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hi, I am a physics student in grade eleven. I do not have much knowledge of physics but I have a doubt:

i am well aware of the theory that all masses dropped at the same height from earth will accelerate at the same speed but i say this is wrong i have come up with a possible contradiction for this:

In the first case, both masses are the same and accelerate toward each other at the same rate. Now in cases 2 and 3, I have reduced the mass (m1), as this happens point of collision moves towards the greater mass. Now let's consider that all 3 collisions take the exact same time for collision. the distance between the objects in all three cases is the same. Now looking specifically at cases 2 and 3 we can see that the m1 object has a larger distance to cover compared to the m2 object. Because of this, I conclude that acceleration has to be different in both cases. Yes, I have kind of magnified this idea in the thought experiment, but even though (if this effect is true) this effect is very very insignificant. it's still present. Yes, this may not affect physical experiments at large, but I still persist that the effect is present.

PS: If this thought experiment is false feel free to challenge me.

use the trailing image for reference.

hope i get a opposition soon!!

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Good morning, Jeff and welcome.

What you are describing in dynamics is known as the two body problem.

This will describe whether the bodies will actuall collide or do a dance around each other.

Note that the thre body problem has not been solved analytically.

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I would suggest to make a drawing of two masses, where one is  m2 = 5.972 × 10^24 kg,  and the m1 = 1 kg, and a second experiment where m1 = 2 kg. Do you think one is able to measure the difference in acceleration of m1? Maybe instead of drawing, you should do the calculation...

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2 hours ago, jeff einstein said:

hi, I am a physics student in grade eleven. I do not have much knowledge of physics but I have a doubt:

i am well aware of the theory that all masses dropped at the same height from earth will accelerate at the same speed but i say this is wrong i have come up with a possible contradiction for this:

In the first case, both masses are the same and accelerate toward each other at the same rate. Now in cases 2 and 3, I have reduced the mass (m1), as this happens point of collision moves towards the greater mass. Now let's consider that all 3 collisions take the exact same time for collision. the distance between the objects in all three cases is the same. Now looking specifically at cases 2 and 3 we can see that the m1 object has a larger distance to cover compared to the m2 object. Because of this, I conclude that acceleration has to be different in both cases. Yes, I have kind of magnified this idea in the thought experiment, but even though (if this effect is true) this effect is very very insignificant. it's still present. Yes, this may not affect physical experiments at large, but I still persist that the effect is present.

PS: If this thought experiment is false feel free to challenge me.

use the trailing image for reference.

hope i get a opposition soon!!

I think you have a good point in principle, regarding the general case of two masses attracted together, but in the case of objects falling towards Earth, the movement of Earth is negligible, as its mass is so much vastly greater than the object that is regarded as "falling" towards it. So if you really want, you might say the usual description is an approximation, but you might run the risk of earning points for pedantry.😀

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deleted - the previous responses are better

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deleted - the previous responses are better

OK

I was going to comment that if you want the equivalent single body problem you need to use what is called the reduced mass., and the relative acceleration and velocity.

I can supply the maths if anybody wants this.

Edited by studiot
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18 minutes ago, studiot said:

OK

I was going to comment that if you want the equivalent single body problem you need to use what is called the reduced mass., and the relative acceleration and velocity.

I can supply the maths if anybody wants this.

Without the math, the simple resolution of the 'contradiction' in the OP is that the momentary acceleration of the falling body does not depend on the falling body's mass, but as the other body moves toward it, the total distance of the fall depends on the falling body's mass as well.

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I have no idea what your thought experiment has to do with gravity, if anything.

The principle of equivalence has to do with two different masses in the presence of a third mass (source of field).

The fact that relative accelerations between mutual objects are equal and opposite is a trivial kinematical fact, and has nothing to do with gravity.

Maybe I have misunderstood the whole thing. If that's the case, I do apologise.

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It is important to distinguish between the acceleration of each object involved and the total closing rate between the two objects.

