Why don't heavy objects fall more SLOWLY?

[Saw]

I’ve been thinking about the simple pendulum and doubt about the validity of the last formula I posted in #88. But really it’s not a question of loading the forum with more tries without knowing whether they are true or not.

Maybe the experts can help.

The question is: The simple pendulum is, after all, a case of “attenuated gravity”, just like in Galileo’s inclined plane example, isn’t it? If so, it seems we should observe in it the usual pattern observed in gravity, only with lesser magnitude. This pattern is that increasing *any* of the involved masses does affect time. If analyzing the situation from the CM frame, acceleration is not affected, because the system's CM changes if you change any of the masses, but the change of time is there. If analyzing the situation from the frame of any of the intervening masses, that becomes more apparent, because the relative acceleration increases. But the mathematical derivation could be done from the perspective of any frame and the impact on time should always shine up. For the case of orbital motion, I have seen those derivations and they actually converge on the conclusion that the orbital period varies as a function of any of the masses. Shouldn’t this apply as well to the simple pendulum experiment? If a bob is set to oscillate on the Moon, its period augments with regard to the same bob oscillating on the Earth, because the Moon is less massive than the Earth. And if you used the Moon as the bob of a pendulum on the Earth, shouldn’t that make the Earth oscillate as well around the CM of the system and shouldn’t both planets complete their cycles more frequently (lower period) than if the bob were a 1-kg mass? Of course, if one mass is the Earth and the other is a much smaller object, the latter is negligible for practical purposes, but it’d be still interesting to know the full formula (accounting for the *two* masses), because that may relevant, for example, for the purpose discussed here (to what extent does this kind of experiments serve to detect lack of proportionality between inertial and gravitational mass?).

 

[TurtleMeister]

Of course, if one mass is the Earth and the other is a much smaller object, the latter is negligible for practical purposes, but it’d be still interesting to know the full formula (accounting for the *two* masses), because that may relevant, for example, for the purpose discussed here (to what extent does this kind of experiments serve to detect lack of proportionality between inertial and gravitational mass?).

I do not know how to do the math you are asking for (nor do I have the time right now), but I can tell you that in order to use this method to detect a lack of proportionality between inertial and gravitational mass you would need to be able to control the composition of the "attractor" (large) mass. A torsion balance experiment can use this technique, where the period, or frequency, of the pendulum is measured. Of course the torsion balance experiment would not use the earth, but a controlled composition attractor mass. The larger this mass is in relation to the pendulum mass the better.

 

[Saw]

I can tell you that in order to use this method to detect a lack of proportionality between inertial and gravitational mass you would need to be able to control the composition of the "attractor" (large) mass. A torsion balance experiment can use this technique, where the period, or frequency, of the pendulum is measured. Of course the torsion balance experiment would not use the earth, but a controlled composition attractor mass. The larger this mass is in relation to the pendulum mass the better.

I agree with all that. It's just that I was not happy with the derivation of Period (T) for the "simple pendulum" that I wrote at the end of post #88, because it suffers from the same problem as the the standard derivation, namely:

- the derivation is done on the basis of the individual acceleration of the mass m of the bob;

- that means the considered reference frame is the centre of mass of the system bob-Earth;

- given so, it's not strange that such individual acceleration turns out to be independent of the bob's mass;

- however, period is a different thing: period SHOULD be affected by the bob's mass and that should shine up, regardless whether you make the derivation in the CM's (inertial) frame or in the (accelerated) frame of the bob or the Earth;

- ok, in practical terms, if m (bob's mass) is negligible wrt to M (Earth's mass), the contribution of the bob's mass to the period is virtually inexistent, but conceptually it's there and should appear in the formula and

- last but not least, the concept that the bob's mass does affect the period becomes fully relevant when you try to use the experiment to discuss on the proportionality between inertial and gravitational mass.

