each other, and if the force of M1 on M2 were not
equal-in-magnitude-and-opposite-in-direction to the force of M2 on
M1, then the center of mass of M1 and M2 would oscillate around at
the orbital period. And, if the M1-M2 orbit were eccentric, then
if the force of M1 on M2 were not
equal-in-magnitude-and-opposite-in-direction to the force of M2 on
M1, the center of mass would have a net acceleration with respect
to a Newtonian inertial reference frame.

We can measure the motion of the center of mass (by timing the radio
signals of binary or millisecond pulsars arriving on Earth) for the
case where M1 is the Earth and M2 is the moon, and for the case where
M1 is the Sun and M2 is Jupiter/Saturn/other planets.

[In fact, my very first published scientific paper was
about this measurement. It was only 2 pages long, and
had exactly one novel idea in it. My Ph.D advisor looked
at it and said "that's one more idea than in some papers
I've seen". :) ]

Experimentally, we find that the M1-M2 center of mass does NOT
oscillate or accelerate in this way, suggesting (strongly) that
the force of M1 on M2 *is* in fact
equal-in-magnitude-and-opposite-in-direction to the force of M2
on M1.

So, in conclusion, the basic answer to Luigi's question "*Why* is the
force of M1 on M2 equal-in-magnitude-and-opposite-in-direction to the
force of M2 on M1?" is "conservation of momentum/energy".
--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
currently on the west coast of Canada
"[I'm] Sick of people calling everything in crypto a Ponzi scheme.
Some crypto projects are pump and dump schemes, while others are pyramid
schemes. Others are just standard issue fraud. Others are just middlemen
skimming off the top. Stop glossing over the diversity in the industry."
-- Pat Dennis, 2022-04-25

Jonathan Thornburg [remove -color to reply] il 08/01/2025 08:49:57 ha
scritto:
> ...
> Let's now imagine that M1 and M2 are held apart by a light stick
> so as be at a fixed distance from each other, forming a "dumbbell"
> (still at reset in a Newtonian IRF, and let's say floating out in
> space far from any other masses). Then (if the force of M1 on M2
> were NOT equal-in-magnitude-and-opposite-in-direction to the force
> of M2 on M1), that nonzero "total gravitational force on M1+M2"
> would accelerate the dumbbell with respect to the Newtonian IRF,
> violating the law of conservation of momentum.

This reasoning of yours is very interesting; after reading it I went
for a walk, which is my favorite way to think.

It was an intense and very pleasant half hour.

If there is a stick that keeps the two bodies at a fixed distance, it
means that there are other forces besides the gravitational ones.

Without the stick there is only the gravitational force of body A on
body B and the opposite gravitational force of B on A.

These two forces add together because they both make up the overall
gravitational force that attracts the two bodies towards each other.

The stick has another function completely opposite to the attractive
force of gravity and, in fact, the stick repels instead of attracting,
it opposes, both on the side of body A and on the other side, opposing
the approach of the two bodies.

We therefore have two attractive forces and two repelling forces.

The set of these four forces is absolutely balanced: the sum of the
forces directed toward the approach is exactly equal and opposite to
the sum of the forces directed toward the separation.

If the stick is not there, bodies A and B approach each other by
accelerating (attractive gravitational forces), if the stick is there,
the two bodies A and B stop approaching each other (repelling reaction
forces).

These are two distinct conditions: free motion (without the stick)
constrained motion (with the stick).

If I jump from the wall, while I am in the air the Earth attracts me
towards it and I also attract the Earth towards me: we both exert an
attractive force.

When I land on the floor, the floor repels me and I repel the floor (we
both repel each other preventing the approach).

I am against very long posts that are dispersive, so I will stop here.

In the next post I will focus on the equality or otherwise of the two
distinct pairs of forces in relation to the conservation of momentum.

The moderators, if they wish, can publish this post in the meantime
while waiting for the other, or they can wait to publish them together
(obviously if they find them acceptable).

Luigi Fortunati

In article <<lu6p15F9qp8U1@mid.dfncis.de> I wrote
> if we have a pair of masses M1 and M2,
> fixed in position (with respect to a Newtonian inertial reference frame
> (IRF), to keep things simple) some distance apart, with M1 not equal to
> M2, is there any good reason to think that the gravitational force of M1
> acting on M2 is equal in magnitude and opposite in direction to the
> gravitational force of M2 acting on M1?
>
> There are actually a couple of useful lines of reasoning, each of which
> suggests that the answer is "yes":
>
> [[...]]
>
> So, to summarize, we've shown that if the force of M1 on M2 were
> NOT equal-in-magnitude-and-opposite-in-direction to the force of
> M2 on M1, then you could violate the laws of conservation of momentum
> and conservation of momentum, and build a perpetual motion machne.

