B A contradiction of the equivalence principle?

[Dale]

different masses dropped at a certain height do reach the ground at different times?

Yes. Different masses accelerate the same in the Earth’s field. But the Earth accelerates differently in different masses fields.

 

[hmmm27]

No, this is wrong. $F = ma$ applies separately to the two masses, so you have $F = m_1 a_1$ and $F = m_2 a_2$. Then you can solve for $a_1$ and $a_2$.

Ahhh..... thanks, I thought it looked too easy. So, closing acceleration is . . . okay, given the length of this thread, I find the math answer to the OP's question hilariously simple.

 

[jeff einstein]

yes sir the maths is simple but the thought is hard

 

[jeff einstein]

so i can now conclude that all masses accelerate to the earth at the same rate but different masses reach the ground at different times when held at a particular distance. am i right @Dale and @PeroK sir?

 

[jeff einstein]

actually, i have high respect for you all as you have so much patience... you are probably way more experienced than me at physics and you actually took your time to explain a misunderstood topic to a confused school boy. thank you very much!!!

 

[jeff einstein]

so i can now conclude that all masses accelerate to the earth at the same rate but different masses reach the ground at different times when held at a particular distance. am i right @Dale and @PeroK sir?

just clarify this then i am done

 

[hmmm27]

so i can now conclude that all masses accelerate to the earth at the same rate but different masses reach the ground at different times when held at a particular distance.

I understand the math involved though I didn't show it nor give you an answer, but your explanation of your understanding is self-contradictory and makes no sense, so we have no way of knowing if you actually "get it".

Care to reciprocate the massive amount of time forum members have put into your query under your conditions, and show us the three lines of math involved ? to prove that you actually do understand the answer to whatever your question was in the first place. Or, was your purpose more along the lines of manipulation.

 

[DaveC426913]

To recap:

1. Objects of different masses will all fall to Earth at the same rate (ignoring air resistance) - with the proviso that the objects' masses are insignificant in comparison to Earth's mass.
2. For objects whose mass is a non-insignificant fraction of Earth's, one must take into account the object's own gravitational effect on Earth.

Technically, these are not two distinct cases. Case 1 is an arbitrary simplification of Case 2, wherein we choose to ignore the miniscule effect of a small object's gravitational influence on the outcome. Which is why you have likely encountered both forms of solution in-the-wild.(Right?)

 

[PeterDonis]

Like considering the size and shape of objects and not just their mass.

I think this is a distraction from the main point of the thread. At the very least, I think we should get clear about the simple case, where the masses are moving in vacuum and tidal effects are ignored, before trying to add complications.

 

[PeterDonis]

just clarify this then i am done

Please read my post #52.

 

[PeterDonis]

different masses dropped at a certain height do reach the ground at different times?

Again, read my post #52.

 

[PeroK]

Here's my answer. We drop a 1 kg mass and 2 kg mass from 10m above the Earth's surface. Making some basic assumptions ... The 2kg mass hits the ground at time $t_2$. After time $t_2$, when the experiment is repeated with the 1kg mass, that mass is still $1 \times 10^{-24} m$ above the ground. It takes a further $1 \times 10^{-25} s$ for the $1 kg$ object to hit the ground.

A measurement of $1 \times 10^{-24} m$ in the position of an object is outside the domain of applicability of classical physics.

To test the statement, we need a theory that is applicable on this scale.

 

[PeterDonis]

so you give up sir? and i can go around with my "unproved" misunderstanding on this topic

No. You need to think more carefully about exactly what the different claims involved mean. I am hoping that my post #52 will help. At the very least, it should give you more of a framework for thinking about the problem.

 

[Dale]

so i can now conclude that all masses accelerate to the earth at the same rate but different masses reach the ground at different times when held at a particular distance. am i right @Dale and @PeroK sir?

Yes.

 

[PeterDonis]

Making some basic assumptions ... The 2kg mass hits the ground at time $t_2$. After time $t_2$, when the experiment is repeated with the 1kg mass, that mass is still $1 \times 10^{-24} m$ above the ground. It takes a further $1 \times 10^{-25} s$ for the $1 kg$ object to hit the ground.

This is the part that you can't just handwave in this scenario. You need to show the OP the actual math, since he is only in 11th grade and probably has not had the opportunity to work such problems in detail. I know he said originally that he wasn't interested in the math, but the fact remains that the math is there and is what we all are relying on in making our posts, and without it we're just asking the OP to take claims on our authority, which he is quite justified in refusing to do, since no one should ever take scientific claims as true just on someone else's authority.

I gave the math for a simpler example in post #52, but the basic ideas I gave there are applicable to this scenario.

 

[PeroK]

I gave the math for a simpler example in post #52, but the basic ideas I gave there are applicable to this scenario.

I missed post #52 entirely in the wave of posts from the OP.

 

[hmmm27]

$F = \frac{Gm_1m_2}{r^2} \Rightarrow$
$a_1+a_2 = \frac{G(m_1m_2)}{r^2m_1} + \frac{G(m_1m_2)}{r^2m_2} = \frac{G(m_1 + m_2)}{r^2}\Rightarrow$
$a \propto m_1 + m_2$

[edit: note that $a$ is the total of the closing accelerations $a_1$ and $a_2$ and that the calculations were performed within the framework necessitated by same, ie: it ain't sloppy math]

 

[PeterDonis]

$F = \frac{Gm_1m_2}{r^2} \Rightarrow$
$a_1+a_2 = \frac{G(m_1m_2)}{r^2m_1} + \frac{G(m_1m_2)}{r^2m_2} = \frac{G(m_1 + m_2)}{r^2}\Rightarrow$
$a \propto m_1 + m_2$

The accelerations will be of opposite signs (assuming we're just looking at radial motion, as we are in this thread, otherwise we have to do full vector addition), so there should be a minus sign in your last equation, not a plus sign.

