The Newtonian gravitational equation seems a bit odd to me
F = G * Mm / r^2
First, the force itself (being the weight) , we are told by Galileo, does not affect the acceleration of a falling body. So while m (small mass) can indeed affect the force, what good is it in calculating the acceleration if all different masses fall at the same rate? It therefore should not be able to influence the orbital speed as well, I would say.
Second, if the g acceleration and orbital speed are calculated using M, then, the question would be what is so special about this particular mass M, relative to m which cannot influence the orbital speed?... unless its just that m is too small to notice a change in g...However, this should make Galileos view as not being general, meaning that m does indeed cause an acceleration making the equation itself in need of something.
F = ma
F = m (G M/r^2)
However, if m is also a cause for acceleration, then,
F is also equal to M (Gm / r^2)
If we add both equations, we get
2F = m(GM/r^2) + M (Gm/r^2)
2F = GMm / r^2 + GMm /r^2
2F = 2 GMm/ r^2
Here, although we seem to have made full circle back to F = GMm/r^2, this time, however, we have added a second acceleration..a small one for a small m, but still greater than g originally.
Thank you in advance for your inputs
Grampa Dee
First, the force itself (being the weight) , we are told by Galileo, does not affect the acceleration of a falling body. So while m (small mass) can indeed affect the force, what good is it in calculating the acceleration if all different masses fall at the same rate? It therefore should not be able to influence the orbital speed as well, I would say.
Second, if the g acceleration and orbital speed are calculated using M, then, the question would be what is so special about this particular mass M, relative to m which cannot influence the orbital speed?... unless its just that m is too small to notice a change in g...However, this should make Galileos view as not being general, meaning that m does indeed cause an acceleration making the equation itself in need of something.
F = ma
F = m (G M/r^2)
However, if m is also a cause for acceleration, then,
F is also equal to M (Gm / r^2)
If we add both equations, we get
2F = m(GM/r^2) + M (Gm/r^2)
2F = GMm / r^2 + GMm /r^2
2F = 2 GMm/ r^2
Here, although we seem to have made full circle back to F = GMm/r^2, this time, however, we have added a second acceleration..a small one for a small m, but still greater than g originally.
Thank you in advance for your inputs
Grampa Dee
Comments (49)
The reason for this is that the mass of a falling object is negligible in relation to the mass of the Earth. the mass of the Earth cannot probably still be measured to the ton, and if it could, some of the mass would been the air above and exert a negative force.
In calculating the obit of the planets, it might become significant.I haven't checked your algebra though, to know if your calculations would be correct.
Different masses accelerate at the same rate towards a reference mass because they also have different inertias, which balances the different forces generated.....
Thank you for your response, unenlightened
Quoting unenlightened
Yes, I would agree with that as well; my concern is the Galileo assumption that the mass of a falling body does not affect the acceleration (g) at all. This opposed the Greek philosopher Aristotle, who was said to claim that heavier objects were falling quicker .
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, in my mind.
Quoting unenlightened
I agree that the air will indeed attract the earth as all bodies will cause gravity......
Since the air has weight, it still will have an effect in being part of the earths gravitational acceleration. However you have pointed out something important, as gravity is caused not only by the mass only, but by its density as well; for example, if the earth would be squashed to the size of the moon, it would have a much greater surface gravity....
Grampa Dee
Thank you for responding Pantagruel:
Quoting Pantagruel
I personally see this as being only part of the equation since the falling bodies should attract the earth as well, and the difference of the earths acceleration towards the falling body will be different with different masses .
Grampa Dee
Thank you for replaying Pantagruel
Quoting Pantagruel
Maybe you could help me in this area. The way I see it would be that Mass (the earth) gravitational energy will be causing an acceleration g at a certain distance ....the whole sphere at that distance will have the same g...
M / 4pi * r ^2 = k g
...in fact, this is all we need in my opinion to calculate the g acceleration.
However, since the same is said for the second mass, the two accelerations will need to be added, and if the second mass is great or small will make a difference in the overall g, it seems.