I'm sorry, I garbled that last quoted sentence. What I meant to write
was this:

So, to summarize, we've shown that if the force of M1 on M2 were
NOT equal-in-magnitude-and-opposite-in-direction to the force of
M2 on M1, then you could violate the laws of conservation of momentum
and conservation of energy, and build a perpetual motion machne.

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
currently on the west coast of Canada
"[I'm] Sick of people calling everything in crypto a Ponzi scheme.
Some crypto projects are pump and dump schemes, while others are pyramid
schemes. Others are just standard issue fraud. Others are just middlemen
skimming off the top. Stop glossing over the diversity in the industry."
-- Pat Dennis, 2022-04-25

Luigi Fortunati il 08/01/2025 03:11:43 ha scritto:
> Jonathan Thornburg [remove -color to reply] il 08/01/2025 08:49:57 ha
> scritto:
>> ...
>> Let's now imagine that M1 and M2 are held apart by a light stick
>> so as be at a fixed distance from each other, forming a "dumbbell"
>> (still at reset in a Newtonian IRF, and let's say floating out in
>> space far from any other masses). Then (if the force of M1 on M2
>> were NOT equal-in-magnitude-and-opposite-in-direction to the force
>> of M2 on M1), that nonzero "total gravitational force on M1+M2"
>> would accelerate the dumbbell with respect to the Newtonian IRF,
>> violating the law of conservation of momentum.
>
> This reasoning of yours is very interesting; after reading it I went
> for a walk, which is my favorite way to think.
>
> It was an intense and very pleasant half hour.
>
> If there is a stick that keeps the two bodies at a fixed distance, it
> means that there are other forces besides the gravitational ones.
>
> Without the stick there is only the gravitational force of body A on
> body B and the opposite gravitational force of B on A.
>
> These two forces add together because they both make up the overall
> gravitational force that attracts the two bodies towards each other.
>
> The stick has another function completely opposite to the attractive
> force of gravity and, in fact, the stick repels instead of attracting,
> it opposes, both on the side of body A and on the other side, opposing
> the approach of the two bodies.
>
> We therefore have two attractive forces and two repelling forces.
>
> The set of these four forces is absolutely balanced: the sum of the
> forces directed toward the approach is exactly equal and opposite to
> the sum of the forces directed toward the separation.
>
> If the stick is not there, bodies A and B approach each other by
> accelerating (attractive gravitational forces), if the stick is there,
> the two bodies A and B stop approaching each other (repelling reaction
> forces).
>
> These are two distinct conditions: free motion (without the stick)
> constrained motion (with the stick).
>
> If I jump from the wall, while I am in the air the Earth attracts me
> towards it and I also attract the Earth towards me: we both exert an
> attractive force.
>
> When I land on the floor, the floor repels me and I repel the floor (we
> both repel each other preventing the approach).
>
> I am against very long posts that are dispersive, so I will stop here.
>
> In the next post I will focus on the equality or otherwise of the two
> distinct pairs of forces in relation to the conservation of momentum.

In my animation https://www.geogebra.org/m/ntefhssz I have visualized
the two bodies A and B with their respective decreasing gravitational
fields.

Body B (smaller) is entirely immersed in the strong red ring of force
10 of the gravitational field of A, while body A (whose center of
gravity is far from body B) is only marginally touched by the weak
gravitational force of body B.

If we reduce body B even further to the minimum of its mass (with the
appropriate slider), the gravitational force that B experiences from
body A is at its maximum, while (on the contrary) the gravitational
force of body B becomes practically non-existent and also acts only in
one point and not on the whole of body A (more or less like the
gravitational force of my body acts very weakly on only one point of
the Earth (my room at most) and not on the whole Earth.

Instead, if we increase the mass of body B to the maximum, it becomes
equal to that of body A and, only at this point, the two opposing
gravitational forces (A towards B and B towards A) become totally
equal, adding their mutual attractive effects.

Luigi Fortunati

Luigi Fortunati il 11/01/2025 08:53:59 ha scritto:
> In my animation https://www.geogebra.org/m/ntefhssz I have visualized
> the two bodies A and B with their respective decreasing gravitational
> fields.
>
> Body B (smaller) is entirely immersed in the strong red ring of force
> 10 of the gravitational field of A, while body A (whose center of
> gravity is far from body B) is only marginally touched by the weak
> gravitational force of body B.
>
> If we reduce body B even further to the minimum of its mass (with the
> appropriate slider), the gravitational force that B experiences from
> body A is at its maximum, while (on the contrary) the gravitational
> force of body B becomes practically non-existent and also acts only in
> one point and not on the whole of body A (more or less like the
> gravitational force of my body acts very weakly on only one point of
> the Earth (my room at most)) and not on the whole Earth.
>
> Instead, if we increase the mass of body B to the maximum, it becomes
> equal to that of body A and, only at this point, the two opposing
> gravitational forces (A towards B and B towards A) become totally
> equal, adding their mutual attractive effects.

