Thank you for replying,noAxioms

I suppose you can use [the two added equations] to compute the rate of change in distance between the two objects, only in a 1-dimensional case, but that rate isn't acceleration.
— noAxioms

I agree with you; I certainly am not trying to give a 3 dimensional situation where one might have some partial derivatives and some 3rd derivatives as these can become too complicated.
I think we can deal with the Newtonian equation using only one dimention...I don’t believe that Newton had a three dimensional frame of reference in mind when he was introducing his equation.
However , to claim that g is not acceleration I still don’t get; Im not saying that you are wrong, I’m saying I don’t understand.

It's not that it's simpler. Adding the equations only produces a useful result if there is no motion except along one axis. So for instance, under Newtonian mechanics, the ISS is continuously accelerating (coordinate acceleration) towards Earth at about 8.7 m/s² and yet its distance from Earth is roughly fixed, and its speed relative to Earth is also roughly fixed. This is because it is not a 1d case. The ISS has motion in a direction other than just the axis between it and Earth.
The ISS example also serves as a wonderful example of the difference between the physics definition of acceleration (rate of change in velocity=8.7 m/s² down) vs street definition (zero change in speed). — noAxioms

I fully agree that anything which has an orbit will have a combination of gravitational acceleration and sidereal velocity at the same time...this, I’m certain can become very complicated if one throws in a frame of reference from which everything is calculated. If the orbit isn’t circular (which most aren’t) there will be a change in acceleration (Jerk) being a 3rd derivative.....but our example is 1 dimensional, the distance need not to be great.
It’s very simple,noAnxiom; what I’m asking is this: the acceleration of a body in freefall is GM/r...
What is it about this mass (M), that is so special that the smaller mass (m) cannot have any influence whatsoever on the acceleration?

Grampa Dee

Thank you for responding PhilosophyRunner

To use your car example, I am standing on the pavement not accelerating. I see one car (A) moving accelerating at 3m/s2 towards my left and another car (B) accelerating at 5m/s2 towards my right. That is simply the acceleration of the cars. It is meaningless (from a classical mechanics sense) to then add these up and say car A is accelerating at 8m/s2 from the point of view of the car B. Neither car is accelerating at 8m/s2. Car A is accelerating at 3m/s2 and car B at 5m/s2. — --`r

I would agree that car A is accelerating at 3m/^s^2 and car B is accelerating at 5m/sec^2 relative to the road, however, when we view this as being relative to themselves, then I can only detect 1 acceleration, not two.
Here, I am strictly speaking changes in velocities, having nothing to do with the g force that "might" be involved, as in freefall, there is no force that is being felt by the accelerated body.

thank you again for your time
Grampa Dee

I think we can deal with the Newtonian equation using only one dimention. — Gampa Dee

Any talk of orbital mechanics has velocity that is not parallel with the acceleration, and thus involves at least two dimensions. Orbits can be described in a plane. Only things falling straight up and down can be described in one dimension.

I don’t believe that Newton had a three dimensional frame of reference in mind

All frames of reference have 3 dimensions of space, and require 3 velocity components to describe.

However , to claim that g is not acceleration I still don’t get

g is acceleration, but is simply a constant scalar. So the gravitational pull on the surface of Saturn is 1.08g. It would be wrong to say Saturn has a slightly larger g.

I fully agree that anything which has an orbit will have a combination of gravitational acceleration and sidereal velocity at the same time.

Yes, and those orthogonal vectors make it at least a 2d situation.

this, I’m certain can become very complicated if one throws in a frame of reference from which everything is calculated.

A frame of reference is a starting point, not something thrown in. Without it, any velocity is completely undefined.

If the orbit isn’t circular (which most aren’t) there will be a change in acceleration

Orbital accelerations are always changing, even in the circular case. Remember that it is a vector.

the acceleration of a body in freefall is GM/r...
What is it about this mass (M), that is so special that the smaller mass (m) cannot have any influence whatsoever on the acceleration?

There's nothing special about M since it works with any M. Galileo actually published a tidy proof that the acceleration is independent of the mass of the thing accelerating.

I would agree that car A is accelerating at 3m/^s^2 and car B is accelerating at 5m/sec^2 relative to the road — Gampa Dee

No. Velocity is relative, so it makes sense to talk about velocity relative to the road, but acceleration is absolute. The cars are accelerating at 3 and 5 m/s² period. This is true in any frame. It is meaningless to talk about acceleration relative to something, including itself.