Also, your $\alpha$ (corrected as above) is the acceleration of one mass in a non-inertial frame in which the other mass is at rest. That makes its physical interpretation more complicated. The case where the motion of both masses is significant is best handled, IMO, in a barycentric frame, as I did (for the case of equal masses) in post #52.

 

[jeff einstein]

I probably have a misconception but my physics teachers say that when an object is dropped at a certain height both objects "pull" on each other with the same "force" meaning that even the very very tiny amount of gravitational force the object has compared to the earth has an effect in the collision. all I am asking is are we ignoring the object's gravity cause it's too small? That is why I initially used larger masses such as mercury and the moon which have their own gravity that is noticeable.

 

[PeterDonis]

my physics teachers say that when an object is dropped at a certain height both objects "pull" on each other with the same "force" meaning that even the very very tiny amount of gravitational force the object has compared to the earth has an effect in the collision

The force on both objects has the same magnitude, which depends on the product of the masses, so yes, the mass of both objects affects the magnitude of the force.

The acceleration of both objects, however, is not the same, because it is the force divided by the mass of the object. If one object is very, very, very much more massive than the other (as with the Earth and an ordinary object like a rock, or you), then that object's acceleration is so miniscule as to not be observable, and can be ignored. But if the masses of the objects are not that different (as with the Earth and Mercury or the Moon), then the acceleration of both has to be taken into account. That is what various posts in this thread, including my post #52, have been trying to explain.

 

[DaveC426913]

objects "pull" on each other with the same "force" meaning that even the very very tiny amount of gravitational force the object has compared to the earth has an effect in the collision. all I am asking is are we ignoring the object's gravity cause it's too small?

Yes.

 

[Cutter Ketch]

OK, maybe I am wrong about the theory of equivalence. what I actually meant was the idea that all masses fall to the earth at the same velocity. and I have been asked about if know the equation relating to this, Yes I do but I clearly mentioned that I do not want equations used for this doubt clearance. after all, I want to clarify one thing, Mercury (larger mass relative to the moon) "would fall to the earth" faster than the moon (smaller mass) both positioned at an equal distance, am I right?

There is a caveat here in that these masses are large enough to make the Earth’s acceleration significant, and the Earth’s acceleration is not the same in the two cases. There is no problem dealing with that, but I think that just clouds your issue (particularly if you don’t like math) and we should stick with comparing bowling balls and feathers so we can ignore the Earth’s response.

To be fair (I almost hate to do this) I think I’ll have to explain. To be more precise, at the moment the two cases are at the same distance from the Earth they both experience the same acceleration (relative to the fixed stars, say). As described in previous posts, both the Force and the inertia are proportional to mass and the mass cancels. So, in that sense, they do not fall at different rates.

However, the Earth is also being pulled and has its own significant acceleration. The Earth’s acceleration is different in the two cases, so Mercury would get to the Earth sooner. So in that sense you could say it falls faster.

Or, to put it another way, in center of mass coordinates the acceleration of the second mass is not independent of the second mass. However, the dependence is $$ \left( \frac {m_1} {m_1 + m_2} \right) ^2 $$ so when m1 is significantly greater than m2 this quickly becomes 1.

 

[DaveC426913]

...the Earth’s acceleration is not the same in the two cases ... I think that just clouds your issue ... and we should stick with comparing bowling balls and feathers so we can ignore the Earth’s response.

On the contrary; it is the very thing the OP is now asking about, if you follow their most recent posts.

 

[jeff einstein]

There is a caveat here in that these masses are large enough to make the Earth’s acceleration significant, and the Earth’s acceleration is not the same in the two cases. There is no problem dealing with that, but I think that just clouds your issue (particularly if you don’t like math) and we should stick with comparing bowling balls and feathers so we can ignore the Earth’s response.

To be fair (I almost hate to do this) I think I’ll have to explain. To be more precise, at the moment the two cases are at the same distance from the Earth they both experience the same acceleration (relative to the fixed stars, say). As described in previous posts, both the Force and the inertia are proportional to mass and the mass cancels. So, in that sense, they do not fall at different rates.

However, the Earth is also being pulled and has its own significant acceleration. The Earth’s acceleration is different in the two cases, so Mercury would get to the Earth sooner. So in that sense you could say it falls faster.

Or, to put it another way, in center of mass coordinates the acceleration of the second mass is not independent of the second mass. However, the dependence is $$ \left( \frac {m_1} {m_1 + m_2} \right) ^2 $$ so when m1 is significantly greater than m2 this quickly becomes 1.

i understand this part really well and you have now further solidified my understanding by confirming it but even if this effect was very very very very microscopic I say that larger masses accelerate either slower or faster compared to smaller masses. I say slower or faster because I have a reason let me post an image just wait

 

[jeff einstein]

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.

 

[jeff einstein]

reference image:

 

[DaveC426913]

Your diagram only shows two cases.

 

[Dale]

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.

They take different times. The variable mass falls with the same acceleration in each case, but the non-variable mass has a different acceleration in each case. If they start with the same separation then the time will necessarily vary with longer times for slower accelerations of the non-variable mass.

 

[jeff einstein]

Your diagram only shows two cases.

sorry my bad I had posted the whole writing but sir @Dale said it was hard to read so i cropped the image and I cropped the wrong

 

[DaveC426913]

...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, you are correct.

You can convert F=ma to a=F/m.

For a given F: If m is small, then a is large. If m is large, then a is small.Looked at another way: Newton's first law says the F is equal and opposite, so
m₁a₁ = m₂a₂

And again, if m₁ is smaller than m₂, then a₁ will have to be larger than a₂