The consequence of all this is that the gravitational force of the
larger body of mass M acts on the entire mass <m> of the smaller body
and this justifies the product m*M of Newton's formula, which
corresponds to the force exerted by the larger mass M on the entire
mass <m>.

Instead, the gravitational force of the smaller body of mass <m> cannot
act on the entire body of mass M because M is larger and therefore acts
only on a part of body A of size compatible with <m> and, therefore,
the force of body B on body A is not proportional to m*M but to m*m.

Consequently, the total gravitational force is proportional to the sum
of m*M plus m*m (mM+mm=m(M+m)).

Newton's formula should contain this small change: from F=GmM/d^2 to
F=Gm(m+M)/d^2 (with m<=M) which returns to being equal to the previous
one when the mass <m> is negligible (as happens here on Earth to any
body with respect to the entire Earth).

In fact, when <m> is negligible, m+M (for all practical purposes) is
equal to M.

Is it possible to carry out an experiment to verify which of the two
formulas (F=GmM/d^2 and F=Gm(m+M)/d^2 with m<=M) is more adherent to
reality?

Luigi Fortunati

Ps. If the formula F=Gm(m+M)/d^2 (with m<=M) turns out to be more
correct, the masses of the planets and stars (calculated with the
formula F=GmM/d^2) will have to be recalculated.

Luigi Fortunati <fortunati.luigi@gmail.com> schrieb:

> The consequence of all this is that the gravitational force of the
> larger body of mass M acts on the entire mass <m> of the smaller body
> and this justifies the product m*M of Newton's formula, which
> corresponds to the force exerted by the larger mass M on the entire
> mass <m>.
>
> Instead, the gravitational force of the smaller body of mass <m> cannot
> act on the entire body of mass M because M is larger

That is a non sequitur if there ever was one. Why should this be the
case?

Think of a mass M as being divided into i smaller submasses (all
with the same mass m_part) and of a mass j of being divided into
m smaller submasses with the same mass m_part. Which submass of M
should not interact all submasses of j?

Thomas Koenig il 14/01/2025 09:06:07 ha scritto:
> Luigi Fortunati <fortunati.luigi@gmail.com> schrieb:
>
>> The consequence of all this is that the gravitational force of the
>> larger body of mass M acts on the entire mass <m> of the smaller body
>> and this justifies the product m*M of Newton's formula, which
>> corresponds to the force exerted by the larger mass M on the entire
>> mass <m>.
>>
>> Instead, the gravitational force of the smaller body of mass <m> cannot
>> act on the entire body of mass M because M is larger
>
> That is a non sequitur if there ever was one. Why should this be the
> case?
>
> Think of a mass M as being divided into i smaller submasses (all
> with the same mass m_part) and of a mass j of being divided into
> m smaller submasses with the same mass m_part. Which submass of M
> should not interact all submasses of j?

You can divide the entire mass M of the Earth into as many sub-masses
as you want and it (as a whole) will continue to exert its
gravitational force on the entire mass <m> of my body (from which M*m
derives).

On the other hand, if you divide the mass <m> of my body into as many
sub-masses as you want, it (as a whole) will never be able to exert its
miserable gravitational force on *all* the immense mass <M> of the
Earth but will limit itself to the mass <m> of my room without going
beyond (so: not m*M but m*m).

Luigi Fortunati

[[Mod. note -- You've made a bunch of statements here. Do you have
any evidence for them? We do have a fair bit of experimental data
on Newtonian gravitation... are your statements consistent with the
experimental data?
-- jt]]

Luigi Fortunati il 14/01/2025 10:12:03 ha scritto:
>...
> [[Mod. note -- You've made a bunch of statements here. Do you have
> any evidence for them? We do have a fair bit of experimental data
> on Newtonian gravitation... are your statements consistent with the
> experimental data?
> -- jt]]

The experimental data that are consistent with Newton's formula are
also consistent with the formula F=Gm(m+M)/d^2 (with m<=M) whenever <m>
is negligible compared to <M>, because the negligible <m> makes <m+M>
equal to <M> and the two formulas coincide.

Consequently, all experiments where <M> is the mass of the Earth and
<m> is any laboratory mass (infinitely small compared to <M>) also
confirm my formula.

