A Flaw In Using Fall?

[Mitch Bass]

Reading over the formulas following the post named Gravity Stacking (OP: 514void) two questions come to my mind that are not associated enough with that post to be asked within that post. The first question has to do with the esssence how the first question of the post was was asked: would something as massive as the earth fall at g or 2g towards the earth? Since we are in the part of the forum entitled Relativity I am curious to know if the way science perceives reality after Relativity became part of the scientific vernacular, would the mass of two objects as massive as Earth cause gravitation to make those objects fall towards each other or would it be said that the two objects would move towards each other because of how some would say mass causes space to "bend"/"curve"?

 

The second question I thought of after reading the Gravity Stacking Post (OP: 514void) might seem too extreme to warrant a response, too outlandish to be taken as a serious point and far from the realm of where anyone would put any effort into considering contempletaing. If I get no feedback as a result of my following words I would understand why. Anyway….a postulate of Galileo...all objects fall to the Earth at the same velocity when "let go" regardless of weight. Galieleo predicted that if the Earth had no atmosphere even a feather would fall at the same speed and on the moon, an experiment was done in which it was demonstrated that a feather and an opject of much greater mass did seem to move to the moon at the same velocity when let go. However, ultimately, this postuate, that two objects, regardless of mass "differential" would "fall" towards the Earth at the same velocity, is not right….right? If there was a marble sized object with the mass of the moon being held two thousand yards away from the Earth and there was a marble sized object with the mass of a bowling ball being held two thousand yards away from the Earth...and they were both let go at the same time…the marble sized object with the mass of the moon would pull the Earth towards itself, the marble size object with the mass of a bowling ball would also pull the Earth as well but unmeasureably to any modern day calculating instruments. If these two marble size objects were dropped at the same time, but far away from each other so that the marble sized object with greater mass did not have a gravitational effect on the lesser mass marble sized object, the result would be that the lesser mass marble sized object would move at a greater velocity even though both marble sized objects would make contact with Earth at the same time?

[Strange]

Yes, the two object would both move - they would both "fall" towards their common center of gravity. This can usually be ignored but, for example, the Moon does not really orbit the Earth, both the Earth and the Moon orbit their "barycenter". (This is nothing to do with relativity, by the way. Except in the most general sense.)

 

This makes no difference to Galileo's point. If you measure the rate of fall assuming the the Earth is stationary, then it makes no difference what the mass of the object is. If you analyse it from a different frame of reference, you will see the Earth move more towards the heavier object but the closing speed between the object and the Earth will still be independent of the mass of the object.

[Delta1212]

Yes, the two object would both move - they would both "fall" towards their common center of gravity. This can usually be ignored but, for example, the Moon does not really orbit the Earth, both the Earth and the Moon orbit their "barycenter". (This is nothing to do with relativity, by the way. Except in the most general sense.)

 

Buh dum tss

 

This makes no difference to Galileo's point. If you measure the rate of fall assuming the the Earth is stationary, then it makes no difference what the mass of the object is. If you analyse it from a different frame of reference, you will see the Earth move more towards the heavier object but the closing speed between the object and the Earth will still be independent of the mass of the object.

Technically isn't the attraction dependent upon both masses, it's just that the Earth is so much more massive than anything that we find on the Earth that the non-Earth's contribution is essentially negligible? So an elephant will actually fall faster than a mouse but by an immeasurable amount because compared to the Earth an elephant and a mouse might as well have the same mass.

 

 

That aside, I think fall is a better word than move. Fall implies an acceleration where move generally doesn't. I conjure a different mental image when I hear that two objects fall toward one another than when I hear that two objects move toward one another, and the image I get from the word fall more closely matches what is happening.

Edited June 10, 2014 by Delta1212

[imatfaal]

...

 

Technically isn't the attraction dependent upon both masses, it's just that the Earth is so much more massive than anything that we find on the Earth that the non-Earth's contribution is essentially negligible? So an elephant will actually fall faster than a mouse but by an immeasurable amount because compared to the Earth an elephant and a mouse might as well have the same mass.

 

...

 

 

The force is dependant on both masses - but the acceleration of object one due to that force is inversely proportional to the mass of object one (they cancel). The acceleration of object one is thus independent of the mass of object one - to actually measure you need no air resistance etc. (qv moon hammer feather).

[Delta1212]

 

 

The force is dependant on both masses - but the acceleration of object one due to that force is inversely proportional to the mass of object one (they cancel). The acceleration of object one is thus independent of the mass of object one - to actually measure you need no air resistance etc. (qv moon hammer feather).

Yes, but you said the closing speed of the Earth and the object, rather than the acceleration of the object. I realize that the acceleration due to Earth's gravity is going to be the same, but the acceleration of Earth due to the object's gravity is not (barely) so the closing speed won't be the same (again, technically though not measurably, and insomuch as two accelerating objects have a closing speed).