Here, I am strictly speaking changes in velocities

Yes, that is what coordinate acceleration is. Change in velocity is absolute, even if velocity itself is not. If you're using a non-inertial frame, then you're taking the absolute coordinate acceleration of the other car and adjusting for the alternative frame you're using (in which Newton's laws do not hold), but the coordinate acceleration of the other car is still the same.

You're also adding wrong. For instance, I have two stationary equal-mass objects (planets say) that I release. If you add the accelerations, you get zero, but the relative velocity between them is changing. Acceleration is a vector and if you add them, you need to use vector addition. But It seems to be a mistake to add them at all in this case.

having nothing to do with the g force that "might" be involved, as in freefall, there is no force that is being felt by the accelerated body.

The force 'felt' would be proper acceleration, also absolute.

thank you for replying noAxioms

I think we can deal with the Newtonian equation using only one dimention.
— Gampa Dee
Any talk of orbital mechanics has velocity that is not parallel with the acceleration, and thus involves at least two dimensions. Orbits can be described in a plane. Only things falling straight up and down can be described in one dimension. — noAxioms

Yes; of course...my mistake in writing 1 dimensional problem as being Newtonian...I was thinking of freefall.

I don’t believe that Newton had a three dimensional frame of reference in mind

All frames of reference have 3 dimensions of space, and require 3 velocity components to describe. — noAxioms

I understand; but Newton developed his equation from Kepler, and Kepler was viewing the problem in a two dimensional fashion.... What time did it take for a planet to complete an orbit, which was viewed as simply an orbital velocity, being 2 dimensional in essence, as the time was proportional to the area . Yet, through mathematical manipulation, his constant R^3 / T^2 .
we can get:

a*r^2 = v^2* r
a =v^2 / r....
and
v^2 = a*r
v = sqrt a*r

We can get both the acceleration towards the sun as well as the orbital velocity.This, one would think,
would need a 3 dimensional coordinate system.

However , to claim that g is not acceleration I still don’t get

g is acceleration, but is simply a constant scalar. So the gravitational pull on the surface of Saturn is 1.08g. It would be wrong to say Saturn has a slightly larger g. — noAxioms

The surface g acceleration is dependent to the mass density...and Saturn, being a gaseous planet,would be much less dense than the earth . Would you have the equation for this?
The reason why I had written M / Area was to have the denominator as being r^2 instead of r^3
as in the case of mass density.

A frame of reference is a starting point, not something thrown in. Without it, any velocity is completely undefined. — noAxioms

I was thinking something like Kepler using only distances and time period to make up his equations, and so not really using a frame of reference although it was implied I am sure.

Orbital accelerations are always changing, even in the circular case. Remember that it is a vector. — noAxioms

I thought that is was constant because the direction is towards the center, but I think I see what you mean. So the orbit magnitude in a circular orbit will remain constant ,while when the orbit is elliptical, the magnitude of acceleration will be changing and a change in acceleration is a 3rd derivative, which makes the calculation more tedious , I would think.

There's nothing special about M since it works with any M. Galileo actually published a tidy proof that the acceleration is independent of the mass of the thing accelerating. — noAxioms

I do have a hard time with this...is it possible to share this proof?

Grampa Dee

The surface g acceleration is dependent to the mass density — Gampa Dee

g is a constant. Saturn doesn't have a different g, it has a different acceleration a. The acceleration is dependent on mass and radius and has little direct connection with density, especially since density of any planet/star varies considerably at different depths. You complicating things needlessly by trying to work with area and/or density.

and Saturn, being a gaseous planet,would be much less dense than the earth . Would you have the equation for this?

Yea, a = GM/r²

A frame of reference is a starting point, not something thrown in. Without it, any velocity is completely undefined.
— noAxioms
I was thinking something like Kepler using only distances and time period to make up his equations, and so not really using a frame of reference although it was implied I am sure.

The equations you referenced mention velocity, and velocity is meaningless without the frame reference.

I thought that [Orbital acceleration] was constant because the direction is towards the center[/quote]That direction changes over time. The ISS acceleration direction changes one degree every 15 seconds for instance. If it didn't, the ISS would have left the solar system a long time ago.

while when the orbit is elliptical, the magnitude of acceleration will be changing and a change in acceleration is a 3rd derivative, which makes the calculation more tedious , I would think.

Yes, but you listed all this Kepler stuff that shows how to do that just nicely.