When you drop an object near the Earth, it accelerates towards the Earth at a rate that is solely dependent on its distance from the center of the Earth and the mass of the Earth.   Simultaneously, the Earth accelerates towards the object at a rate that is determined by the same distance, and the mass of the object.

For the everyday type of objects you are likely to drop, the Earth out-masses  them by so many magnitudes that its acceleration is imperceptibly small, and can be ignored for all practical purposes, meaning we can treat the acceleration of the object towards the Earth and the closing rate between the two as being one and same.

As you move to larger and larger objects, the acceleration of the Earth starts to make up a greater proportion of the closing rate, and at some point cannot be ignored( where this point is depends on how accurate you need your solution to be)

So all objects dropped from a given height from the Earth do accelerate at the same rate regardless of their mass, but the Earth's acceleration towards them does change according to their mass.

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3 hours ago, Janus said:

t is important to distinguish between the acceleration of each object involved and the total closing rate between the two objects.

Whatever you are distinguishing it is important to realsie that whatever you observe will depend upon the frame of reference you are working in.

Acceleration and velocity are frame dependent, even in Galilean relativity.

I'm also not sure what an 11th grader would know about this, Jeff is clearly assuming the greater mass is fixed in space or a one body analysis.

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no, the larger mass is not fixed in space. and as many of you have told yes even though my effect is true it is very very very very very very small that it can be ignored

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44 minutes ago, jeff einstein said:

no, the larger mass is not fixed in space. and as many of you have told yes even though my effect is true it is very very very very very very small that it can be ignored

The issue is that you are mixing two distinct effects. One is the acceleration of freefall, the other is closing speed, as Janus has explained

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2 hours ago, jeff einstein said:

no, the larger mass is not fixed in space. and as many of you have told yes even though my effect is true it is very very very very very very small that it can be ignored

It is good that you told us you are 11th grade.

We don't have 11th grade in the UK so I looked up the syllabus.

~Are you just starting or just finishing 11th grade ?

This is important because the 11th grade syllabus contains what you need to properly ananlyse yourthought experiment viz Newton's Laws of Motion and the Laws of Kinematics.
I don't think frames are introduced at this stage.

So it is important to set the response a

You have not stated a most important starting condition about the masses.

Are they at rest, and if so what holds them there ?

Why do they accelerate towards each other ?

Edited by studiot
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good sir I do have medium knowledge about frames of reference and I do study outside the school syllabus and I know more about physics than i should for my grade. anyway, the two questions you asked are more out of topic if I am not wrong cause this is a thought experiment. just to clear things up both objects mentioned have no motion and have no external forces acting on them second they accelerate towards each other because of gravity.

and about the masses being at rest, yes they are at rest but empty space, where as i said there is no external force on them and the initial masses are not in motion and arent spinning or anything like that.

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1 hour ago, jeff einstein said:

good sir I do have medium knowledge about frames of reference and I do study outside the school syllabus and I know more about physics than i should for my grade. anyway, the two questions you asked are more out of topic if I am not wrong cause this is a thought experiment. just to clear things up both objects mentioned have no motion and have no external forces acting on them second they accelerate towards each other because of gravity.

and about the masses being at rest, yes they are at rest but empty space, where as i said there is no external force on them and the initial masses are not in motion and arent spinning or anything like that.

~So the bodies, call them A and B, are 'at rest', yet they feel the influence of gravity.

Since you can't turn gravity on and off, there must be something holding them in place.

So at time t = 0 that something is removed and the force of gravity begins to pull them together.

Both Newton't 3rd Law (N3)  and Newton's Law of gravitation say that the force of B upon A is equal in magnitude but opposite in direction to the foce of A upon B.

Neweton's 2nd Law (N2) says that this force is proportional to the product of the mass and the acceleration of that mass.  (F = ma)

So to maintain the equality the smaller mass will have a larger acceleration.