The use of the "modified" Law of Gravitation initially seemed to solve the issue, but as I said above it's not sufficient. Another correction is needed. I think I know now how to do it. To be consistent, if one plays with individual acceleration, one also has to play accordingly with the "individual radius" of the bob's mass. Alternatively, one can reason from the accelerated frame but then play with relative acceleration and the whole radius. Both routes should converge on the same period of oscillation. I'll try to post the derivations later (I am also short of time now) but the results should look as follows:

<br /> T = \frac{{2\pi }}<br /> {{\sqrt {\frac{{G(m_g + M_g )}}<br /> {{R^2 l}}} }} = 2\pi \sqrt {\frac{{R^2 l}}<br /> {{G(m_g + M_g )}}} = 2\pi \sqrt {\frac{l}<br /> {{\frac{{Gm_g }}<br /> {{R^2 }} + \frac{{GM_g }}<br /> {{R^2 }}}}} = 2\pi \sqrt {\frac{l}<br /> {{\frac{{Gm_g }}<br /> {{R^2 }} + g}}} <br />

 

[TurtleMeister]

Ok, I see what you're trying to do. And I think you're on the right track. You need to consider the relative acceleration instead of the COM based one.

 

[Saw]

Ok, I see what you're trying to do. And I think you're on the right track. You need to consider the relative acceleration instead of the COM based one.

Yes, judging from the accelerated frame (of the Earth, for example) and hence considering the relative acceleration, everything looks easier:

<br /> \begin{gathered}<br /> a_{relative} = \frac{{G(m_g + M_g )}}<br /> {{R^2 }}\frac{x}<br /> {l} = \omega ^2 x \hfill \\<br /> \omega ^2 = \frac{{G(m_g + M_g )}}<br /> {{R^2 l}} \hfill \\<br /> \omega = \sqrt {\frac{{G(m_g + M_g )}}<br /> {{R^2 l}}} \hfill \\<br /> T = \frac{{2\pi }}<br /> {{\sqrt {\frac{{G(m_g + M_g )}}<br /> {{R^2 l}}} }} = 2\pi \sqrt {\frac{{R^2 l}}<br /> {{G(m_g + M_g )}}} = 2\pi \sqrt {\frac{l}<br /> {{\frac{{Gm_g }}<br /> {{R^2 }} + \frac{{GM_g }}<br /> {{R^2 }}}}} = 2\pi \sqrt {\frac{l}<br /> {{\frac{{Gm_g }}<br /> {{R^2 }} + g}}} \hfill \\ <br /> \end{gathered} <br />

The second form is similar to the accepted formula for the period of orbital motion (with just R^2•l instead of R^3) and the last one shows how little different the outcome is from what you would obtain with the standard pendulum formula.

But what worried me is that judging from the inertial reference of the COM you should also obtain the same outcome, since in Newtonian mechanics, period (time), unlike acceleration, is frame-independent. Finally, I looked carefully at the derivation, from the COM frame, of Kepler’s 3nd Law (Newton’s version) that I found in http://www.vikdhillon.staff.shef.ac.uk/teaching/phy105/celsphere/phy105_derivation.html. The key is that in the COM frame you must consider both individual acceleration and individual radius.

Starting with your “modified” Law of Gravitation (with the standard one the result is similar, although with the difference we can comment at the end), the expected result is obtained as follows:

<br /> \begin{gathered}<br /> F_m = m_i \frac{{M_i }}<br /> {{(m_i + M_i )}}\frac{G}<br /> {{R^2 }}(m_g + M_g ) \to \hfill \\<br /> a_m = \frac{F}<br /> {{m_i }} = \frac{{GM_i }}<br /> {{R^2 }}\frac{{(m_g + M_g )}}<br /> {{(m_i + M_i )}} \hfill \\<br /> F_{net} = m_i \frac{{GM_i }}<br /> {{R^2 }}\frac{{(m_g + M_g )}}<br /> {{(m_i + M_i )}}\frac{{x_m }}<br /> {{l_m }} = m_i a_m \to \hfill \\<br /> a_m = \frac{{GM_i }}<br /> {{R^2 }}\frac{{(m_g + M_g )}}<br /> {{(m_i + M_i )}}\frac{{x_m }}<br /> {{l_m }} = \omega ^2 x_m \to \hfill \\<br /> \omega ^2 = \frac{{GM_i }}<br /> {{R^2 }}\frac{{(m_g + M_g )}}<br /> {{(m_i + M_i )}}\frac{1}<br /> {{l_m }} = \frac{{GM_i }}<br /> {{R^2 }}\frac{{(m_g + M_g )}}<br /> {{(m_i + M_i )}}\frac{1}<br /> {{l\frac{{M_i }}<br /> {{(m_i + M_i )}}}} = \frac{{G(m_g + M_g )}}<br /> {{R^2 l}} \to \hfill \\<br /> \omega = \sqrt {\frac{{G(m_g + M_g )}}<br /> {{R^2 l}}} \hfill \\<br /> T = \frac{{2\pi }}<br /> {{\sqrt {\frac{{G(m_g + M_g )}}<br /> {{R^2 l}}} }} = 2\pi \sqrt {\frac{{R^2 l}}<br /> {{G(m_g + M_g )}}} = 2\pi \sqrt {\frac{l}<br /> {{\frac{{Gm_g }}<br /> {{R^2 }} + \frac{{GM_g }}<br /> {{R^2 }}}}} = 2\pi \sqrt {\frac{l}<br /> {{\frac{{Gm_g }}<br /> {{R^2 }} + g}}} \hfill \\ <br /> \end{gathered} <br />