I do have a hard time with this...is it possible to share this proof?

Nothing in science is actually a proof, but the reasoning goes like this: You have two identical balls of mass M each. They presumably fall at the same rate. Now you connect them by a thin thread or spot of glue, creating one object of mass 2M. Either the connection makes some sort of magical difference, or the 2M mass should fall at the same rate as before. Hence the rate is independent of mass.

One can reason that the connection cannot apply a downward force to both objects (violating Newton's laws), but Newton's laws were not available at that time yet. F=ma was unknown, and the theory of impetus and such was still a thing.

Thank you for responding noAxioms

g is a constant. Saturn doesn't have a different g, it has a different acceleration a. — noAxioms

Ok; I’ll try to remember to use “a” instead :) ...but I did see some charts which identify planets using g in the same way they use distance in terms of earth distance units (AU).

https://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html

The acceleration is dependent on mass and radius and has little direct connection with density, especially since density of any planet/star varies considerably at different depths. You complicating things needlessly by trying to work with area and/or density. — noAxioms

Well, for me personally, I ask myself whether there is a meaning behind r^2 besides simply being a distance squared. It would seem logical, for me, to view gravity as being, for example, some sort of energy emanating from the massive body (probably at light`s speed), which would continuously be reducing it`s energy density as it travelled away from the mass source, This would, in my mind, constitute a form of diverging medium which could be responsible for the acceleration. This would mean that r^2 would have something to do with the surface area....

I do have a hard time with this...is it possible to share this proof?

Nothing in science is actually a proof, but the reasoning goes like this: You have two identical balls of mass M each. They presumably fall at the same rate. Now you connect them by a thin thread or spot of glue, creating one object of mass 2M. Either the connection makes some sort of magical difference, or the 2M mass should fall at the same rate as before. Hence the rate is independent of mass. — noAxioms

The 2 M already side by side(even without the string) while accelerating towards the earth, I believe also accelerates the earth towards themselves equally, so no difference will be seen.... I have also mentioned this in one of the posts to unenlightened.

I would in fact surely agree to viewing the system as a whole(earth and falling bodies)as being invariant in terms of energy “if” the fallen body was taken from the earth (did not come from outer space), for in this case, the falling body was formally on the ground being part of the cause for the g acceleration, and since while the falling body will attract the earth as well, then,for the whole system, the overall acceleration will not change.[/b]


But the equation would still remain a = GM/r^2 + Gm / r^2 .... since what was taken from the ground no longer is part of the earth’s gravity but is now attracting the earth instead.

One can reason that the connection cannot apply a downward force to both objects (violating Newton's laws), but Newton's laws were not available at that time yet. F=ma was unknown, and the theory of impetus and such was still a thing. — noAxioms

Well, I think that Galileo was simply claiming that if a heavier object accelerates faster, then, the two masses tied together ought to accelerate faster than the two masses when separate.
Since, like you said, the idea of every mass attracting every other mass was not yet developed,
it wasn't a bad argument.

Grampa Dee..

Ok; I’ll try to remember to use “a” instead :) ...but I did see some charts which identify planets using g in the same way they use distance in terms of earth distance units (AU). — Gampa Dee

Those charts would be correct. Gravity (a) on Saturn is ~1.08g and it orbits at about 9.5 AU. Both g and AU are constants. Saturn does not define a different AU any more than it defines a different g.

https://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html

Interesting that it quotes the Saturn gravity at 0.92g, less than that of Earth. Their definition of where the surface is must be considerably higher than the more common altitude. It's not like it actually has a surface like 'sea level' or anything, and Earth gravity is not measured where the gas density becomes negligible. Its number of moons is also considerably out of date.

Well, for me personally, I ask myself whether there is a meaning behind r^2 besides simply being a distance squared. It would seem logical, for me, to view gravity as being, for example, some sort of energy emanating from the massive body (probably at light`s speed), which would continuously be reducing it`s energy density as it travelled away from the mass source,

That's one way of looking at it. A light shining at intensity proportional to the mass would decrease in brightness at the square of the distance from the light. But gravity doesn't travel, so you can take the analogy only so far. Gravitational waves travel at c, but gravitational waves are not responsible for the attraction between masses. They're only responsible for carrying the changes to the field, which involves energy expenditure only when the field is changing. For instance, Earth's orbit radiates about 200 watts of gravitational waves into space.

