A Flaw In Using Fall?

[Mordred]

I am asking you to show me how you got to your equation obviously you don't want to do that. Yes I know the above is basic level and yes I specified the above doesn't have the dimension of force I specified that in the post above

This is basic introductory physics. It simply says [math]m \frac{d^2x}{dt^2}=NetForce[/math]

 

thank you that was all you had to explain

 

lol though now I feel 100% stupid for now realizing this

Edited June 15, 2014 by Mordred

[xyzt]

I am asking you to show me how you got to your equation obviously you don't want to do that. Yes I know the above is basic level and yes I specified the above doesn't have the dimension of force I specified that in the post above

 

thank you that was all you had to explain

 

lol though now I feel 100% stupid for now realizing this

Glad that I managed to get you all squared.

[xyzt]

If two test probes of massses [math]m_1,m_2[/math] are dropped one at a time from the same altitude [math]D[/math] above a massive body of mass [math]M>>m_{1,2}[/math] then the test probe of larger mass will hit first. This is due to the fact that the "time to collision" is given by:

 

[math]t_i=\frac{D^{3/2}}{\sqrt{2G(M+m_i)}}(arctan \sqrt{\frac{R+r_i}{D-(R+r_i)}}-\frac{\sqrt{(R+r_i)(D-(R+r_i)}}{D})[/math]

 

where [math]R[/math] is the radius of the massive bode and [math]r_i[/math] is the radius of the test probe.

If the two test probes have the same radius [math]r_1=r_2[/math], then the "time to collision" depends only on the mass of the respective test probes, the larger the mass, the shorter the time.

The derivation of the above formula is not trivial but I could post it if there is enough interest.

Working out the solution through GR (the Schwarzschild solution) one obtains :

 

[math]t_i=\frac{D^{3/2}}{\sqrt{2GM}}(arctan \sqrt{\frac{R+r_i}{D-(R+r_i)}}-\frac{\sqrt{(R+r_i)(D-(R+r_i)}}{D})[/math]

 

After the initial shock, one needs to remember that the Schwarzschild solution is obtained for a testprobe of zero mass, the "disagreement" is due to GR using [math]m=0[/math].

Edited June 15, 2014 by xyzt

[imatfaal]

I am very interested in this and would like to see the exploration continue - can I therefore ask some of disputants to confirm or deny the basic parameters; firstly so I know I haven't got the wrong end of the stick, and secondly so we don't end up with endless changing of set-up to avoid ever having to admit that another might be more right

 

The setup is as follows

 

1. Two test objects of same radius ( r ) (to avoid silly answers that only concern a fat ball versus a tiny one) - but one of these is significantly more massive than the other - m_1 and m_2

 

2. These two objects are a distance above earth - each objects CofM is exactly the same distance (D) as the other's CofM from the CofM of the Earth (which is mass M_e and radius R_e)

 

3. We can assume spherical symmetry, arbitrary precision, no air etc (aka full spherical cow mode)

 

4. Whilst the objects are at exactly the same distance from CoM_Earth - they are seperate. The two objects and the CofM_Earth make an isoceles triangle. the angle between the two similar sides (ie at CofM_Earth) can be called theta.

 

5. The Objects are released at the same time.

 

------------------------

 

I initially agreed with Tom - as I could see nothing that changed the similarity of each objects, they start together and I cannot see where the situation would arise when this similarity would change.

 

But if I consider a situation where theta equals 180 degrees - then it is obvious that we now have a complex problem and my simplistic idea fails. Then if I narrow the angle - surely it cannot go back to being simple until theta equals zero; but then there is no lateral separation and it is a new problem.

 

NB XYZT - please dont just post the same equation again and say you explained this in post two or something. You have failed to convince quite a few posters; a little less exasperation and a more explanation please.

[MigL]

It seems everyone is trying to outdo each other with the math and losing any semblance of common sense/

 

As imatfaal has stated, when angular separation between the two test probes is zero, it is easy to see that swansont is right. However the situation does not change or become more complex when the angular separation is 180 deg ( or anywhere in between ). This is because although the large earth mass is accelerated towards the heavier test probe, the lighter probe has, in this case, a larger acceleration since it is drawn to the larger COMBINED mass of the earth and heavier test probe. I.E. they hit the ground simultaneously ! Is this not simple perturbation ?

You cannot have different situations arising when the action ( as in Lagrangian ) is the same. This is dictated by conservation laws and symmetry.

Edited June 15, 2014 by MigL

[xyzt]

 

NB XYZT - please dont just post the same equation again and say you explained this in post two or something. You have failed to convince quite a few posters; a little less exasperation and a more explanation please.

I have posted the exact equations of motion. From these equations it is not possible to decide whether the two test probes fall together or separate. Since none of the "doubters" posted any math, none of them proved anything one way or another so far.

Nevertheless, the fact that when dropped "alone", the heavier test probe hits in a shorter time than the lighter test probe is undisputable and undisputed. I established this very clearly. Until someone manages to integrate (wrt z) the equations of motion I have posted, the other case will remain unresolved. "Prose" arguments don't count.

It seems everyone is trying to outdo each other with the math and losing any semblance of common sense/

 

As imatfaal has stated, when angular separation between the two test probes is zero, it is easy to see that swansont is right.