If we started instead with the official version of Newton’s formula…

<br /> \begin{gathered}<br /> F_m = \frac{{GM_g^a m_g^p }}<br /> {{R^2 }} \to \hfill \\<br /> F_{net} = \frac{{GM_g^a m_g^p }}<br /> {{R^2 }}\frac{{x_m }}<br /> {{l_m }} = m_i a_m \to \hfill \\<br /> a_m = \frac{{GM_g^a }}<br /> {{R^2 }}\frac{{m_g^p }}<br /> {{m_i }}\frac{{x_m }}<br /> {{l_m }} = \omega ^2 x_m \to \hfill \\<br /> \omega ^2 = \frac{{GM_g^a }}<br /> {{R^2 }}\frac{{m_g^p }}<br /> {{m_i }}\frac{1}<br /> {{l_m }} = \frac{{GM_g^a }}<br /> {{R^2 }}\frac{{m_g^p }}<br /> {{m_i }}\frac{1}<br /> {{l\frac{{M_i }}<br /> {{(m_i + M_i )}}}} = \frac{{G(m_i + M_i )}}<br /> {{R^2 l}}\frac{{M_g^a }}<br /> {{M_i }}\frac{{m_g^p }}<br /> {{m_i }} \to \hfill \\<br /> \omega = \sqrt {\frac{{G(m_i + M_i )}}<br /> {{R^2 l}}\frac{{M_g^a }}<br /> {{M_i }}\frac{{m_g^p }}<br /> {{m_i }}} \hfill \\<br /> T = \frac{{2\pi }}<br /> {{\sqrt {\frac{{G(m_i + M_i )}}<br /> {{R^2 l}}\frac{{M_g^a }}<br /> {{M_i }}\frac{{m_g^p }}<br /> {{m_i }}} }} \hfill \\ <br /> \end{gathered} <br />

… then the result is awkward, because –apart from some funny need to cancel M active grav. with M inertial but m passive grav. with m inertial- the period is determined by inertial masses, while it seems it should depend on the gravitational masses. The same problem would arise in the derivation of Kepler’s Third law. So, if true, this would be good feedback for the “modified” version of Newton’s Law of Gravitation.

It'd also confirm, as you hold, that the pendulum experiment is not apt for validating the WEP, because any change in the composition of the bob would have an undetectable impact on the period... even without any modification of Newton's Law of Gravitation.

 

[TurtleMeister]

… then the result is awkward, because –apart from some funny need to cancel M active grav. with M inertial but m passive grav. with m inertial- the period is determined by inertial masses, while it seems it should depend on the gravitational masses. The same problem would arise in the derivation of Kepler’s Third law. So, if true, this would be good feedback for the “modified” version of Newton’s Law of Gravitation.

Yes, the modified version handles the separation of gravitational mass and inertial mass much better.

It'd also confirm, as you hold, that the pendulum experiment is not apt for validating the WEP, because any change in the composition of the bob would have an undetectable impact on the period... even without any modification of Newton's Law of Gravitation.

