• Re: Newton's Gravity

    From Jonathan Thornburg [remove -color t@21:1/5 to Luigi Fortunati on Wed Jan 1 08:08:45 2025
    In article <vl0q35$28cau$[email protected]> 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]" <[email protected]>
    on the west coast of Canada

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jonathan Thornburg [remove -color t@21:1/5 to Luigi Fortunati on Fri Jan 3 22:18:17 2025
    In article <vl0q35$28cau$[email protected]> 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 <[email protected]>, 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$2sdba$[email protected]>, 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]" <[email protected]>
    "[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

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jens Schweikhardt@21:1/5 to All on Mon Jan 6 15:01:29 2025
    Luigi Fortunati <[email protected]> wrote
    in <vlejo4$15i8k$[email protected]>:
    [...]
    # 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)

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jonathan Thornburg [remove -color t@21:1/5 to Luigi Fortunati on Wed Jan 8 07:49:57 2025
    In article <vlejo4$15i8k$[email protected]>, 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$28t5l$[email protected]>, 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]" <[email protected]>
    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

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jonathan Thornburg [remove -color t@21:1/5 to [email protected] on Thu Jan 9 08:33:08 2025
    In article <<[email protected]> 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]" <[email protected]>
    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

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Thomas Koenig@21:1/5 to Luigi Fortunati on Tue Jan 14 08:06:07 2025
    Luigi Fortunati <[email protected]> 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?

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jonathan Thornburg [remove -color t@21:1/5 to Luigi Fortunati on Thu Jan 16 08:38:21 2025
    [Moderator's note: That's a repost of the previous contribution of the
    author with a slight slight edit]

    In article <vm2d31$1mbqr$[email protected]>, 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$2acss$[email protected]>, 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]" <[email protected]>
    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.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Jonathan Thornburg [remove -color t@21:1/5 to Luigi Fortunati on Thu Jan 16 08:17:10 2025
    In article <vm2d31$1mbqr$[email protected]>, 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]" <[email protected]>
    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.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)