Grampa Dee
But you are conflating a generalized equation for force (which includes two masses) and acceleration (which includes two masses but in which the inertia of the much smaller of the two masses always balances whatever additional force it contributes to the overall system, and so can be factored out). Yes, for any given large body there is an acceleration equation which disregards the mass of the "falling" smaller object. But also yes, the overall force realized does vary if the mass of the smaller object varies.
I'm sorry, Pantagruel, I might very well be misunderstanding what you're saying.
Quoting Pantagruel
I do agree that the overall force will vary with different masses.
My post is about reconciling gravity with Galileos concept of different masses having exactly the same acceleration in freefall;... this is the problem Im having.right now.
Grampa Dee
Thank you for responding Pantagruel
Quoting Pantagruel
I don't understand how they can have the same acceleration, Pantagruel.
That's the problem that I have.
Grampa Dee
Initial acceleration of an object due to gravity of a primary is mass independent. I mean, F=ma, which if substituted directly into F=GMm/r² gets you A=GM/r², something independent of m altogether.
Careful though. This expresses coordinate acceleration relative to some inertial frame. This means that if you drop a 10 cm diameter ball of iron (10 kg say, just guessing) from the leaning tower, it hits the ground in around 3 1/3 seconds. But now if you increase the density of that ball to say the mass of the moon, it will hit the ground in less time, not because its initial coordinate acceleration is any greater, but because it has enough mass to significantly cause the ground to come up and meet it. So R decreases quicker, and as it does so, coordinate acceleration also increases quicker, a secondary effect barely measurable with our dropped moonlet. But the ground coming up is a primary effect and the decreased time to impact would definitely be measurable (not to mention the damage to the planet from dropping something that heavy, I mean the Pisa tower itself would collapse just from having that thing nearby)
Yep.
Quoting unenlightened
Not quite; as noAxiom shows, the mass of the falling object is irrelevant.
Thank you for responding noAxiom
Quoting noAxioms
I fully agree with your description noanxiom, but to claim that the increased acceleration due to the earth moving upwards is separated from the downward moving mass, I dont quite get. If someone drives towards you at 20 km/hr and you towards him/her at 20 km/hr, the velocity is indeed 40 km/hr relative to the two cars... we add the two velocities
Quoting noAxioms
Here is the reason why I have a problem with the equation. As it is written it follows Galileos axiom, for it doesnt matter what you put as the small mass, the acceleration will continue as being (GM/r^2).
But then, what do we do for the mass equal to the mass of the moon? The upper equation will calculate the acceleration still as being (GM / r^2)...
It is here that I personally believe that another acceleration needs to be added onto the first.
being A is also = Gm/r^2... if we add both of them, then we get A = (GM / R^2) + (Gm / r^2)
However , this equation does not agree with Galileo since a change in mass for the small m will indeed change the total acceleration of the system.
Keep in mind that all this is Newtonian physics, which doesn't describe reality. I squashed the mass of the moon into a grapefruit, and that pushes the limits of that simple formula. For instance, one has to start asking about the location of the clock used to measure this acceleration, because unlike in Newtonian physics, it makes a difference.
Under Newtonian physics, yes. But we're not adding speeds here, we're computing coordinate acceleration.
Initial acceleration of an object due to gravity of a primary is mass independent. I mean, F=ma, which if substituted directly into F=GMm/r² gets you A=GM/r², something independent of m altogether.
noAxioms
Until r starts changing...
Still works, at least under Newtonian physics. Same coordinate acceleration.
That would be wrong. The acceleration of the moon would not have that component. Earth does. Remember, I was stating that the formula gives acceleration relative to some inertial frame. I think you are trying to use the accelerating frame of Earth when adding them like that. But the force is given by F=GMm/r², and since acceleration is A = F/m, the moon accelerates by the simple formula, not adding the Earth part to it. Either that or F=ma is wrong, which is a denial of some pretty basic laws.