I assume you meant to say: any change in the active gravitational mass of the bob as a result of a change in it's composition would have an undetectable impact on the period. Yes, I have realized this for a long time. But not until this equation was I able to relate it to others. The bob could have no active gravitational mass at all and it would still be undetectable. That is why I am so puzzled by experiments such as STEP.

I appreciate all the mathematical experimenting you've done with the equation Saw. It all looks good to me. But as I have already stated, I am poor with math. So you should not rely solely on my opinion.

 

[Saw]

I assume you meant to say: any change in the active gravitational mass of the bob as a result of a change in it's composition would have an undetectable impact on the period. Yes, I have realized this for a long time. But not until this equation was I able to relate it to others. The bob could have no active gravitational mass at all and it would still be undetectable. That is why I am so puzzled by experiments such as STEP.

Right. Maybe you could explain STEP more in detail.

I appreciate all the mathematical experimenting you've done with the equation Saw. It all looks good to me. But as I have already stated, I am poor with math. So you should not rely solely on my opinion.

Well, the math itself, the algebra may be correct. What may be wrong is the assumptions. The first part (taking the standard formula but applying to it the modified Law of Gravitation) looks less objectionable. About the second part (deriving on the basis of relative acceleration or individual acceleration with an individual radius), I am far less sure now. Maybe it's wrong because it relies on a bad understanding of the physical situation. (I am thinking that the mass of the bob affects the tension, perhaps that's the key...). It's a pity that the experts do not pop into comment, but probably that would require a specific question, in a more concise manner. I may post it at some time.

 

[TurtleMeister]

Right. Maybe you could explain STEP more in detail.

http://www.sstd.rl.ac.uk/fundphys/step/"
http://www.npl.washington.edu/eotwash/intro/intro.html"
Both of the above are modern tests of the equivalence principle, similar to Newton's pendulum experiment that you mentioned earlier. And both have/will produce the same null result for the same reason you mentioned earlier (even if the gravitational mass of one object is <> the other). The Eot-Wash experiment could work if the controlled composition mass were the attractor instead of the pendulum. And the satellite experiment can never work because we cannot control the composition of the earth. Can anyone tell me if I am missing something here?

 

[Saw]

Before making any further progress… I’ve realized that in post #73 I wrote a formula for free-fall time that is an over-simplification. I think the underlying idea posted there is still valid, but anyhow the good formula, as written in Wikipedia (http://en.wikipedia.org/wiki/Free-fall_time), which assumes mass m is negligible, would be:

<br /> t_{orbit} = \frac{{2\pi }}<br /> {{\sqrt {GM} }}\left( {\frac{R}<br /> {2}} \right)^{3/2} = \frac{{\pi R^{3/2} }}<br /> {{\sqrt {2GM} }}<br />

<br /> t_{ff} = t_{orbit} /2 = \frac{{\pi R^{3/2} }}<br /> {{2\sqrt {2GM} }}<br />

 

[TurtleMeister]

I recently noticed an interesting link between the modified universal law of gravitation equation and Newton's second law.

F = \frac{G(m^{g}_1 + m^{g}_2)}{r^2(\frac{1}{m^{i}_1}+\frac{1}{m^{i}_2})}

The equation above can also be expressed as:

F = \mu a

where,

\mu = \frac{m^{i}_1m^{i}_2}{m^{i}_1+m^{i}_2} = \text{reduced mass}

a = \frac{G(m^{g}_1+m^{g}_2)}{r^2} = \text{relative acceleration}

 

[Saw]

That’s a good point, bringing in the concept of reduced mass! It converges with the progress I was making in the domain of collisions (see the thread “Acceleration in an elastic collision”), where the conclusion seems to be that the force intervening in the collision can also be defined as:

F = \mu (1 + \varepsilon )v_{rel}^{initial}

where:

<br /> \begin{gathered}<br /> \mu = \frac{{m_1 m_2 }}<br /> {{m_1 + m_2 }} = {\text{reduced mass}} \hfill \\<br /> \varepsilon = \frac{{v_{rel}^{final} }}<br /> {{v_{rel}^{initial} }} = {\text{coefficient of restitution}} \hfill \\<br /> a = (1 + \varepsilon )v_{rel}^{initial} = {\text{relative acceleration}} \hfill \\ <br /> \end{gathered} <br />

On the other hand, if you want to consider the possibility that gravitational masses (included in the relative acceleration term) differ from inertial masses (those included in the reduced mass term), then you have to set an upper limit to the former.