I would in fact surely agree to viewing the system as a whole(earth and falling bodies)as being invariant in terms of energy

Energy is conserved in a closed system, yes. So is momentum. Earth/moon system is not particularly a closed system for energy since so much of it comes in and also leaves, both by EM radiation.

For example, the moon's orbital distance increases by several cm a year as Earth transfers its angular momentum to it. It also transfers about 3% of its angular energy to it, and radiates the remainder away as heat. So energy is lost in the process of the moon's orbital changes.

“if” the fallen body was taken from the earth (did not come from outer space), for in this case, the falling body was formally on the ground being part of the cause for the g acceleration, and since while the falling body will attract the earth as well, then,for the whole system, the overall acceleration will not change.[/b]

If the falling body was taken from the ground, that reduces M, and it will thus accelerate less than a similar object falling from space. For a small rock, the difference is immeasurable since the mass of Earth changes more per second than any nitpicking about where you got the rock.

The moon is like that. It is a big rock that was essentially taken from the ground, not having arrived from space. When they brought back the moon rocks, they were quite surprised to find it was a chunk of Earth and not something else like every other moon out there. But it accelerates at Σ GM/r² where Σ is the summation of the accelerations due to every bit of mass out there, mostly the due to the sun and also Earth (what's left of it after the moon was scooped out), but everything else pulls at it as well.

But the equation would still remain a = GM/r^2 + Gm / r^2

This cannot be right. For two equal stationary masses, it says they will not accelerate towards each other since the sum is zero. This is not the case. Even if you fixed that, the equation there does not express an acceleration, so the 'a=' part in front is blatantly wrong. Nothing accelerates at that rate.

Thank you for responding noAxioms

Ok; I’ll try to remember to use “a” instead :) ...but I did see some charts which identify planets using g in the same way they use distance in terms of earth distance units (AU).
— Gampa Dee
Those charts would be correct. Gravity (a) on Saturn is ~1.08g and it orbits at about 9.5 AU. Both g and AU are constants. Saturn does not define a different AU any more than it defines a different g.

https://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html — noAxioms

While I do understand that “g” as such is a constant, g is used as a form of measurement for other planets as well. Should have I rather written how many “g” s a planet’s surface gravity possess?

Well, for me personally, I ask myself whether there is a meaning behind r^2 besides simply being a distance squared. It would seem logical, for me, to view gravity as being, for example, some sort of energy emanating from the massive body (probably at light`s speed), which would continuously be reducing it`s energy density as it travelled away from the mass source,
That's one way of looking at it. A light shining at intensity proportional to the mass would decrease in brightness at the square of the distance from the light. But gravity doesn't travel, so you can take the analogy only so far. — noAxioms

For me personally, my upper statement would not constitute gravity as such...if we continue...

“to view gravity as being, for example, some sort of energy emanating from the massive body (probably at light`s speed), which would continuously be reducing it`s energy density as it travelled away from the mass source, this would, in my mind, constitute a form of diverging medium which could be responsible for the acceleration. This would mean that r^2 would have something to do with the surface area....”

I would rather identify the sort of diverging medium as having the quality of gravity.

For example, the moon's orbital distance increases by several cm a year as Earth transfers its angular momentum to it. It also transfers about 3% of its angular energy to it, and radiates the remainder away as heat. So energy is lost in the process of the moon's orbital changes. — noAxioms

Interesting... I heard the moon was moving away from the earth, but didn’t know the cause.

If the falling body was taken from the ground, that reduces M, and it will thus accelerate less than a similar object falling from space — noAxioms

I agree, and the way I view this is simply that the mass from outer space will increase the total gravity of the system,

For a small rock, the difference is immeasurable since the mass of Earth changes more per second than any nitpicking about where you got the rock. — noAxioms

The idea of nitpicking, noAxioms, is to try to explain a system’s physical laws...
For example, let us assume that I am correct (just for argument’s sake)...
I understand very well, that measurements have been made many times supporting Galileo’s law ( all bodies fall at the same rate).
However, if this is due to the difference of acceleration between two unequal masses in freefall as being too small to measure, then, the whole law would be incorrect even if our measurements couldn’t prove it.
We do know that stars orbital velocities at the edge of galaxies don’t agree with our laws, unless we add more “dark mass”....we have never detected any dark mass yet...so while the theory of dark matter is still valid, it remains that a possible different gravitational law ought to possible as well.