 

I probably shouldn't have said the elephant will 'fall' faster in that case, but I was looking at it from an Earth-stationary perspective.

 

Am I mistaken in all this?

Edited June 10, 2014 by Delta1212

[swansont]

Yes, but you said the closing speed of the Earth and the object, rather than the acceleration of the object. I realize that the acceleration due to Earth's gravity is going to be the same, but the acceleration of Earth due to the object's gravity is not (barely) so the closing speed won't be the same (again, technically though not measurably, and insomuch as two accelerating objects have a closing speed).

 

I probably shouldn't have said the elephant will 'fall' faster in that case, but I was looking at it from an Earth-stationary perspective.

 

Am I mistaken in all this?

It depends on how the problem is stated. If the objects are dropped at the same time, they will hit at the same time; any motion of the earth is common to both and may be ignored. If you are looking at individual events, the larger mass (still assuming << M_e) strikes first (albeit immeasurably) because of the imperceptible increase in acceleration of the earth. It won't fall "faster", though, since its acceleration is the same as the smaller mass as measured by an observer at rest with the CoM. The smaller mass would actually attain a higher speed, in that tiny length of time it falls the minuscule extra distance.

[xyzt]

It depends on how the problem is stated. If the objects are dropped at the same time, they will hit at the same time; any motion of the earth is common to both and may be ignored. If you are looking at individual events, the larger mass (still assuming << M_e) strikes first (albeit immeasurably) because of the imperceptible increase in acceleration of the earth. It won't fall "faster", though, since its acceleration is the same as the smaller mass as measured by an observer at rest with the CoM. The smaller mass would actually attain a higher speed, in that tiny length of time it falls the minuscule extra distance.

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.

Edited June 11, 2014 by xyzt

[swansont]

If two test probes of massses [math]m_1,m_2[/math] are dropped 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 they are dropped at the same time, the earth will respond to the sum of their masses and they will hit at the same time, assuming only the mass is different. If they are dropped individually, the time to impact is different, as your equation shows. The two situations are not equivalent, so the statement of the problem matters.

[xyzt]

 

If they are dropped at the same time, the earth will respond to the sum of their masses and they will hit at the same time, assuming only the mass is different. If they are dropped individually, the time to impact is different, as your equation shows. The two situations are not equivalent, so the statement of the problem matters.

The solution posted is for one test probe being dropped at a time. I think that dropping both test probes simultaneously is a much more complicated problem (one test probe at a time is already very complicated) , I do not know the solution (I only know that this would be a variant of the "Three-body problem") and I do not know if your claim is true or false, can you post the math supporting your claim? I would be very interested in learning.

Edited June 11, 2014 by xyzt

[swansont]

The solution posted is for one test probe being dropped at a time. I think that dropping both test probes simultaneously is a much more complicated problem (one test probe at a time is already very complicated) , I do not know the solution (I only know that this would be a variant of the "Three-body problem") and I do not know if your claim is true or false, can you post the math supporting your claim? I would be very interested in learning.

 

The math is identical to what you posted; the interaction between the test masses can be safely ignored, as can their small separation, so the probe mass is the sum of the two objects being tested. Newtonian gravity, like all forces, obeys vector summation.

 

IOW, the earth will respond to two co-located masses m the same way it will respond to a mass of 2m.

[xyzt]

 

The math is identical to what you posted; the interaction between the test masses can be safely ignored, as can their small separation, so the probe mass is the sum of the two objects being tested. Newtonian gravity, like all forces, obeys vector summation.

 

IOW, the earth will respond to two co-located masses m the same way it will respond to a mass of 2m.

What makes you think that the two test probes do not separate during the fall? You are making an assumption that is not backed up by any math. Actually, the more I think about it, the more I become convinced that your assumption has no basis.

Edited June 11, 2014 by xyzt

[dimreepr]

So what if they did? The only way that would be significant is if they separate horizontally.

 

 

Edit/ Depending, of course, on the height of the drop.

Edited June 11, 2014 by dimreepr

[xyzt]

So what if they did? The only way that would be significant is if they separate horizontally.

I am talking about radial separation, meaning that they do NOT hit the Earth at the same time. Contrary to swansont's claim above that started it all.

Edited June 11, 2014 by xyzt

[dimreepr]

I'm sorry, my claim? Starting what?

[xyzt]

I'm sorry, my claim? Starting what?

Sorry, swansont's claim.

Edited June 11, 2014 by xyzt

[swansont]

I am talking about radial separation, meaning that they do NOT hit the Earth at the same time. Contrary to swansont's claim above that started it all.

 

What force would cause this? The only radial force on each test mass is from the earth, and the acceleration will be identical.

[xyzt]

 

What force would cause this? The only radial force on each test mass is from the earth, and the acceleration will be identical.