In the collision formula, that limit, derived from the conservation of kinetic energy, is that relative acceleration cannot exceed 2 times relative velocity, since the coefficient of restitution can range from 0 to 1. Certainly, if you know that coefficient based on the ratio final/initial relative velocity, the formula is not very useful. But the idea is that you could also obtain it empirically as a coefficient inherent to the material that each body is composed of. In particular, one would have to use, together with “reduced or effective mass”, the “effective coefficient” of the two materials, as if they were one = k₁k₂/(k₁+k₂). I still haven’t thought, however, how to translate this to the gravity formula.

 

[Saw]

I've realized that what I called above, referring to collisions, the relative (or total) acceleration and force would actually be velocity increase and momentum of the system, respectively. To really get the acceleration and hence the force I should have divided (1+e)v_rel by the collision time.

 

[Saw]

I've realized that what I called above, referring to collisions, the relative (or total) acceleration and force would actually be velocity increase and momentum of the system, respectively. To really get the acceleration and hence the force I should have divided (1+e)v_rel by the collision time.

Oops. I correct again: where I said momentum, I should have said change of velocity and hence of momentum in a given time (collision time) = impulse, isn't it?

 

[Saw]

I have been thinking that the concept of “gravitational mass” may be somehow misleading. Really the underlying issue is: does the composition of the bodies involved in a gravitational interaction have any bearing in the outcome of a gravitational interaction? And, well, there is no need to mess that up with the concept of mass. One could perfectly let the sum of the masses of the “relative acceleration” term be divided out by the sum of the masses of the “reduced mass” term and still ask that question.

Of course, then the issue is, how could you make that factor (composition of the bodies) relevant, if (speculating about the idea) you wished to? In the context of collisions (what we could call contact force), such factor is no doubt relevant and it shines up mathematically in the coefficient of restitution, which is a dimensionless number. In the context of springs (elastic force), it appears as the spring constant, which is measured in N/m. In the context of gravity, we have the constant G, measured in Nm^2/kg^2. Thus… if we considered the possibility that G were multiplied by a certain (dimensionless) coefficient, would that be simply illogical, would it clash against physics principles?

It seems it would have to be a “reductive coefficient”, ranging between 0 and 1, like the coefficient of restitution (let’s call it “e”). Thus we could have less but not more relative acceleration than in the standard formula. On top of that, it would be empirically constructed, like I gather that it happens with “e”, on the basis of the properties (composition) of the two involved bodies. In fact, it would be “e”.

For example, if we have two bodies “dropped” from a relative distance R and acquiring a certain relative velocity at collision time and if the collision is partially elastic and partially plastic, the bodies would bounce off each other with outgoing relative velocity = (1+e) times their incoming relative velocity… Then they separate until running out of outgoing kinetic energy due to the gravitational force between them. Since that outgoing KE is now lower (part of it has been dedicated and is kept busy heating or deforming the bodies), gravity will stop them and revert their direction sooner, before they reach the initial distance R. Thus they’ll return with lower velocity, which will diminish even more after collision. And so on until the bodies are brought to rest with each other.

But does all that mean that the gravitational interaction is fully “elastic” (it only transfers to the bodies energy that is useful for motion, that is either kinetic or potentially convertible into kinetic energy)? In principle, I do not se why it should be necessarily so. Note this: when the bodies are separating from each other, they are being stretched, due to the non-uniformity of the gravitational field. That is the well-known phenomenon of tides. And when they descend or get closer to each other, I think that they progressively lose that condition, because (as the force increases due to the inverse square law) the difference of strength of the field between the two extremes becomes narrower. Given this, should the bodies perfectly recover their original form? Why should they? They will not when they are later compressed due to the collision. Should the gravitational stretching be different? Shouldn’t it also generate, due to the composition of the bodies, some conversion of the energy transferred by gravity into other non-mechanical forms? I wonder if there is any logical, not purely experimental reason banning that.

