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?
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 <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)?
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?
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.
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."
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?
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!
...
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.
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.
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.
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
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?
...
[[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 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) [[...]]
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?
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) [[...]]
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?
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?
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?
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