But the forces exerted by the Earth on the two test probes are NOT identical, as a matter of fact they are different so the test probes are highly likely to SEPARATE in flight, such that they are NO LONGER subjected to the same acceleration. You need to solve a very complicated "three body problem" in order to get the equations of motion. I can post the solution for the two body case, if you wish, you can try to generalize to the case of the three bodies, the generalization is not trivial, you are oversimplifying things.

[swansont]

The velocity will be from the acceleration, which will be the same, since a = F/m. Unless it's being offered up as a GR test where you can't assume this, but that's an entirely different problem.

[xyzt]

The velocity will be from the acceleration, which will be the same, since a = F/m. Unless it's being offered up as a GR test where you can't assume this, but that's an entirely different problem.

I think that it is a LOT more complicated. You need to solve for [math]X(t), x_1(t),x_2(t)[/math] the system of coupled differential equations:

 

[math]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2})[/math]

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

[math]m_2\frac{d^2x_2}{dt^2}=-(\frac{Gm_2M}{(X-x_2)^2}+\frac{Gm_1m_2}{(x_1-x_2)^2})[/math]

 

with the initial conditions:

 

[math]x_1(0)=x_2(0)=D[/math]

[math]X(0)=0[/math]

[math]\frac{dX}{dt}=\frac{dx_1}{dt}=\frac{dx_2}{dt}=0[/math] at [math]t=0[/math]

The "times to collision" are obtained from the conditions [math]X(t)-x_i(t)=R+r_i[/math], [math]i=1,2[/math]

Edited June 11, 2014 by xyzt

[md65536]

Technically isn't the attraction dependent upon both masses, it's just that the Earth is so much more massive than anything that we find on the Earth that the non-Earth's contribution is essentially negligible? So an elephant will actually fall faster than a mouse but by an immeasurable amount because compared to the Earth an elephant and a mouse might as well have the same mass.

Skimming the thread I didn't see this last question answered.

 

Yes, the elephant falls faster because the Earth also falls toward the elephant. But don't forget the caveat "If you measure the rate of fall assuming the the Earth is stationary", in which case the answer is No. They fall at the same rate.

 

This makes more sense in context of the other thread where the falling mass is of the same magnitude as Earth's, where it more significant. Consider two large masses, and you're keeping one of them stationary, so that it is resisting the gravitational attraction of the other. What is the force that would be required to keep it stationary? (Ans: it's equal and opposite of the fictitious force of the other mass's gravitational pull on the stationary object.) Suppose that you kept the one mass stationary using rockets. What is the acceleration provided by those rockets? (Ans: Equal and opposite of the gravitational acceleration toward the other mass.) If you keep one mass stationary, and sum up the accelerations, you find that the other mass indeed accelerates at a rate independent of its own mass (though the force involved in keeping the first mass stationary depends on it).

 

In your example, the increased closing acceleration due to the elephant's mass would be negated by the acceleration provided by the rockets (or the nail on which Earth is pinned, or the acceleration of your frame of reference,* or whatever means you use to keep Earth stationary).

 

 

 

* Err... nah... I guess you'd have to physically counteract the pull of gravity to negate the increased closing acceleration. But if you can keep Earth stationary by calling the pull on it "negligible", then the increased closing acceleration is negligible.

Edited June 12, 2014 by md65536

[swansont]

I think that it is a LOT more complicated. You need to solve for [math]X(t), x_1(t),x_2(t)[/math] the system of coupled differential equations:

 

[math]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2})[/math]

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

[math]m_2\frac{d^2x_2}{dt^2}=-(\frac{Gm_2M}{(X-x_2)^2}+\frac{Gm_1m_2}{(x_1-x_2)^2})[/math]

 

with the initial conditions:

 

[math]x_1(0)=x_2(0)=D[/math]

[math]X(0)=0[/math]

[math]\frac{dX}{dt}=\frac{dx_1}{dt}=\frac{dx_2}{dt}=0[/math] at [math]t=0[/math]

The "times to collision" are obtained from the conditions [math]X(t)-x_i(t)=R+r_i[/math], [math]i=1,2[/math]

 

Now simplify

 

We only need to care about the x₁-x₂ terms. Those are the only terms that make any difference anyway. The motion of the earth doesn't matter, since it's common to both. So tell me, how will we get any radial direction separation if they are dropped at the same time and from the same distance?