 

[boit]

If I lived in a planet where more massive objects were falling much slower, I will create a perpetual motion energy generator in no time. How? Since by implication more massive object will have less potential energy, I'll rig the contraption to haul stones in a single bag and then throw them down sepatately. On Earth surely 1/2(1+1+1)10 m^2/s^2 can not be less than 1/2(3)10 m^2/s^2 [from E=1/2mv^2]. This is because whatever you do the centre of mass will be a point source. The trick could be to make your more massive object so tall that the COG is in a zone where accelaration is say half 'g'. But unfortunately you can't make them rigid enough.

 

[Saw]

If I lived in a planet where more massive objects were falling much slower

Boit, I understand that this is a little confusing, since the OP is actually that, but in the end we are not discussing that, we drifted to a related issue, namely equivalence between inertial and gravitational mass. In this context, I simply asked something like: if a collision between planets is imperfectly elastic (so that the outgoing relative v of separation < the incoming relative v of approach), we assume nevertheless that the subsequent gravitational interaction is perfectly elastic (the relative v of return after the path closes = the relative velocity of departure), but does it HAVE TO be so? Of course, the relative return v can't be higher than relative departure v, but can't it be lower, even in the absence of external forces?

If it were so and that deceleration happened in inverse proportion to the masses, the velocity of the CoM of the system would not change = conservation of momentum would be respected. Yes, Kinetic Energy, at the end of the cycle, would not be conserved but that is not unsual: that happened in the collision and it's not dramatic. The sacred principle is that Total Energy must be conserved, but that is also respected if the KE converts into any another form, even if it is not potential energy that is reusable for motion. The only principle that would be breached is that gravity ensures 100% conservation of mechanical energy. And... although this principle is quite useful for calculations and true in almost all practical situations where gravity is the only intervening force, is it necessarily always applicable or rather a practical assumption, which could however break at large scales?

 

[Flustered]

If I jump in the air and come back down due to gravity, you guys are saying that my speed in which I fall back to the ground is the same speed in which satellites orbit the Earth? Both bodies have gravity acting upon it so wouldn't it be the same speed?

 

[Saw]

If I jump in the air and come back down due to gravity, you guys are saying that my speed in which I fall back to the ground is the same speed in which satellites orbit the Earth?

Not really. The principle refers to acceleration, not to speed. All objects falling towards the Earth, regardless their different masses, suffer the same acceleration, as long as you leave air resistance aside. This is true as measured from or with regard to an inertial reference frame or (if you disregard as negligible the acceleration caused onto the Earth by the object in question) with regard to the Earth itself. And as you know acceleration means change of speed or direction per time unit.

You suffer the same acceleration as the satellite but your speed is lower. You started falling (at the point of maximum height) with speed zero and have fallen a short path and hence have accelerated (your speed has grown) during a short time. Instead, the satellite had to be launched with a high speed so as to enter into orbit. It is true that once it is in orbit, its speed does not change much; actually, if the orbit were circular, its speed would not change at all, because all the work of gravity woud be used changing its direction; yet the satellite's speed, for the reason explained above, will always be higher than yours, if you just jump and fall back.

 

[Dali]

This is a simple question, and I'm sure it has a trivial answer, but this thought occurred to me just now.

How is it that gravity is able to accelerate things at the same rate regardless of its mass, while one definition of mass is the measure of resistance to acceleration? If two objects are 100 miles from Earth, I would expect that the Earth's gravity would be able to accelerate the lighter object more quickly, while the heavier object would resist the acceleration and fall more slowly. Why is this not the case?

Well - it can't be so. Just think of the situation if heavier objects really had fallen slower, what would have happened if you divided the heavy object in two parts? Should those parts start falling quicker then? Why? Or divide it in 1000 small light parts! Should speed depend on if those parts where held together or not? How tight do the parts need to sit together to fall slower than they would individually? It obviously could not work that way.

Best way to think about it is that gravity acts equally on each atom in every object. No matter how many atoms there are, or if they are bound together or not.

 

[Buckleymanor]

If you are in a rocket ship out in space undergoing acceleration and you "drop" a marble and a bowling ball in your ship, they will "fall" with the same acceleration. Einstein's equivalence principle is saying that gravity is the same thing as the "force" that pushes you back in your car seat when you step on the gas.