The moon is like that. It is a big rock that was essentially taken from the ground, not having arrived from space. When they brought back the moon rocks, they were quite surprised to find it was a chunk of Earth and not something else like every other moon out there. But it accelerates at Σ GM/r² where Σ is the summation of the accelerations due to every bit of mass out there, mostly the due to the sun and also Earth (what's left of it after the moon was scooped out), but everything else pulls at it as well. — noAxioms

But it accelerates at Σ GM/r²

If there is no mistake in this observation , then this will certainly prove that I am in error and that Galileo’s law is indeed valid.....

where Σ is the summation of the accelerations due to every bit of mass out there, mostly the due to the sun and also Earth (what's left of it after the moon was scooped out), but everything else pulls at it as well

However, this part does seem to show that there are some “unknown” elements as well.

But the equation would still remain a = GM/r^2 + Gm / r^2

This cannot be right. For two equal stationary masses, it says they will not accelerate towards each other since the sum is zero. This is not the case. — noAxioms

Yes...true; but again, we are dealing with semantics...I did not go to university to learn science, and my mathematical background is indeed limited. I guess we need to add vectors...
I will explain it the way I understand it... if a car comes towards you as you are driving your car, the measured velocity relative to both of you is v..., if you knew your velocity relative to the road as being .5v, then you would say that the other car is coming towards you at –.5v....the total v will not be 0...

Even if you fixed that, the equation there does not express an acceleration, so the 'a=' part in front is blatantly wrong. Nothing accelerates at that rate. — noAxioms

Ok,... maybe this is where I’m off the rail, although I don’t yet understand why...
Are you saying that instead of velocity, in the upper example I wrote, if we used acceleration instead , it wouldn’t work?

... if a car accelerates towards you as you are accelerating towards it, the total acceleration will “not” be a + a ?? I will think about that...

Thank you for your time

Grampa Dee

if a car comes towards you as you are driving your car, the measured velocity relative to both of you is v..., if you knew your velocity relative to the road as being .5v, then you would say that the other car is coming towards you at –.5v....the total v will not be 0... — Gampa Dee

We may be mixing up speed & velocity. Speed is a scalar but velocity is a vector.

In your example, both cars are traveling at a speed of 0.5v relative to the road (again no direction). In order to calculate the relative velocity we need to do vector calculations (which can get complicated). In the most general situation the two objects may be moving in arbitrary directions and may or may not ever collide. In your simplified one dimensional example, the velocity of one car is .5v and the velocity of the other car is -5v. But these are vectors, so to calculate the relative velocity between the two you cannot simply add the .5v & -5v and get 0. You have to factor in the direction and do vector math. The end result will be a relative speed of v - and the vector math will show a collision between the two cars at some point in time depending on the values of v and the starting distance between the two cars.

.Galileo’s law ( all bodies fall at the same rate). — Gampa Dee

That's not the law, and you wording is trivially falsified. I can drop a rock off a building and simultaneously throw another one downward. The thrown one will fall at a greater rate and arrive first.

So the law is something like "the coordinate acceleration of a freefalling body is independent of the mass of the freefalling body". I'm sure Galileo didn't word it that way, but it disambiguates between that and different kinds of acceleration.

However, this part does seem to show that there are some “unknown” elements as well.

Well yea. The alternative is some klnd of solipsism where things are gravitationally attracted only to known object, which makes the knower very special.

For two equal stationary masses, it says they will not accelerate towards each other since the sum is zero. This is not the case.
— noAxioms
I guess we need to add vectors...

Adding the vectors is what totals zero. The acceleration of neither object was correctly expressed since neither is zero.

if a car comes towards you as you are driving your car, the measured velocity relative to both of you is v...,

That's impossible. If a car is moving relative to me at v, then I am moving relative to it at it at -v by definition.

if you knew your velocity relative to the road as being .5v, then you would say that the other car is coming towards you at –.5v....the total v will not be 0.

Total v? You want to add velocity of me relative to the road to the velocity of them relative to me? The total of that is zero, and yes, that would give the velocity of them relative to the road. Their car is parked. If it isn't, then your figures can't be right. They are moving relative to me at -.5v and the road is also moving relative to me at -.5v, so the two are relatively stationary since they have identical velocity relative to me.

All this is illustrative of vector arithmetic, but seems otherwise unrelated to gravity and acceleration.