Total acceleration of the system is zero by conservation of momentum. So don't add them like that. It would be wrong to do so.
In reality, it would take not 3.3 seconds to hit the ground, but probably just a fraction of a second in the dense moon case. You're not going to compute that by just adding the accelerations, which treat both objects as point masses or rigid spheres, which they are not.
I understand the case for the dense moon would be extreme....would you have a problem with the equation I have written to Pantagruel?
M / 4pi * r ^2 = k g
Here, if r is small then g can become extremely large.
However, I would also claim that this is only half of the equation.
Any other mass g acceleration would need to be added at every point in space...
...just a thought.
Grampa Dee
Context is needed for that. This seems to come from here:
Quoting Gampa Dee
First of all, gravity isn't energy. I have no idea what you might consider the 'gravitational energy' of Earth. Mass divided by 4 ? r² gives you, well, I don't know what. It seems vaguely related to area of a circle. You equate this to 'k g', but no idea what 'k' is (kilo?). 'g' is the constant 9.8 m/sec², so you're seemingly equating this function of mass and radius to the constant 9800 m/sec²
Earth does have a sort of potential energy (negative) which represents the energy required to separate all the mass to infinity. I don't think you're talking about that.
You seem to be equating g with A. 'g' is a constant magnitude of acceleration (a scalar), so it cannot be smaller or larger. A is a variable, and a vector, not a scalar. A = GM/r², so yes, 'A' becomes quite large if r is small enough. Saying 'g' can be quite large is like saying a meter can be larger if my table is wide enough.
'mass g' doesn't parse. I don't know what you mean by this. Are you now adding random objects here and there? Then you need to separately compute the acceleration of each and add those accelerations. Newton showed (via shell theorem) that any spherical distribution of mass of radius r can be treated as a point mass by objects outside of r.
Sorry, but I kind of came in late to this discussion and I don't know what you're trying to do beyond what is described by these very simple equations. For small objects, they all fall at the same rate, and the only reason the feather falls slower is due to air friction. They brought a feather to the moon to show it falling at the same rate as a rock. PR stunt..,. tax dollars at work.
I'm not sure where your formula comes from, but let me see if I can understand and answer your question.
Take a rock of mass m, on earth of mass M, a distance r from the center of mass of the earth, and let us ignore air resistance.
The force on the rock:
F=G*Mm/r^2 (1)
Also
F=ma (2)
substituting 2 into 1 gives
a=G*M/r^2
So the acceleration of the rock in earth's gravitational field is not dependent on the mass of the rock. It is dependent on the distance of the rock from the center of mass of the earth, and on the mass of the earth.
You seem to be asking, what if M were different? If M were different, then the acceleration due to gravity will also be different. This can obviously be seen in that the same rock will have a lower acceleration dropped on the moon than dropped on earth. Galileo was talking about objects of different masses being dropped in the same gravitational field and at the same distance.
-1kg and 100kg dropped from the same height on earth in a vacuum, will have the same acceleration. -1kg and 100kg dropped from the same height on the moon in a vacuum, will have the same acceleration.
-1kg dropped from the same height on earth and on the moon will NOT have the same acceleration.
Is this what you are getting at?
Quoting noAxioms
I view gravity as a potential for movement observed as the weight of a body, if the mass is stationary, located at a certain height above the ground, or the increase of k.e. when it accelerates towards the ground.
Quoting noAxioms
For me, g is the potential gravitational acceleration given to a test mass caused by some other mass (usually a large one), not necessarily the earth; it could be the moon, mars or Jupiter. I was not using the letter a because we cannot speak of a force as being the cause. However, if you want me to write down A, instead of g, then, no problem, noAxiom,, I will use the letter A.
So , you wrote: . A = GM/r², so yes, 'A' becomes quite large if r is small enough.
This is all that I was saying.
Now, concerning the example of the high density ball pulling the earth towards itself ; How would you write the equation, if A = GM/r² is the acceleration caused by the earth?
