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Physics Question - ma = mg
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RJSV:
Yup..., BostonMan;
   I think we are in pretty good shape, today. I had wondered if 'g' expressed more explicitly would maybe be; g(subscript Mars) for Martian surface, g(subscript Moon) for moon surface measurements, etc. Thanks for clarification.
   This is all voluntary, remember the eevblog hosts might say "Participate, if you please, but please keep it positive."

   And timfox providing some 'sane' relief.
   And as to veryveryveryvery small extra elements to any parameters under discussion: They are just that, and mostly 'negligable' in everyday live. Now, with modern growth, we know about, and we somewhat know which effects take effect strongly, and when such effects (while still there) can be ignored.
That guidance would maybe fill out the scope, of OP's general question.
    Does ma = mg ? Yes, under some certain circumstances, that being adjusting your applied force, and then additional math, if relativistic conditions dictate, (strong, or very weak).
   It's a volunteer task, blogging, and so makes any frustrations a whole order of magnitude more tolerable!!
Nominal Animal:

--- Quote from: RJHayward on June 28, 2021, 02:08:11 am ---the 'g' is called 'acceleration constant' for our case on Earth sized object.
--- End quote ---
Sure; I called it acceleration due to gravity, which I intended to be understood in exact same sense as 'acceleration constant'.

Perhaps this helps:

What is the speed of light, if you are in a dark chamber without a single photon? Does the speed of light still exist?  Yes, of course.

Does the speed of light in vacuum, c, exist outside photons?  Yes, of course.  We now define c as exactly 299,792,458 m/s, using that to derive the exact length of one meter.  (One second is defined as being equal to the time duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the fundamental unperturbed ground-state of the caesium-133 atom.)

Can acceleration exist with zero velocity?  Yes, of course.  Jump up into the air.  At the peak of your trajectory, your velocity is exactly zero, but have about 9.80m/s² of downward acceleration (and that stays pretty much constant from the moment your legs are in fully extended position and no longer touch the ground, until they touch the ground again; as air resistance affects the acceleration a bit).

In calculating forces using formulae like F = m a, one must remember that this does not necessarily involve any movement, because only the sum total of such forces acting on an object are really physically measurable.  Indeed, the box sitting still on the floor is subject to force F1 = m g due to gravity, and to force F2 due to the surface it is sitting on resisting deformation; and from the box being still, we know these forces must cancel out, and are therefore equal but opposite.

If we pick a subset of forces, we can sometimes get very unrealistic ideas about the situation.  For example, one could estimate the gravitational force exerted on the box by the northern hemisphere of the universe, and then claim that the result somehow must be balanced by the static force due to the ground not deforming under the box.  Uh, no, if you pick a silly set of forces, the model itself is then silly, and does not describe reality.

Another way to define force is via an energy potential and the vector gradient operator:
$$\vec{F} = -\nabla V$$
The scalar field \$V\$ is not any kind of energy emanation or all-pervading field that binds us together; it is simply a measure of energy.  On isosurfaces where \$V\$ is constant, you do not need to transfer any energy.  (If we have a frictionless surface here on Earth, in a vacuum chamber so there is no air resistance, an object on that surface would keep their velocity, until they bump into a wall or something.)

For gravitational potential energy between two point-like masses, it is \$V = -G M m / r\$, where \$G\$ is the gravitational constant, \$M\$ and \$m\$ are their masses, and \$r\$ is their distance.  Along the line between those point-like masses, gradient is just partial derivative with respect to \$r\$, and \$F = \partial (G M m / r) / \partial r = - G M m / r^2\$.

Now, if we combine a few of those constants, mainly via \$g = G M / r^2\$, we end up with \$F = m g\$.

What is this magic constant \$g\$?  Why, nothing but the gravitational constant [m³ kg⁻¹ s⁻2] times mass [kg] divided by the squared mean radius of Earth [ m⁻² ].  Dimensional analysis shows that the resulting units are [ m s⁻2 ] = m/s².  Acceleration.  What about the numerical value?  6.674×10⁻¹¹×5.972×10²⁴/(6.371×106)² ≃ 9.82.

Ah-ha!  If we treat gravity as just one form of potential energy, described as \$V = -G M m / r\$ between point-like particles having masses \$M\$ and \$m\$ separated by distance \$r\$, we end up with each of them seeing a force \$F = m g\$ towards their shared center of mass.

So, in this case, \$g\$ has nothing to do with acceleration per se; it is just some constant we derived from Earth's mass and radius, and the gravitational constant \$G = 6.674\cdot10^{-11} \, m^3 \, kg^{-1} \, s^{-2}\$.