The marble and the ball only fall with the same acceleration if you "drop" them.If you placed the marble and ball on the floor of the spaceship when it was accelerating and then decelerated or stopped the craft, the bowling ball and marble would accelerate away from the floor at different rates.Gravity is similar to the force that pushes you back in car seat when you step on the gas but not the same.

 

[TurtleMeister]

If you placed the marble and ball on the floor of the spaceship when it was accelerating and then decelerated or stopped the craft, the bowling ball and marble would accelerate away from the floor at different rates.

If what you mean by "stop the craft" is to simply turn off the thrusters then that's not true. If you disregard the elasticity of the objects then both balls will remain on the floor. To bring them off the floor you would need to reverse the thrusters. And even then they would accelerate away from the floor at the same rate. That's because the balls will not really accelerate at all. The ship will accelerate from the balls.

 

[Buckleymanor]

If what you mean by "stop the craft" is to simply turn off the thrusters then that's not true. If you disregard the elasticity of the objects then both balls will remain on the floor. To bring them off the floor you would need to reverse the thrusters. And even then they would accelerate away from the floor at the same rate. That's because the balls will not really accelerate at all. The ship will accelerate from the balls.

No it would not, the craft would just have to be accelerating or traveling at a rate less than maximum speed the bowling ball had attained when it was first accelerated by the spacecraft .The objects are elastic I don't see how you can disregard this unless they are made of unobtanium.
You might as well conclude that the balls are stationary and the craft is doing all the movement and acceleration around them.
If the balls were placed on your car seat and accelerated and the car crashed the marble would be accelerated faster and travell further than the bowling ball if it's exit through the winderscrean was unristricted.

 

[Jasso]

.If you placed the marble and ball on the floor of the spaceship when it was accelerating and then decelerated or stopped the craft, the bowling ball and marble would accelerate away from the floor at different rates.

How do you manage to get this? Ignoring elastic rebound, they would appear to accelerate in the opposite direction as the rocket, with the same magnitude.

Gravity is similar to the force that pushes you back in car seat when you step on the gas but not the same.

And how would you tell the difference between constant accelerating and gravity in a soundless, windowless car?

No it would not, the craft would just have to be accelerating or traveling at a rate less than maximum speed the bowling ball had attained when it was first accelerated by the spacecraft .

Unless the direction of acceleration were to change, then the objects would not come off of the floor, even if the magnitude of acceleration were to decrease. That is because the spaceship will always be moving faster at a time t+dt and the only way the object will do that is if the floor of the spaceship is in contact with it.

The objects are elastic I don't see how you can disregard this unless they are made of unobtanium.

Ah, I see how you managed to get that earlier. It's an idealization, used to ignore whatever is irrelevant to the situation. And it's irrelevant since once the object loses contact with the floor, its elasticity doesn't matter; it will continue with constant velocity because of Newton's First Law.

You might as well conclude that the balls are stationary and the craft is doing all the movement and acceleration around them.

That's pretty much the entire point of the equivalence principle, that there is no difference between being on a spaceship with constant acceleration, or being on the surface of a planet with the gravitational acceleration.

If the balls were placed on your car seat and accelerated and the car crashed the marble would be accelerated faster and travell further than the bowling ball if it's exit through the winderscrean was unristricted.

No, they would continue on with the same velocity they had right before they lost contact with the seat.

 

[TurtleMeister]

Buckleymanor, I agree with Jasso's objections to your comments. An analogy to ignoring the elasticity of the objects is in the thought experiment for the universality of free fall. In that scenario we commonly ignore the effects of air resistance and buoyancy when we say that two objects will accelerate at the same rate regardless of there difference in mass. We do that because, as Jasso pointed out, those effects are irrelevant to the point of the thought experiment.

To demonstrate, here's another thought experiment. Imagine two balls of different mass sitting on a table in an accelerating spacecraft (or sitting on the surface of earth). Because of the elasticity of the objects, the two balls, and the table they are sitting on will deform from the force of acceleration. When the thrusters are turned off (or Earth's gravity is turned off) the balls and the table will resume their normal shape (rebound), causing the balls and the table/ spacecraft (or earth) to accelerated away from each other. But once the balls and the table are no longer in contact, the acceleration stops, and they continue on away from each other at a constant speed. This is the effect that we want to ignore.

