General > General Technical Chat
Physics Question - ma = mg
RJSV:
But STILL:
I'm saying to the OP 'Boston' question that
mg in f=mg
is NOT movement, the 'g' is called 'acceleration constant' for our case on Earth sized object.
That's a static force equation, a FIELD having potential
to accelerate IF that's a free body to be moved, a la 'Newton'.
The other way; ' f=ma', is for the case when you are pushing something: Your force, again, on a body free to move can cause an accelerated movement. And I believe the physics folks had stated, that the nature of the motion, such as including E= 1/2 mv(squared), cannot be distinguished, as either case has same equations and outcomes.
That's why I would tend to suggest that 'g' be called an acceleration constant, for specifying a static situation, like weighing your gold bar.
AND, 'a' is a more general term, signifying how (your) hand pushing a free body out in space in any single isolated case, is going to, YES, actually start moving that sucker, accordingly. You can change that by pushing harder or less hard. And modulate the resulting acceleration alongside.
? See any mention, of photons, above?
? Rest Mass ?
? Year 2021 ?
No, I'm presenting 'Year 1740' arguments, and that approach allows for all the advanced physics that we have today, as well.
bostonman:
--- Quote --- That's why I would tend to suggest that 'g' be called an acceleration constant, for specifying a static situation, like weighing your gold bar.
AND, 'a' is a more general term, signifying how (your) hand pushing a free body out in space in any single isolated case, is going to, YES, actually start moving that sucker, accordingly. You can change that by pushing harder or less hard. And modulate the resulting acceleration alongside.
--- End quote ---
This is what I initially questioned. I hate to return to the same question, and, in a way, I'm asking rhetorically.
ma can't equal mg all the time. If F=ma, and, to compliment your hand pushing a free body example, the a can be anything. Only in the case of objects (or free bodies) where gravity is acting on it can ma = mg.
Little 'g' should just be called a constant, however, it can't be a constant since I can be standing on a mountain and gravity is less (by a small amount) than someone else standing at sea level.
From what I understood before initiating this thread is that them being equal has little meaning and isn't anything to "brag" about because we live on Earth. Anywhere in the universe ma can equal mg.
Now as for the in depth, informative, and long replies, what I find is that very smart people dislike providing simple answers because much more exists. If I'm pushing a box, I may just consider the pushing factor, whereas, friction should also be included.
Obviously my physics knowledge is low, so I look for the general answers based on simple problems. If I jump right into free bodies, friction, the actual gravity at my height, etc... that will just confuse things causing less focus on the actual question.
I tend to provide long, in depth answers too when my friends ask me simple questions. It's never easy for me to say the sky is blue without explaining that it's a shade of blue because.... blah blah
RJSV:
Original Post person: BostonMan:
That last looked good, but I want to ask, was the original 'g' a variable, such as in saying "The 'g' in the Mars example will be smaller.
OR, is 'g' just a solid constant, for Earth sized cases ?
Gives even more context.
Going forward from here, I got no problem, accumulating ever more science 'depth', such as the photon related stuff in posts above. Like the guy said, any innovation in theories must take account what we already know from classical-only.
bostonman:
--- Quote --- That last looked good, but I want to ask, was the original 'g' a variable, such as in saying "The 'g' in the Mars example will be smaller.
OR, is 'g' just a solid constant, for Earth sized cases ?
--- End quote ---
I assume you meant 'that last post looked good'.
Little 'g' being a constant in the sense that it's understood we are talking about gravity on Earth. Now 9.8m/s^2 is a general rule for physics 101, but, if I dug a five-mile deep hole, gravity would be less because it becomes negligible (???) at the center of Earth. Also, if I stood on a five-mile high mountain, it would be less because I'm further from the surface of Earth.
This means that to argue gravity on Earth is 9.8m/s^2 would be wrong because it can change depending on your height on Earth. Unfortunately, little 'g' can also mean gravity on Mars (or anywhere in the universe) because if I want to know my Weight on mars, I need to use W=mg. To call little 'g' a constant would be wrong, but, I hate to critique TBBT, but I think saying ma = mg is taken out of context.
It's like saying your age is equal to my age, but if your birthday is in March and mine is December, we are the same age for a short period of time. So ma = mg only in cases involving the discussion of Weight on Earth and a free falling object on Earth. Otherwise, a free body may have different acceleration, or, maybe the free body takes place on Mars and now ma = mg thus throwing out 9.8 as little 'g'.
From what I learned about Mass, it's not always the same throughout the universe, however, physics 101 tells me it is. From what I learned from this thread is: mass can change due to small tiny factors; none of which I was aware of.
If anything, I knew gravity is different at different heights and 9.8 is the general number to use for general calculations, but I didn't know mass changed, but, in any case, the fact this was mentioned reminds me to watch how I speak (and type).
My pet peeve is people who argue with statements like 'it's a fact', or 'best ever'. Nothing is a fact unless it can be proven and best ever is just silly.
It's a fact I'm sitting at the keyboard typing, it's not a fact that anyone reading this can argue it's a fact I typed this. I could have had someone type it for me, used speech to text, etc...
Anyway, my point is that whether this thread went much, much deeper, or I (and others) were completely wrong to state things in such a fashion, just reminded me to speak correctly, and, really, it helped me understand how much more exists to Weight, Mass, etc... Now at least I can say 9.8 is good, but know if I were to become a rocket scientists, more exists than just 9.8.
TimFox:
g is a real acceleration. Since the gravitational force on a dropped mass is proportional to that mass, the acceleration of a falling body will be independent of its mass.
Simple algebra: the acceleration of mass m1 is a = F/m1 = (1/m1) x G m1m2/R2 = G m2/R2, which we define as g.
Here, G is Newton's universal gravitational constant (not g), m1 is the mass of the dropped object, m2 is the other mass (Earth or moon or whatever), and R is the distance between the two objects, specifically between their centers of gravity. In our case, that is the radius of the Earth or moon.
In a given region where the gravity is uniform, such as a cubic meter at the earth's surface, any object whatsoever, of whatever mass (neglecting air resistance) when dropped freely will be accelerated by an acceleration equal to g.
In another region, such as a cubic meter at the moon's surface, the value of g is different than the value on earth, due to the different mass and radius of the large attractor (moon vs. earth).
I gave the equation for the period of a pendulum, which is independent of the mass, and has g explicitly in the formula: the period of a given pendulum will therefore be different on the moon or on earth.
There are further details, such as the fact that during our drop experiment the motion of the Earth can be neglected, since it is much heavier.
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version