Even if you fixed that, the equation there does not express an acceleration, so the 'a=' part in front is blatantly wrong. Nothing accelerates at that rate.
— noAxioms

if a car accelerates towards you as you are accelerating towards it, the total acceleration will “not” be a + a ??

Correct. Nothing is accelerating at a+a (which is zero) nor a-a which is twice something. Doing would violate Newton's laws, F=ma in particular.
Each car is accelerating at 1a, presumably in opposite directions.

Take once again a pair of equal mass planets X&Y orbiting each other with X at coordinate -1, 0 and Y at 1,0..The both trace a unit circle about the origin at 0,0. X has a coordinate acceleration of 1,0 and Y has a coordinate acceleration of -1,0. Nothing is accelerating at 2 anything. The distance between them isn't changing over time, so adding the acceleration magnitudes does not in any way express the rate of change of their separation.

I suggest you read up on inertial and non-inertial frames of reference. I think your misunderstanding stems from not knowing the difference between them and treating both as if there is no difference.

Thank you for replaying PhilosophyRunner

↪Gampa Dee I suggest you read up on inertial and non-inertial frames of reference. I think your misunderstanding stems from not knowing the difference between them and treating both as if there is no difference. — PhilosophyRunner

Yes, I will do that;
and thank you again for your time.

Grampa Dee

Thank you for responding Erich

We may be mixing up speed & velocity. Speed is a scalar but velocity is a vector. — EricH

I may have; although I did mention the direction as being "towards each other"..

In your example, both cars are traveling at a speed of 0.5v relative to the road (again no direction). In order to calculate the relative velocity we need to do vector calculations (which can get complicated). In the most general situation the two objects may be moving in arbitrary directions and may or may not ever collide. In your simplified one dimensional example, the velocity of one car is .5v and the velocity of the other car is -5v. But these are vectors, so to calculate the relative velocity between the two you cannot simply add the .5v & -5v and get 0.[/quote]

I agree, this is indeed what I was saying; I did claim the total velocity as being v.
However,I was actually talking of accelerations, and the velocity example was only to make a point. PhilosophyRunner wrote saying that I will need to read up on inertial and non-inertial frames because this is the part that confuses me ...and this is what I'll be doing.
— EricH

You have to factor in the direction and do vector math. The end result will be a relative speed of v - and the vector math will show a collision between the two cars at some point in time depending on the values of v and the starting distance between the two cars. — EricH

Yes; I will need to look up vector math as well, because I just realized that addition of "tip to tail"
vectors will also give me 0 in this case.

Thank you for your time, Erich
Grampa Dee

Thank you for responding noAxioms

.Galileo’s law ( all bodies fall at the same rate).
— Gampa Dee
That's not the law, and you wording is trivially falsified. I can drop a rock off a building and simultaneously throw another one downward. The thrown one will fall at a greater rate and arrive first. — noAxioms

Maybe it’s not a law, I don’t know. but I did hear of Galileo’s law of falling bodies... if Newton has three laws of motion, I personally don’t see why Galileo’s “description” of freefalling bodies can’t be viewed in terms of a “law”. To throw a rock downwards will make the rock fall at a greater rate, as you mentioned... but we were talking about a freefall caused strictly by gravity.

However, this part does seem to show that there are some “unknown” elements as well.

Well yea. The alternative is some klnd of solipsism where things are gravitationally attracted only to known object, which makes the knower very special. — noAxioms

What I was saying, noAxiom was this:

. But it accelerates at Σ GM/r² ( talking about the moon)

If there is no mistake in this observation ,then this will certainly prove that Galileo’s law is indeed valid.....

If they can indeed account for every other mass in the sigma factor, then it should prove ( I think) that
Galileo was right ... and it should be considered a law, if it isn’t already.

if a car comes towards you as you are driving your car, the measured velocity relative to both of you is v..
.,
That's impossible. If a car is moving relative to me at v, then I am moving relative to it at it at -v by definition. — noAxioms

Sorry; I meant the relative velocity measured by you, the driver, was v

Total v? You want to add velocity of me relative to the road to the velocity of them relative to me? — noAxioms

...the car coming towards me I measure as being v; I also measure my velocity relative to the road as being -.5v...(sorry, I had this wrong...-.5v since I claimed the car coming towards me as having a velocity of +v already) then I calculate (I do not measure) the car as moving at velocity of .5v towards me(now that I know that I also am moving at -. 5v towards him/her)....I don’t know what is wrong with this example apart from my - sign error ...sorry

if a car accelerates towards you as you are accelerating towards it, the total acceleration will “not” be a + a ??
Correct. Nothing is accelerating at a+a (which is zero) nor a-a which is twice something. Doing would violate Newton's laws, F=ma in particular.
Each car is accelerating at 1a, presumably in opposite directions. — noAxioms

So how should we calculate the acceleration between two cars if...