If we ignore the elasticity of the objects then the balls will remain on the floor when the thrusters are turned off. The only way to bring them off the floor would be to reverse, or change direction of, the thrusters. If we reverse the thrusters, then the spacecraft will accelerate away from the balls.

 

[Buckleymanor]

How do you manage to get this? Ignoring elastic rebound, they would appear to accelerate in the opposite direction as the rocket, with the same magnitude.

That's the point you can't ignore elastic rebound.

And how would you tell the difference between constant accelerating and gravity in a soundless, windowless car?

By using elastic rebound to note the difference speeds of acceleration upon different objects.

Unless the direction of acceleration were to change, then the objects would not come off of the floor, even if the magnitude of acceleration were to decrease. That is because the spaceship will always be moving faster at a time t+dt and the only way the object will do that is if the floor of the spaceship is in contact with it.

Would that not depend on the speed the magnitude of acceleration decreased.

Ah, I see how you managed to get that earlier. It's an idealization, used to ignore whatever is irrelevant to the situation. And it's irrelevant since once the object loses contact with the floor, its elasticity doesn't matter; it will continue with constant velocity because of Newton's First Law.



That's pretty much the entire point of the equivalence principle, that there is no difference between being on a spaceship with constant acceleration, or being on the surface of a planet with the gravitational acceleration.

There is.

No, they would continue on with the same velocity they had right before they lost contact with the seat.

Which would be different for the bowling ball and marble.Substitute car and seat for catapult.
Fire bowling ball, then marble, with the same amount of force, which travels furthest and fastest.

 

[russ_watters]

You're confusing acceleration and force.

 

[Buckleymanor]

You're confusing acceleration and force.

Don't forces come into play when onboard accelerating spacecraft or cars.
I ain't sure if it's just a question of blanking out the windows and making them soundproof.

 

[Buckleymanor]

Buckleymanor, I agree with Jasso's objections to your comments. An analogy to ignoring the elasticity of the objects is in the thought experiment for the universality of free fall. In that scenario we commonly ignore the effects of air resistance and buoyancy when we say that two objects will accelerate at the same rate regardless of there difference in mass. We do that because, as Jasso pointed out, those effects are irrelevant to the point of the thought experiment.

To demonstrate, here's another thought experiment. Imagine two balls of different mass sitting on a table in an accelerating spacecraft (or sitting on the surface of earth). Because of the elasticity of the objects, the two balls, and the table they are sitting on will deform from the force of acceleration. When the thrusters are turned off (or Earth's gravity is turned off) the balls and the table will resume their normal shape (rebound), causing the balls and the table/ spacecraft (or earth) to accelerated away from each other. But once the balls and the table are no longer in contact, the acceleration stops, and they continue on away from each other at a constant speed. This is the effect that we want to ignore.

If we ignore the elasticity of the objects then the balls will remain on the floor when the thrusters are turned off. The only way to bring them off the floor would be to reverse, or change direction of, the thrusters. If we reverse the thrusters, then the spacecraft will accelerate away from the balls.

Completly agree as to ignoring the air resistance and buoyancy effects on different objects in free fall.To enable us to come to the conclusion that they accelerate at the same rate regardless of there different mass.
The difference between that and the anology you are making is that it is possible to conduct an experiment with two different objects free falling within a vacuum to check your results.
Ignoring elasticity in a thought experiment is not possible experimentally so might as well substitute it for green cheese.

 

[russ_watters]

Don't forces come into play when onboard accelerating spacecraft or cars.

Of course - whenever you have one (net force, that is) you have the other. But you defined your scenarios such that in one case the accelerations were equal (thus, forces different) and in the other the forces were equal (thus accelerations different).

 

[Buckleymanor]

Of course - whenever you have one (net force, that is) you have the other. But you defined your scenarios such that in one case the accelerations were equal (thus, forces different) and in the other the forces were equal (thus accelerations different).

If I were to hit a bowling ball with a tennis bat and then a marble, both with the same amount of force won't one travell further than the other.
The marble and the bowling ball are separated on the floor of the craft.