Quoting noAxioms
...I am sorry for having used the letter g, noAxiom...just replace it by A instead, from now on. Now you mentioned that when adding masses we also need to add accelerations... this is all that I am saying in this post.
The Newtonian gravitational equation, identifying two masses, has only one acceleration , the one caused by the earth.
Quoting noAxioms
I am privileged to have any response at all, noAxiom....I have merely a layman viewpoint on this issue, and am simply trying to explain the problems that I have...
Thank you for your time
Grampa Dee.
Thank you for replying,PhilosophyRunner
Quoting PhilosophyRunner
No, I am asking how is it that we do not include the acceleration of the earth caused by the second mass?
Grampa Dee
Quoting noAxioms
4 ? r², being the surface area of a sphere, would be dependent to the surface Acceleration
M / 4 ? r² = A * constant....It was meant to what you have written concerning a high density mass.
While, this is not directly the density of mass, it seems to identify a mass having a certain spherical area as having a certain gravitational surface acceleration... if the radius is small the acceleration will be greater...
Of course, the equation could be erroneous, but I can't see why....that is why I asked you what you thought about the equation
Grampa Dee.
If you want to model a system where both bodies are accelerating you will have to use the two-body equations of motion, and set up a inertial frame of reference as the origin (eg: the barycenter). In the simplified calculations, the larger body can be taken as the inertial frame of reference as it is not accelerating, this is not true if it is accelerating.
Secondly why would we include the acceleration of the earth caused by the second mass? That is not what we were calculating - we were calculating the acceleration of the rock. If you wanted to calculate the acceleration of the Earth caused by the rock, that is a separate calculation
Thank you for replaying, PhilosophyRunner
Quoting PhilosophyRunner
ok; I think we might be going somewhere. When we measure the acceleration of the rock towards the earth, aren't we not measuring, at the same time, the acceleration of the earth towards the rock? How could you know the difference?
Quoting PhilosophyRunner
ok; so, what would be this equation?
Grampa Dee
No, because you have to measure the acceleration from an inertial frame of reference, which is neither the earth nor the rock in our example if we are including the earth's acceleration.
"Acceleration of the rock towards the earth" implies measuring the acceleration of the rock from earth's frame of reference. If the earth is accelerating towards the rock, this is not a valid frame of reference.
The same goes for " the acceleration of the earth towards the rock" as the rock is not a valid frame of reference.
Instead, what we should be doing is measuring the acceleration of the earth in 3d coordinates from a point in space that is not accelerating (any non accelerating point), and measuring the acceleration of the rock from this same point. And from the point, the rock will be seen to accelerate quickly towards the earth, while the earth will barely accelerate towards the rock.
This way of analysis is not just correct for gravity, but for anytime you are using classical mechanics to analysis motion. Think of a stationary passerby watching a car hitting a tennis ball. The ball will have a high acceleration as a result of the collision, while the car will hardly accelerate at all. From an inertial frame of reference, the car and tennis ball do NOT both have the same acceleration from the collision.
Quoting Gampa Dee
a1 = Gm2r/r3
Where a1 is the acceleration of body 1 from an inertial frame of reference. Note this is a vector in 3d.
For example see: https://orbital-mechanics.space/the-n-body-problem/two-body-inertial-motion.html
The symbol for that is 'a', not 'g'. 'a' is a vector variable acceleration. g is a scalar constant acceleration. Neither are a force. Force is measured in Newtons and uses the symbol F. Try to use standard symbols when discussing such things, as personal preferences only lead to confusion.
We've been doing that, but it's actually lowercase. 'A' is used for Area (mathematics) and electrical current (physics). So I'm committing the same offense; :sad:
Well, I would make A lowercase to fix that problem, and the rest is correct. The acceleration of the moon when it is at radius r is exactly that in Newtonian physics. That does not mean it will hit the ground in 3 1/3 seconds like the 10 kg ball. As I said, a small fraction of a second is more likely.