Car 1 .... a1 = 2 m / s^2 .relative to the road
Car 2.... a2 = - 2 m /s^2 relative to the road

At time 1sec the first car has a velocity of 2 m / s , the second car will have a velocity of -2m/s
What is the relative “speed” between the two cars? I see 4m/s as measured by car1
at time 2 sec.the first car has a velocity of 4 m/s the second car will have a velocity of - 4 m/s
at time 3 sec the first car has a velocity of 6m/s the second car will have a velocity of -6 m/s
It seems the acceleration relative to both cars should then be, in my opinion 4 m/s^2 ...as measured by car 1

Having said this, I will also look up for the proper manner to do this right.

Take once again a pair of equal mass planets X&Y orbiting each other with X at coordinate -1, 0 and Y at 1,0..The both trace a unit circle about the origin at 0,0. X has a coordinate acceleration of 1,0 and Y has a coordinate acceleration of -1,0. Nothing is accelerating at 2 anything. The distance between them isn't changing over time, so adding the acceleration magnitudes does not in any way express the rate of change of their separation. — noAxioms

..No,.not in this case, because the acceleration is due completely to the change in direction...
But we were talking about the case of a freefalling body.
However, I don’t understand why one has the negative acceleration of the other..wouldn’t they crash in this case?
Also, I know that I’m mixed up when we’re dealing with vectors...
But, what we are discussing could be discussed in terms of speeds, if you will...we are trying to discuss whether a more massive body falls faster than a less massive one..


Grampa Dee.

.I don’t know what is wrong with this example apart from my - sign error ...sorry — Gampa Dee

You seem bent on adding or subtracting velocity or acceleration values, and if the sign is wrong on one of them, you get very incorrect results. So it's important.

So how should we calculate the acceleration between two cars

Yet again, acceleration is absolute. There is no 'acceleration between cars' since acceleration isn't a relation.

That said, you seem to be wanting to compute the Newtonian change in coordinate velocity of one car relative to the accelerating frame of the other car. Change in coordinate velocity is acceleration only for inertial coordinate systems.

Anyway, yes, that change in coordinate velocity of the object (the other car) is [acceleration of that object] minus [acceleration due to the coordinate system you're using].

Car 1 .... a1 = 2 m / s^2 .relative to the road
Car 2.... a2 = - 2 m /s^2 relative to the road

The road doesn't matter since acceleration is not a relation. Velocity is, but not acceleration.

At time 1sec the first car has a velocity of 2 m / s , the second car will have a velocity of -2m/s
What is the relative “speed” between the two cars? I see 4m/s as measured by car1

As measured by anybody actually.

It seems the acceleration relative to both cars should then be, in my opinion 4 m/s^2 ...as measured by car 1

Except it isn't acceleration. It is simply a change in coordinate speed of one car relative to the other car. Acceleration is something else, and is not a relation. But yes, relative speed between the cars (in Newtonain mechanics) changes at a rate of 4 m/s², assuming they started at a stop. It doesn't work in all cases if they don't start mutually stationary.

You can see why finding a reference about non-inertial frames might be useful. Accelerating frames are one kind, but there are several other kinds.

No,.not in this case, because the acceleration is due completely to the change in direction...
But we were talking about the case of a freefalling body.

I assure you that planets are freefalling. The term means that they're being acted upon by nothing but gravity. Under relativity theory, it means that their worldlines are straight, that is, they trace a geodesic through spacetime. But we're talking Newtonian mechanics here where gravity is a force.

However, I don’t understand why one has the negative acceleration of the other..wouldn’t they crash in this case?

In the case of masses in a mutual circular orbit , each mass has a tangential velocity relative to the other, so as they accelerate towards each other, they miss, maintaining a constant separation.

Also, I know that I’m mixed up when we’re dealing with vectors...
But, what we are discussing could be discussed in terms of speeds

Just use 'speed' instead of velocity if you mean the scalar. But careful, since addition and subtraction of speeds gives ambiguous results. Car A is moving at speed 5 relative to me and car B at 7 relative to me. What is the speed of A relative to B? Answer: not enough information supplied. Could be anywhere from 2 to 12. If velocity was used, there'd be just the one answer.

noAxioms — noAxioms

Thank you noAxioms, and good post;
I think that I'm not sufficiently prepared when discussing vectors, and so I'll
read more on the issue..