How are we doing so far ?

Edited by studiot
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ok isn't that just further proving my point Are you opposing my theory or supporting it?

Edited by jeff einstein
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You should understand that "being at rest" are meaningless words in physics. Motion is relative.

You should understand that what you are picturing as your "point of collision" is frame-dependent.

You should understand that comparing how two different test masses accelerate towards a certain third mass which is source of a gravitational field acting upon those masses (context of the equivalence principle) is a different situation than the one you propose (two masses that collide under their mutual gravity).

It's perhaps worth saying that gravity alone very rarely results in collisions in classical physics, because of centrifugal barriers, so the situation you propose is not nearly as general as it should be, especially considering the extraordinary character of your claim.

You should have an understanding of relative motion*, vs centre of mass motion, vs motion described in a more general inertial frame.

There are many important things that you misunderstand here, and so far you don't seem to be willing to understand, in spite on good efforts by several members.

Please, be aware that you do not have a theory. You do seem to have several important misconceptions on basic Newtonian physics that lead you to believe a widely accepted theory, fastidiously checked, both experimentally and theoretically, forwards and backwards, is wrong.

You should clarify your position, before anybody can prove you wrong. Nonsense cannot be proven wrong.

@Janus and @studiot in particular have emphasized this.

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5 hours ago, jeff einstein said:

ok isn't that just further proving my point? Are you opposing my theory or supporting it?

As I am not sure what your theory is I can't say.

By the way (BTW) please use the correct word which is' hypothesis' for your proposition, 'theory' has a special meaning in Science.

Dr Swanson has pointed out that your hypothesis about falling bodies (presumably to Earth) is quite independent of the gravitational attraction of two isolated bodies of similar small size,so that they may be considered as 'point masses'.

So let us continue through some calculations that can be covered in 11th grade mathematics.

First we will look at the statement you queried

On 9/15/2023 at 8:17 AM, jeff einstein said:

am well aware of the theory that all masses dropped at the same height from earth will accelerate at the same speed but i say this is wrong i have come up with a possible contradiction for this:

Let mE be the mass of Earth, m1 be the mass of A and m2 be the mass of B

For any single mass N2 say that

$F = am$    where a is the acceleration and F is the force.

Also we have for any pair of masses Newton't law of gravitation says that

$F = G\frac{{{m_1}{m_2}}}{r}$  where r is the distance between their centres and G is a universal constant.

So for masses A and B, both at distance r above the surface (or therefore the centre)

${a_1} = \frac{{{F_1}}}{{{m_1}}} = G\frac{{{m_1}{m_E}}}{{{m_1}r}} = G\frac{{{m_E}}}{r}$

and

${a_2} = \frac{{{F_2}}}{{{m_2}}} = G\frac{{{m_2}{m_E}}}{{{m_2}r}} = G\frac{{{m_E}}}{r}$

Therefore we have that the accelerations due to their gravitational attraction to the Earth are the same.

${a_1} = {a_2}$

Therefore they fall at the same rate.

Once we have agreed this we can move on to the question of A and B attracted to each other in isolation, where the accelerations are not the same but depend upon distance r.

This is an altogether more difficult question.

Edited by studiot
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OOps what amistake

The equations for gravity should  be proportional to inverse square of the distance.

$F = G\frac{{{m_1}{m_2}}}{r^2}$

and

${a_1} = \frac{{{F_1}}}{{{m_1}}} = G\frac{{{m_1}{m_E}}}{{{m_1}r^2}} = G\frac{{{m_E}}}{r^2}$

${a_2} = \frac{{{F_2}}}{{{m_2}}} = G\frac{{{m_2}{m_E}}}{{{m_2}r^2}} = G\frac{{{m_E}}}{r^2}$

My apologies

Edited by studiot
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ok whats next

so a1 = a2 as in m1 and m2 respectively?

so basically m1 and m2 have the exact same acceleration regardless of size?

or is this when both masses are equal of course it seems more obvious in this way

apart from the equation and the maths, in the diagram I have drawn if the center of "collision" is towards the greater mass then the larger mass shall accelerate towards the collision point but as we see the distance between the larger masses distance from the collision point is larger compared to the smaller mass and thus if both accelerate towards each other then the larger mass should accelerate slower or this would result in the collision point occurring in the middle which is not true. so i say a1 is not equal to a2

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*regardless of mass

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On 9/18/2023 at 2:30 PM, jeff einstein said:

*regardless of mass

Indeed.