I'm assuming that the dense super-mass is rigid, so yes, the acceleration applies to every point in the moon-ball. I am not assuming Earth is sufficiently rigid to not deform under the ungodly tidal stress the ball would apply to it. The sidewalk slab 60 meters below will be yanked up without waiting for Earth to catch up with it.
Acceleration is absolute, not relative, so adding them seems to result in a fairly meaningless value. I suppose you can use it to compute the rate of change in distance between the two objects, only in a 1-dimensional case, but that rate isn't acceleration. For instance, if an apple detaches from a tree and the distance between me and it decreases at 9.8 m/sec², that in no way suggests that I am accelerating at that rate, and in fact I'm accelerating (coordinate acceleration, not proper acceleration) away from it a little bit.
If we switch to proper acceleration, both the apple and I have a proper acceleration of about 1g until the apple detaches, at which point I have 1g proper acceleration upward and the apple has none. So it is me that hits the apple, not the other way around. But that isn't really the Newtonian way of looking at it.
But there is an equation for each object, each dependent on only that mass, and not on the mass of the thing accelerating towards it.
Quoting Gampa Dee
OK, but the surface area seems to be a needless complication. We know the acceleration as a=GM/r². Where the surface is is irrelevant so long as it is below r.
That would not work. A grapefruit has a similar area, but far less gravitational acceleration at its surface. So acceleration is not a function of just area.
Quoting Gampa Dee
As P-R points out, the coordinate acceleration described by Newton's equations is relative to any inertial coordinate system, and the equations don't work when used with a non-inertial coordinate system such as an accelerating or rotating one (Earth is both).
Mind you, plenty of books treat things like the Earth-moon system in isolation, which is to reference an accelerating coordinate system. It works to an extent, but adds confusion. For instance, few are aware that the moon always accelerates towards the sun since the sun exerts more force on it that does Earth. That means that when the moon is between the two (solar eclipse), the moon is accelerating directly away from Earth.
From the accelerating frame of Earth, the moon seems to maintain a fairly constant distance that isn't much a function of which direction the sun is. Yes the orbit of the moon (like any orbit) is eccentric to a degree, but that again isn't due to the sun.
Great posts all around. :up:
Thank you PhilosophyRunner
Quoting PhilosophyRunner
Yes, I will....
Thank you again.
Grampa Dee
Thank you for replying, noAxioms
Quoting noAxioms
I will try to do that noAxiom...Im not well versed in the scientific terminology . Quoting noAxioms
I agree, its Einstein who recognized that a freefall object didnt experience any force.
Quoting noAxioms
Ok; Im not sure what youre saying but this might be where Im not thinking it through.
Quoting noAxioms
Ill ponder on this, noAxiom...thanks again for your time
Grampa Dee
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).
No. Under Newtonian mechanics, relative to any inertial coordinate system, it is the apple and only the apple that is accelerating.
Only under Einstein does it become you that experiences acceleration, as measured by an accelerometer that you carry with you. The accelerometer on the apple would read zero. But this is now proper acceleration, not coordinate acceleration. Coordinate acceleration becomes dependent on the arbitrary coordinate system of choice and it even varies from one inertial coordinate system to the next. The equations become quite different.
That's what the accelerometer does. Multiply what it says by your mass and you get your weight, which is why one is weightless on the ISS.
Quoting Gampa Dee
I went through the example, and I will try to explain how I have understood to the best of my ability.what was written.....I'm not claiming to be correct, only that this is how I have interpreted the text.
Also beware of my use of symbols...they're awful as I don't have the use of the correct symbols.
For example R(double dot) simply means acceleration, however, use the text beside you and you will be able to follow......Here goes nothing :)
[b]Two-Body Equations of Motion in an Inertial Frame
The position vectors R1 and R2 of two point masses in this reference frame are:
R1 = X1Ihat + Y1Jhat + Z1Khat
R2 = X2Ihat + Y2Jhat + Z2khat[/b]
Here, we are given a 3 dimensional reference frame X,Y,Z where two bodies are positioned at (x1,y1 z1) and (x2,y2,z2)...I dont understand how they could be identified as vectors though, since position is not a vector.