Instead, observations of planets and stars are consistent only with the
formula F=GmM/d^2 because their masses (obviously never measured
directly) were obtained precisely with the formula F=GmM/d^2.

If we recalculated the masses of planets and stars with the formula
F=Gm(m+M)/d^2 (with m<=M), the observational data would be consistent
only with this formula and not with the other.

Finally, there are the experimental data with torsion balances on
bodies of small and verified mass, but these forces (as far as I know)
are too small to be measured in newtons.

And precisely for this reason, I asked if there was an experiment of
this last type suitable for verifying which of the two formulas is more
adherent to reality.

Is there anyone well-informed about torsion balances who can answer?

Luigi Fortunati

In article <vm2d31$1mbqr$1@dont-email.me>, Luigi Fortunati wrote:
> The consequence of all this is that the gravitational force of the
> larger body of mass M acts on the entire mass <m> of the smaller body
> and this justifies the product m*M of Newton's formula, which
> corresponds to the force exerted by the larger mass M on the entire
> mass <m>.
>
> Instead, the gravitational force of the smaller body of mass <m> cannot
> act on the entire body of mass M because M is larger and therefore acts
> only on a part of body A of size compatible with <m> and, therefore,
> the force of body B on body A is not proportional to m*M but to m*m.
>
> Consequently, the total gravitational force is proportional to the sum
> of m*M plus m*m (mM+mm=m(M+m)).
>
> Newton's formula should contain this small change: from F=GmM/d^2 to
> F=Gm(m+M)/d^2 (with m<=M) [[...]]

Suppose we have 3 similar masses A, B, and C, arranged like this:

B
A C

with B and C touching so as to form a compound object B+C. What is the
horizontal gravitational force between A and the compound object B+C?

If we follow Luigi's formula, we'd get
G m_A (m_A + m_BC)/d^2 = G m_A (m_A + m_B + m_C)/d^2 (1)

But another way to calculate this same force is that it's just the
sum of the horizontal gravitational force between A and B, and the
horizontal gravitational force between A and C. (The vertical
gravitational forces between B and C don't matter.) Again using
Luigi's formula, this gives
G m_A (m_A + m_B)/d^2 + G m_A (m_A + m_C)/d^2
= G m_A (2m_A + m_B + m_C)/d^2 (2)

Clearly, calculating the horizontal gravitational force via (1) gives
a different answer from calculating it via (2).

In other words, if we consider decomposing the larger mass into pieces,
Luigi's formula gives two different results for the same quantity, i.e.,
the formula is self-contradictory.

Here's a related problem: what if m_B = m_C = m_A/2 so that
m_A = m_B + m_C, i.e. A has the same mass as B+C? How do we decide
which body (A or B+C) should be "m" and which should be "M" in Luigi's
formula?

Newton's formula always gives the same result (G m_A (m_B + m_C)/d^2)
now matter how the masses are decomposed.

Luigi also asked:

> Is it possible to carry out an experiment to verify which of the two
> formulas (F=GmM/d^2 and F=Gm(m+M)/d^2 with m<=M) is more adherent to
> reality?

Yes. With a torsion pendulum it's fairly easy to directly measure
the gravitational forces between laboratory masses. See the Wikipedia
article
https://en.wikipedia.org/wiki/Cavendish_experiment

Here's nice collection of reprint articles on these and similar
measurements:

G. T. Gillies, editor
"Measurements of Newtonian Gravitation"
American Association of Physics Teachers, 1992
ISBN 0-917853-46-6

I should also note the conference "Testing Gravity 2025" being held
Jan 29-Feb 2 in Vancouver, Canada,
https://www.sfu.ca/physics/cosmology/TestingGravity2025/
I'll be attending this conference, and I'll try to post a synopsis
of some of the presentations to s.p.r. The conference program includes
a talk by someone from the Eot-Wash group discussing tortion-pendulum
and similar exeriments (Michael Ross, "New experimental tests of gravity
from Eot-Wash group").

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
on the west coast of Canada
The Three Laws of Thermodynamics:
1) You can't win, only lose or break even.
2) You can only break even at absolute zero.
3) You can't get to absolute zero.