But this is an analysis of Newtonian gravity.

Have you given up on this subject?

We can now make a start on analysing your thought experiment, from first principles.

We will need a coordinate system with an origin that we can define along with a sign convention.

This situation is very common in mechanics and the sign convention is often not mentioned.

But it is very useful  as it tells us which way thing are moving an also since we don't necessarily know all the directions at the outset, it tell us if our initial guesses (assignments) are correct or not.

OK so the attachment offers all this for two particles moving only along the x axis.

The sign convention is that left to right is positive.

I have drawn the most general situation in which both particles have a velocity and two forces acting on them.
One of these forces is an external force the other shows an interaction force between the particles for each particle.
If our assumed directions turn out to be not as shown then the particular forces or velocities will be negative, not positive.

First we write Newton's second law equation for each particle in the form force = rate of change of momentum.

Remembering back to simple differentiation we can add these equations up as shown, since the derivative of a sum is the sum of the derivatives.

After adding them up we bring Newton's third law into play to show a very important fact.

The total momentum change due to the total external forces is unaffected by the mutual interaction.

This is of course conservation of momentum.

Note a similar argument applies in statics to allow us to discount all the internal forces of a system and only work with the external loads.

I have also included the position of the Centre of Mass, at a distance from the origin,  in preparation for the next stage in the analysis.

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• 2 months later...

You are wrong. What you missed is the shapes of the masses and the initial and changing angles of the forces. From the start, your masses m1, m2, and E are all pulling towards each other's center of mass in straight lines. From center to center to center the lines form a triangle. As m1 moves towards m2, the direction of the pull of E changes. The angle between the two lines of force from the center of E to m1 and m2 reduces. The linear distance from m1 to center of E, and from m2 to center of E remains equal as they accelerate. The line from m1 to m2 remains tangential to sphere E throughout.

Watching the triangle, on the plane of the triangle, as the masses move, the triangle tilts and grows narrower while remaining isosceles. Both m1 and m2 are always the same distance from the center of E at every moment of falling no matter how disproportionately they accelerate towards each other. Their paths on the plane will be different in length, but as long as m1 and m2 are equal diameter spheres, they will always reach a spherical  E simultaneously. If E was flat, they would reach the surface at different times. Are you living on a flat Earth? 🤔

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1 hour ago, OptimisticCynic said:

You are wrong. What you missed is the shapes of the masses and the initial and changing angles of the forces. From the start, your masses m1, m2, and E are all pulling towards each other's center of mass in straight lines. From center to center to center the lines form a triangle. As m1 moves towards m2, the direction of the pull of E changes. The angle between the two lines of force from the center of E to m1 and m2 reduces. The linear distance from m1 to center of E, and from m2 to center of E remains equal as they accelerate. The line from m1 to m2 remains tangential to sphere E throughout.

Watching the triangle, on the plane of the triangle, as the masses move, the triangle tilts and grows narrower while remaining isosceles. Both m1 and m2 are always the same distance from the center of E at every moment of falling no matter how disproportionately they accelerate towards each other. Their paths on the plane will be different in length, but as long as m1 and m2 are equal diameter spheres, they will always reach a spherical  E simultaneously. If E was flat, they would reach the surface at different times. Are you living on a flat Earth? 🤔

Who is wrong?

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I was replying to jeff einstein. The original post author.

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