Re: Newton's Gravity

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