The reason it turns out to 9.82 m/s² and not 9.80 m/s², is that because of inertia, and Earth spinning around on its axis, objects near the equator have an inertial force (or fictitious force, as it is caused by inertia and not by any potential energy) away from the axis of rotation of about 0.03 m/s² or so.  This force (again, which is due to inertia, so not caused by any potential energy) is \$F = m \omega^2 r\$, where \$m\$ is the mass of the object rotating, \$r\$ is the radius of rotation, and \$\omega\$ is the angular velocity.  (You can convert \$x\$ RPM to radians per second using \$\omega = 2 \pi x [\text{rad}/\text{s}] / 3600 [\text{RPM}].  \$\omega = 2\pi [\text{rad}/\text{s}]\$ is one turn per second.)

Because the gravity field of the Earth is not consistent with sea level, it is just much simpler to fuzz it all and include that inertial effect in the mess for simplicity (because the other stuff like gravitational "anomalies" – just a word meaning the effects due to Earth being lumpy and fluid inside, not smooth and evenly distributed – mean this approximation is already off by so much that including the inertial effects won't make it any worse).

But, as shown above, just the fact that we have a term that both has the units of acceleration, and in some cases can be physically measured (by observing a weight falling in vacuum), does not mean that actual motion has to be involved.  Just like speed of light in vacuum is a constant, that has uses related to photons and stuff, but is not dependent on there being something zipping along at that velocity.  Or any velocity: remember \$E = m c^2\$.

(I cannot help myself: "But what about electrons orbiting atomic nuclei? Don't they experience centripetal forces?" No, because electrons are delocalized in that fuzzy quantum mechanical way, and not sharp points in orbits around the nuclei like you often see pictured.  The shape, or probability density function, varies depending on the orbital.  Each electron thus delocalized does have a property we call angular momentum, which is deeply involved in how those orbitals get their shapes; and although this angular momentum is split into different parts (spin angular momentum which kinda works out as if the particle was spinning on its axis, but in key aspects isn't; and orbital angular momentum which kinda works as if the delocalized electron had the properties of rotating around the nucleus but doesn't) some confuse the real-world-sounding terms (which even sometimes behave like the properties with those same names in classical, non-quantum mechanics), and end up being the graphics artists I mentioned, asking the physicist why they don't just use pastel tones to reduce gravity.)

It pains me to see intelligent people struggle to comprehend ideas, only because the terms we use are so damn misleading/unintuitive/confusing.

Make no mistake, they are that to me too.  I have a hard time associating any specific "law" or property or constant with just the name of the person who discovered it, because although of historical interest, those names do not help at all in grasping the concepts they represent; you just have to learn them by rote.  And I don't.  At least "invariant mass"/"rest mass" versus "relativistic mass" is descriptive; if they were called "Lorenzian mass" and "Einsteinian mass", I think I'd have to quit physics altogether.
RJSV:
That last post a good one. A Nominal animal PHYSICIST beats an EE Everytime.
   Especially the parts explaining about moving or non-moving, in a world having forces and acceleration; framing for the reader the distinctions.
   AND the side-bits on electrons operating in non-intuitive ways, I mean, I had to turn off TV set so I could focus on reading clearly. Nice, and way over my head, but a learning experience.
bostonman:
Without getting too deep with math, or an explanation, how do we know an object has X kg of mass, and gravity (rounding off) is 9.8m/s^2 based on the beginning of time?

We know a 100kg mass is 100kg, but that's based off 9.8m/s^2, but gravity is derived from a Force, and the Force is derived from a known mass and gravity.

If we went back hundreds of years (or a twenty-six-hundred year journey as with TBBT - and I'm not trying to be funny), scientists needed a known gravity to calculate mass; or a known mass to calculate gravity.

I'm sure the math has been worked out several times over hundreds of years, but it seems because each factor is derived from the other, any one can be wrong throwing off the other.

Same with calculating the mass of planets. I believe it's done by looking at how gravity pulls on surrounding stars (?) and thus mass is calculated, however, we needed to start off with the mass of one planet which would be Earth (I should say I assume we started with Earth). If we got the mass of Earth wrong based off using a mass we thought was X kg, then everything derived since is wrong.
TimFox:
The mass of an object is completely independent of the local acceleration of gravity g.
The weight of an object is directly proportional to the local acceleration of gravity g.
It is an interesting philosophical question, which requires serious astrophysical investigation, if the universal gravitational constant G has changed over time.
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