The "angular separation" is zero AT START. You have no proof that the angular separation is zero throughout the fall. You are simply stating your conclusion as "proof". Needless to say, this doesn't form a valid proof. I spent quite a bit of time with swansont explaining to him this failure, it seems he understood and abandoned this flawed line of argumentation. We seem to have found a more rigorous way of finding out the truth, based on the analysis of the differential equations describing the equations of motion.

 

 

 

However the situation does not change or become more complex when the angular separation is 180 deg ( or anywhere in between ). This is because although the large earth mass is accelerated towards the heavier test probe, the lighter probe has, in this case, a larger acceleration since it is drawn to the larger COMBINED mass of the earth and heavier test probe. I.E. they hit the ground simultaneously ! Is this not simple perturbation ?

You cannot have different situations arising when the action ( as in Lagrangian ) is the same. This is dictated by conservation laws and symmetry.

 

can you put the above in mathematical language? Please.

Edited June 15, 2014 by xyzt

[md65536]

As imatfaal has stated, when angular separation between the two test probes is zero, it is easy to see that swansont is right. However the situation does not change or become more complex when the angular separation is 180 deg ( or anywhere in between ). This is because although the large earth mass is accelerated towards the heavier test probe, the lighter probe has, in this case, a larger acceleration since it is drawn to the larger COMBINED mass of the earth and heavier test probe. I.E. they hit the ground simultaneously ! Is this not simple perturbation ?

You cannot have different situations arising when the action ( as in Lagrangian ) is the same. This is dictated by conservation laws and symmetry.

No, you're wrong. It's true that swansont and xyzt are speaking of different cases that rule out the possibility of the other's.

 

If you choose to use your reasoning alone, then you must also factor in (among other possible complications) that the lighter test mass has a different distance to travel to the center of gravitation of the COMBINED mass, than the initial distance to Earth. For example, with theta=180 degrees, the "combined mass" shrinks in size over time as its two components attract, increasing the total distance the lighter mass needs to fall.

 

There are cases where they hit at the same time, and cases where the heavier mass hits first, and it should be possible to contrive cases where the lighter mass hits first.

[Mordred]

The "angular separation" is zero AT START. You have no proof that the angular separation is zero

 

Ok I can offhand think of a few situations that can be used to support this, so I agree with you on that aspect. Might help if you provide a few examples

Edited June 15, 2014 by Mordred

[md65536]

The setup is as follows [...]

Thanks imatfaal, I think this should clear up a lot of issues.

 

However I think something needs to be decided about the interaction of the test masses. Whether they're allowed to intersect or stick together gravitationally should make a difference.

 

For example, here is a situation that approaches swansont's conditions while keeping xyzt's condition that the masses can't intersect.

 

Place Earth at the origin (0,0). Place the heavier test mass at (x,r), and the lighter test mass at (x,-r). According to swansont's solution, the motion of the test masses in the y direction is irrelevant, and their x position should remain the same. They should remain at a fixed offset relative to each other, gravitationally stuck together.

 

Can we not conclude then that the Earth must end up with a positive y-value, and thus make contact with the heavier mass first? For example if one test mass has a mass equal to Earth's while the other's is near zero, Earth will end up around (x/2,r/2).

 

What assumptions (other than allowing the test masses to intersect) could change this result? Or how could swansont's solution work without some different assumption? If they're allowed to intersect they can be placed on the x-axis with no interaction (shell theorem), and certainly impact Earth simultaneously.

Edited June 15, 2014 by md65536

[swansont]

 

I initially agreed with Tom - as I could see nothing that changed the similarity of each objects, they start together and I cannot see where the situation would arise when this similarity would change.

 

But if I consider a situation where theta equals 180 degrees - then it is obvious that we now have a complex problem and my simplistic idea fails. Then if I narrow the angle - surely it cannot go back to being simple until theta equals zero; but then there is no lateral separation and it is a new problem.

 

 

I had assumed the objects were co-located, which I have noted more than once. The separation between them is assumed negligible.

[imatfaal]

 

I had assumed the objects were co-located, which I have noted more than once. The separation between them is assumed negligible.

 

Co-located - I must agree with you. For anything to change - ie for one object to be subject to an altered (additional or reduced) force then their must be a radial separation; but there can only be a radial separation if something has already changed. If they start together there is never any point at which they have any pre-existing imbalance of forces that will cause them to separate radially.

 

 

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[latex] m_1\frac{d^2x_1}{dt^2}=-(\frac{Gm_1M}{(X-x_1)^2}+\frac{Gm_1m_2}{(x_1-x_2)^2}) [/latex]

 

My problem with this equation (and its partner for m2) is that it provides the same answer for x1>x2 and x1<x2 .

 

You are squaring the difference between the two x coordinates and thus removing any directionality. If x1>x2 then the force on x1 from x2 is towards the earth, but if x2>x1 then the force on x1 from x2 is away from the earth.

that's alongside the other objection that - at t(0) x_1=x_2 and the second term must go away - and at any tiny instant after t=0 the two objects will have the same separation (remember that since t=0 their initial displacement, velocity, and acceleration have been the same) and thus the force on the two objects must remain the same