[DimaMazin]

I think that it is a LOT more complicated. You need to solve for [math]X(t), x_1(t),x_2(t)[/math] the system of coupled differential equations:

 

[math]M\frac{d^2X}{dt^2}=+(\frac{GMm_1}{(X-x_1)^2}+\frac{GMm_2}{(X-x_2)^2})[/math]

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

[math]m_2\frac{d^2x_2}{dt^2}=-(\frac{Gm_2M}{(X-x_2)^2}+\frac{Gm_1m_2}{(x_1-x_2)^2})[/math]

 

with the initial conditions:

 

[math]x_1(0)=x_2(0)=D[/math]

[math]X(0)=0[/math]

[math]\frac{dX}{dt}=\frac{dx_1}{dt}=\frac{dx_2}{dt}=0[/math] at [math]t=0[/math]

The "times to collision" are obtained from the conditions [math]X(t)-x_i(t)=R+r_i[/math], [math]i=1,2[/math]

Excuse me. I don't know the math, but only inequality can be decision of the problem.

[xyzt]

 

Now simplify

 

We only need to care about the x₁-x₂ terms.

OK, you will need to solve the system of equations, if I understand correctly, you claim that [math]x_1(t)-x_2(t)=0[/math]. I am skeptical. So , solve the system and prove your point.

 

Here, subtract the two equations:

 

[math]\frac{d^2x_1}{dt^2}=-(\frac{GM}{(X-x_1)^2}+\frac{Gm_2}{(x_1-x_2)^2})[/math]

[math]\frac{d^2x_2}{dt^2}=-(\frac{GM}{(X-x_2)^2}+\frac{Gm_1}{(x_1-x_2)^2})[/math]

 

You get:

 

[math]\frac{d^2(x_1-x_2)}{dt^2}=-(\frac{GM}{(X-x_1)^2}-\frac{GM}{(X-x_2)^2}+\frac{G(m_2-m_1)}{(x_1-x_2)^2})[/math]

 

Actually, the problem is even more complicated because we must take into consideration that the centers of the test probes are laterally separated as well:

 

[math]\frac{d^2x_1}{dt^2}=-(\frac{GM}{(X-x_1)^2}+\frac{Gm_2}{(x_1-x_2)^2+(r_1+r_2)^2})[/math]

[math]\frac{d^2x_2}{dt^2}=-(\frac{GM}{(X-x_2)^2}+\frac{Gm_1}{(x_1-x_2)^2+(r_1+r_2)^2})[/math]

 

So, the correct equation is:

 

[math]\frac{d^2(x_1-x_2)}{dt^2}=-(\frac{GM}{(X-x_1)^2}-\frac{GM}{(X-x_2)^2}+\frac{G(m_2-m_1)}{(x_1-x_2)^2+(r_1+r_2)^2})[/math]

 

Looking at the above, it becomes immediately apparent that it is impossible to have [math]x_1-x_2=0[/math] because it would result into:

 

[math]0=-\frac{G(m_2-m_1)}{(r_1+r_2)^2}[/math]

 

So, contrary to your claim, the test probes hit the ground simultaneously if and only if [math]m_1=m_2[/math], a trivial conclusion.

While it has become clear that it is difficult, if not impossible to find the equations of motion [math]X(t),x_1(t),x_2(t)[/math] we can safely conclude that the two probes hit the ground simultaneously if and only if they are identical, they must not only have the same mass (see above) but the also must have the same radius (otherwise, the one with larger radius touches down first).

Edited June 12, 2014 by xyzt

[imatfaal]

Skimming the thread I didn't see this last question answered.

 

Yes, the elephant falls faster because the Earth also falls toward the elephant. But don't forget the caveat "If you measure the rate of fall assuming the the Earth is stationary", in which case the answer is No. They fall at the same rate....

 

 

SwnsnT answered it - but now I am getting confused. I have put my thoughts in separate sentences to allow delineation of where the fault arises.

 

The scenario with the elephant has a greater attractive force (M_elephant*M_earth>M_mouse*M_earth) and thus the earth accelerates more than in the scenario with the mouse.

 

From an outside perspective the elephant and the mouse both accelerate at exactly the same rate (a_animal = GM_earth/R^2) - but the earth's acceleration is greater in the elephant scenario (GM_elephant/R^2>GM_mouse/R^2)

 

The elephant will hit the earth in a slightly shorter time as the earth is moving towards it quicker than the earth would have moved towards the mouse.

 

From outside perspective; At any time after release the elephant and the mouse will have the same speed - acceleration is the same.

 

From outside perspective; Mouse travels a greater distance in a greater time (cos the earth isnt rushing up to greet it ) and thus has a higher speed on impact

 

 

If however you measure from the earth (and thus assume the earth is stationary) - you are by default including the added acceleration of the earth in the elephant scenario to the acceleration of the elephant and you will record a higher acceleration for the elephant.

 

The elephant hits first if viewed from an outside perspective - therefore it must hit first from all perspectives.

[xyzt]

 

 

The elephant hits first if viewed from an outside perspective - therefore it must hit first from all perspectives.

Correct. See the rigorous analysis in this post.

Now, if one drops both the elephant and the mouse simultaneously, contrary to Swansont's claim , they do not hit the Earth simultaneously.

Edited June 12, 2014 by xyzt