[b]Let r be the vector pointing from m1 to m2, which we also phrase as m2 relative to m1. Then:
r = r2 r1
r = (x2 x1)Ihat + (y2 y1) Jhat + (Z2 Z1)Khat[/b]
Since the mass is said to be pointing towards the other mass, then ok, r is then a vector.
[b]We also define a unit vector pointing from m1 toward m2:
Uhat*r = r / r[/b]
This, being a unit vector, will not do anything to the equation as such except maybe to identify r^2(denominator), which is not a vector,....as being identified as a vector....in my opinion.
where r (denominator) is the magnitude of r ( nominator), or the distance between the two masses.
[b]Forces in the Two-Body System#
The two masses are acted upon only by their mutual gravitational pull. F12 is the force exerted on m1 by m2 and F21 is the force exerted on m2 by m1. By Newtons third law:
F12 = -F21[/b]
We have two forces coming from the two bodies...1 being + and the other -.
[b]Newtons second law says that the force is equal to the mass times the acceleration:
(17)#
F12 = m1 R(double dot)1
F21 = m2 R(double dot)2
where R¨ is the absolute acceleration of the subscripted mass. Absolute means that the acceleration is taken relative to an inertial reference frame. This is important because Newtons second law only applies for absolute accelerations[/b].
This is simply, in my opinion,
F1 = m1*a1
F2 = m2*a2
F1 =- F2
[b]Since the only force in this system is the gravitational attraction, the force is also equal to Newtons law of gravitation, (1). The force of m2 on m1, F12, points in the positive direction of u^r. Because of Newtons third law, as represented by (16), the force of m1 on m2, F21, points in the negative direction of u^r. This is shown in (18):
F12 = Gm1m2 / r^2 (uhatr)
F21 = - Gm1m2 / r^2 (uhatr)[/b]
Here we have two of the Newtonian gravitational equations... one of them is negative identifying the reverse direction of the second force.
Finding the Equations of Motion *********
Heres where our focus of discussion lies.
[b]Combining (17) and (18), we find:
m1R1(double dot) = Gm1m2 / r^2 (uhatr)
m2R2(double dot) = - Gm1m2 / r^2 (uhatr)[/b]
This would mean, in my opinion, m1*a1 = Gm1m2 / r^2 ... u(hat)r is simply a unit vector...I think we can ignore it.The same goes for m2R2(double dot)
[b]Finally, we divide through by the mass on the left side of each equation and replace u^r with its definition, (15) to arrive at the two-body inertial equations of motion:
R1(double dot) = Gm2 r / r^3
R2(double dot) = -Gm1r /r^3[/b]
Notice here that there are two accelerations, which are not going to have the same magnitude.
a1 = Gm2 / r^2 ( one of the r in r^3 cancels out with the r in the numerator)
a2 = - Gm1 / r^2
So, now we have 2 accelerations; one is negative, the other positive. What do we do at this point?
If we have two cars having velocities of v1 and v2(since it is going towards v1), relative to the road, what will the observers within the car measure the relative velocity between themselves?
Its not going to be v1 v2, but v1 + v2....I see the same problem for the two accelerations.
I hope this post wasn't too long.
Grampa Dee
There you have done an excellent job, and now have acceleration from an inertial frame of reference. Great. You wanted the acceleration, now you have them.
Then you try to move that to the frame of reference of the earth and of the rock, at which point Newton's laws are no longer valid. This step is wrong. You can change the coordinate system, that is fine, but when you change it so that the origin is an accelerating body, Newton's laws no longer work correctly.
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.
[b][/b]
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 dont 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 dont get; Im not saying that you are wrong, Im saying I dont understand.
[b]Quoting noAxioms[/b]
I fully agree that anything which has an orbit will have a combination of gravitational acceleration and sidereal velocity at the same time...this, Im certain can become very complicated if one throws in a frame of reference from which everything is calculated. If the orbit isnt circular (which most arent) 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.