Thank you for your time.

Grampa Dee

I looked up on how to solve vector addition problems
Already,I don't seem to understand this addition of vectors

http://www.lon-capa.org/~mmp/kap6/cd149.htm

Adding the velocity vectors yields vt + v c = 0 mi/h.

For me, velocity is having a speed,in a certain direction. relative to something else,
So,the addition of vectors in this case is 0...but relative to what?

Since the velocities add up to 0, will the same be true for the momentum vectors? Of course not!
The momentum of the car is c = mc c= 20,000 kg m/s
The momentum of the truck is t = mt t= -131,000 kg m/s (negative because v is to the left)
The total momentum is therefore = c + t = -111,000 kg m/s

Again,we have a momentum,of -111 kgm/s....what does that even mean?

http://www.lon-capa.org/~mmp/kap6/cd149.htm

Adding the velocity vectors yields vt + v c = 0 mi/h. — Gampa Dee

Yea, but I find it very deceptive to add those two vectors since it doesn't produce a meaningful result. There's no such thing as 'the total velocity of a system'. If the car was inside the truck trailer and moving at vc relative to the truck, then adding vc to vt would yield the car velocity relative to the road. But that's not what's going on here.

The page never asks what the velocity difference is between the two, but the key word there is 'difference', in which case, to get the car velocity relative to the truck, one would subtract the vectors: vc - vt which would yield 130 mi/hr or about 58.1 m/sec

So,the addition of vectors in this case is 0...but relative to what?

As I said, it doesn't yield anything meaningful. I don't like the example text. It obfuscates more than it clarifies anything.
It also mixes metric (kg mass) with American units (mi/hr). That's unforgivable in a physics text.
Their momentum vectors seem to have a ratio of about 4, but the text says the ratio is closer to 7. That's poor graphics.

The total momentum is therefore = c + t = -111,000 kg m/s

Yes, there is such a thing as total momentum of the system. That addition is meaningful.
I also protest 111,000. I got a figure a bit lower than that, but they seem to lose accuracy when they don't use a consistent precision. The car momentum for instance is over 20300, but they round that to 20k.

Again,we have a momentum,of -111 kgm/s....what does that even mean?

Just what it says. For instance, if, in space (no friction with road), the car were to hit the truck and stick to it in a tangled wreck, the new 5200 kg mass would be moving left at about 21 m/sec to the left, the total momentum / total mass. Momentum is conserved in a closed system. I put them in space to keep it closed since the road would very much be exerting forces if it was there.

Thank you for replying,noAxioms

So,the addition of vectors in this case is 0...but relative to what?
As I said, it doesn't yield anything meaningful. I don't like the example text. It obfuscates more than it clarifies anything.
It also mixes metric (kg mass) with American units (mi/hr). That's unforgivable in a physics text.
Their momentum vectors seem to have a ratio of about 4, but the text says the ratio is closer to 7. That's poor graphics. — noAxioms

That’s exactly what I feel; there’s nothing meaningful...I don’t have any problems with the addition of the two vectors (tip to tail) being 0...but I don’t see anything relevant to the problem itself. I do understand , for example, that a force vector acting on a mass will cause an acceleration in the direction of the force, and adding an opposite force having the same magnitude will give an acceleration of 0.However, this is about two opposite forces acting on a single body.

Again,we have a momentum,of -111 kgm/s....what does that even mean?
Just what it says. For instance, if, in space (no friction with road), the car were to hit the truck and stick to it in a tangled wreck, the new 5200 kg mass would be moving left at about 21 m/sec to the left, the total momentum / total mass. Momentum is conserved in a closed system. I put them in space to keep it closed since the road would very much be exerting forces if it was there. — noAxioms

Very good; thank you....it made sense but only when you brought both masses together with the new velocity, the total having preserved the conservation of momentum....

Car 700kg * 29.05m/s = 20,340kgm/s
Truck 4500kg * -29.05m/s = -130,725 kgm/s
-110,385kgm/s being the new momentum after the crash....I think I understand.

I saw this as being a potential scenario instead, where the car and the truck were about to collide...but had not yet collided.

Grampa Dee