[Moderator's note: That's a repost of the previous contribution of the
author with a slight slight edit]

In article <vm2d31$1mbqr$1@dont-email.me>, Luigi Fortunati wrote:
> The consequence of all this is that the gravitational force of the
> larger body of mass M acts on the entire mass <m> of the smaller body
> and this justifies the product m*M of Newton's formula, which
> corresponds to the force exerted by the larger mass M on the entire
> mass <m>.
>
> Instead, the gravitational force of the smaller body of mass <m> cannot
> act on the entire body of mass M because M is larger and therefore acts
> only on a part of body A of size compatible with <m> and, therefore,
> the force of body B on body A is not proportional to m*M but to m*m.
>
> Consequently, the total gravitational force is proportional to the sum
> of m*M plus m*m (mM+mm=m(M+m)).
>
> Newton's formula should contain this small change: from F=GmM/d^2 to
> F=Gm(m+M)/d^2 (with m<=M) [[...]]

In article <vm536m$2acss$1@dont-email.me>, Thomas Koenig pointed out
a crucial ambiguity with Luigi's suggested formula. His argument may
be easier to follow if we consider a simple special case: Suppose we
have 3 similar masses A, B, and C, arranged like this:

B
A
C

with B and C actually touching (hard to represent in ASCII-art) so as to
form a compound object B+C. What is the horizontal gravitational force
between A and the compound object B+C?

If we follow Luigi's formula, we'd get
G m_A (m_A + m_BC)/d^2 = G m_A (m_A + m_B + m_C)/d^2 (1)

But another way to calculate this same force is that it's just the
sum of the horizontal gravitational force between A and B, and the
horizontal gravitational force between A and C. (The vertical
gravitational forces between B and C don't matter.) Again using
Luigi's formula, this gives
G m_A (m_A + m_B)/d^2 + G m_A (m_A + m_C)/d^2
= G m_A (2m_A + m_B + m_C)/d^2 (2)

Clearly, calculating the horizontal gravitational force via (1) gives
a different answer from calculating it via (2).

In other words, if we consider decomposing the larger mass into pieces,
Luigi's formula gives two different results for the same quantity, i.e.,
the formula is (depending on your taste in words) either ambiguous or
self-contradictory.

Here's a related problem: what if m_B = m_C = m_A/2 so that
m_A = m_B + m_C, i.e. A has the same mass as B+C? How do we decide
which body (A or B+C) should be "m" and which should be "M" in Luigi's
formula?

Newton's formula always gives the same result (G m_A (m_B + m_C)/d^2)
now matter how the masses are decomposed.

Luigi also asked:

> Is it possible to carry out an experiment to verify which of the two
> formulas (F=GmM/d^2 and F=Gm(m+M)/d^2 with m<=M) is more adherent to
> reality?

Yes. With a torsion pendulum it's fairly easy to directly measure
the gravitational forces between laboratory masses. See the Wikipedia
article
https://en.wikipedia.org/wiki/Cavendish_experiment

Here's nice collection of reprint articles on these and similar
measurements:

G. T. Gillies, editor
"Measurements of Newtonian Gravitation"
American Association of Physics Teachers, 1992
ISBN 0-917853-46-6

I should also note the conference "Testing Gravity 2025" being held
Jan 29-Feb 2 in Vancouver, Canada,
https://www.sfu.ca/physics/cosmology/TestingGravity2025/
I'll be attending this conference, and I'll try to post a synopsis
of some of the presentations to s.p.r. The conference program includes
a talk by someone from the Eot-Wash group discussing tortion-pendulum
and similar exeriments (Michael Ross, "New experimental tests of gravity
from Eot-Wash group").

--
-- "Jonathan Thornburg [remove -color to reply]" <dr.j.thornburg@gmail-pink.com>
on the west coast of Canada
The Three Laws of Thermodynamics:
1) You can't win, only lose or break even.
2) You can only break even at absolute zero.
3) You can't get to absolute zero.

Jonathan Thornburg [remove -color to reply] il 16/01/2025 09:38:21 ha
scritto:
> In article <vm536m$2acss$1@dont-email.me>, Thomas Koenig pointed out
> a crucial ambiguity with Luigi's suggested formula. His argument may
> be easier to follow if we consider a simple special case: Suppose we
> have 3 similar masses A, B, and C, arranged like this:
>
> B
> A
> C
>
> with B and C actually touching (hard to represent in ASCII-art) so as to
> form a compound object B+C. What is the horizontal gravitational force
> between A and the compound object B+C?

Is this diagram of yours the one on the left side of my animation
https://www.geogebra.org/m/yvvc2rap ?

If not, can you tell me what changes I need to make to make it the way
you thought?

>
> Here's a related problem: what if m_B = m_C = m_A/2 so that
> m_A = m_B + m_C, i.e. A has the same mass as B+C? How do we decide
> which body (A or B+C) should be "m" and which should be "M" in Luigi's
> formula?

If m=M there is no decision to make because (replacing M with m) the
formula becomes F=G2m^2/d^2.

Luigi Fortunati