Its very simple,noAnxiom; what Im 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
Quoting --`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
All frames of reference have 3 dimensions of space, and require 3 velocity components to describe.
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.
Yes, and those orthogonal vectors make it at least a 2d situation.
A frame of reference is a starting point, not something thrown in. Without it, any velocity is completely undefined.
Orbital accelerations are always changing, even in the circular case. Remember that it is a vector.
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.
Quoting Gampa DeeNo. 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.
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.
The force 'felt' would be proper acceleration, also absolute.
[b]Quoting noAxioms[/b]
Yes; of course...my mistake in writing 1 dimensional problem as being Newtonian...I was thinking of freefall.
[b]Quoting noAxioms[/b]
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.
[b]Quoting noAxioms[/b]
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.
Quoting 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.
Quoting 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.
Quoting noAxioms
I do have a hard time with this...is it possible to share this proof?
Grampa Dee
Yea, a = GM/r²
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.
Yes, but you listed all this Kepler stuff that shows how to do that just nicely.
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.
Quoting noAxioms
Ok; Ill 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
Quoting 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....
Quoting 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.
[b]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 earths gravity but is now attracting the earth instead.
Quoting 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..
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.
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.
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 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 ? GM/r² where ? 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.
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.
Quoting 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 planets surface gravity possess?
Quoting 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.
Quoting noAxioms
Interesting... I heard the moon was moving away from the earth, but didnt know the cause.
Quoting 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,
Quoting noAxioms
The idea of nitpicking, noAxioms, is to try to explain a systems physical laws...
For example, let us assume that I am correct (just for arguments sake)...
I understand very well, that measurements have been made many times supporting Galileos 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 couldnt prove it.
We do know that stars orbital velocities at the edge of galaxies dont 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.
Quoting noAxioms
But it accelerates at ? GM/r²
If there is no mistake in this observation , then this will certainly prove that I am in error and that Galileos law is indeed valid.....
where ? 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.
Quoting 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...
Quoting noAxioms
Ok,... maybe this is where Im off the rail, although I dont yet understand why...
Are you saying that instead of velocity, in the upper example I wrote, if we used acceleration instead , it wouldnt 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
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.
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.
Well yea. The alternative is some klnd of solipsism where things are gravitationally attracted only to known object, which makes the knower very special.
Adding the vectors is what totals zero. The acceleration of neither object was correctly expressed since neither is zero.
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.
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
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.
Thank you for replaying PhilosophyRunner
Quoting PhilosophyRunner
Yes, I will do that;
and thank you again for your time.
Grampa Dee
Thank you for responding Erich
Quoting EricH
I may have; although I did mention the direction as being "towards each other"..
Quoting 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
Quoting noAxioms
Maybe its not a law, I dont know. but I did hear of Galileos law of falling bodies... if Newton has three laws of motion, I personally dont see why Galileos description of freefalling bodies cant 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.
Quoting noAxioms
What I was saying, noAxiom was this:
. But it accelerates at ? GM/r² ( talking about the moon)
If there is no mistake in this observation ,then this will certainly prove that Galileos 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 isnt already.
Quoting noAxioms
Sorry; I meant the relative velocity measured by you, the driver, was v
Quoting 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 dont know what is wrong with this example apart from my - sign error ...sorry
Quoting 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.
Quoting 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 dont understand why one has the negative acceleration of the other..wouldnt they crash in this case?
Also, I know that Im mixed up when were 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.
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].
The road doesn't matter since acceleration is not a relation. Velocity is, but not acceleration.
As measured by anybody actually.
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.
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.
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.
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.
Quoting 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?
[b]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[/b]
Again,we have a momentum,of -111 kgm/s....what does that even mean?
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
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.
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.
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
Quoting noAxioms
Thats exactly what I feel; theres nothing meaningful...I dont have any problems with the addition of the two vectors (tip to tail) being 0...but I dont 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.
Quoting 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