A contradiction of the equivalence principle?

[PeroK]

so this is exactly what I am saying Objects with larger mass (the moon in your case) move towards the earth faster than the smaller mass when dropped at the same height.

Of course. Smaller objects (in a vaccuum chamber) fall at the same rate because they are too small to move the Earth any significant distance.

 

[jeff einstein]

so even at a micro level, larger masses fall towards the earth in less time than smaller masses both held at the same distance.

 

[PeroK]

so even at a micro level, larger masses fall towards the earth in less time than smaller masses both held at the same distance.

Not if you drop them at the same time!

 

[jeff einstein]

why not if i drop at the same time.

 

[PeroK]

why not if i drop at the same time.

Because the Earth moves (if at all) in response to their combined mass.

Can you think of at least two other reasons why not?

 

[jeff einstein]

you mean combined mass as binding both objects together?

 

[jeff einstein]

Because the Earth moves (if at all) in response to their combined mass.

Can you think of at least two other reasons why not?

even though why would it matter if the earth moves?

 

[jeff einstein]

cause you do agree that different masses reach the ground at different times when held at same distance. then what does this have to do about dropping them at the same time

 

[PeroK]

you mean combined mass as binding both objects together?

No. I mean even if you considered the floor of a vacuum chamber to be moving upwards towards the dropped objects (by some immesurably small distance), the floor moves towards both objects.

 

[Dale]

now i am getting a clearer view. but why exactly do MErcury and moons gravity have no affect here

They do have an effect. They make the earth accelerate. Their mass affects how fast the earth accelerated towards them, not how fast they accelerate towards the earth

 

[jeff einstein]

They do have an effect. They make the earth accelerate. Their mass affects how fast the earth accelerated towards them, not how fast they accelerate towards the earth

so still they reach each other at different rates dont they?

 

[jeff einstein]

cause i am aware that earth moves ever so slightly when an object is dropped

 

[PeroK]

cause you do agree that different masses reach the ground at different times when held at same distance. then what does this have to do about dropping them at the same time

You need to take a step back and think about what you are asking. Here's one reason: objects have a physical size. The above equation refers to the distance $r$ between the centres of mass. A physically larger object will hit the ground first (if dropped in a vacuum chamber). It's much more about size than mass.

Shape also matters, as the above equation $F = \frac{GMm}{r^2}$ only applies to hypothetical point masses and perfect spheres.

For small objects, size and shape are many orders of magnitude more important than mass.

 

[Dale]

so still they reach each other at different rates dont they?

Yes.

 

[jeff einstein]

No. I mean even if you considered the floor of a vacuum chamber to be moving upwards towards the dropped objects (by some immesurably small distance), the floor moves towards both objects.

ok then lets consider that we drop these objects at different times. so in theory the object with more mass will reach the ground first (this was agreed by you earlier). so doesn't contradict the idea that all masses reach the earth's surface at the same time when held at the same distance. how does this make sense Either the theory is wrong (which it probably isn't) or I am wrong which I probably am and I want some one to disprove me. I adress this to sir @Dale as well.

 

[jeff einstein]

You need to take a step back and think about what you are asking. Here's one reason: objects have a physical size. The above equation refers to the distance $r$ between the centres of mass. A physically larger object will hit the ground first (if dropped in a vacuum chamber). It's much more about size than mass.

Shape also matters, as the above equation $F = \frac{GMm}{r^2}$ only applies to hypothetical point masses and perfect spheres.

For small objects, size and shape are many orders of magnitude more important than mass.

OK consider that both objects have the same size but different masses. then would they both reach the earth's surface at the same time?

 

[PeterDonis]

If you dropped an object of mass 1kg from the same distance as the Moon, then it would take a certain time to fall to half that distance.

If the Moon were released from rest at the same distance from Earth, then:

a) It would move half that distance in less time than the 1kg object.
b) The Earth would also have moved significantly, so the overall distance between the Moon and the Earth would be less than half the original distance at that time.

This way of describing it might be confusing since there are multiple "distances" being mixed together.

Let's try a simpler scenario: two equal masses $m$ starting from rest separated by a distance $r$, using a barycentric coordinate system, so the origin of coordinates is halfway between the masses. This spatial origin is also where the masses will collide with each other, ending the experiment. The question is, what time will it take for them to collide?

Since the situation is symmetric, we can just consider one mass. Its acceleration, from previously given equations, is $G m / r^2$. Notice that this acceleration depends on the mass $m$ (since both masses are identical and the mass appears twice in the gravitational force formula). So the acceleration of each mass towards the barycenter will be larger the larger the mass $m$ is. That means the time to collision, which is the time for each mass to move half the distance between them, will be shorter the larger the mass $m$ is. This last statement is independent of our choice of coordinates.

If we look at this in coordinates centered on the other mass, the coordinate acceleration at the start does not change, it's still $G m / r^2$; but now there is a correction that has to be added because the origin of these new coordinates is itself accelerating towards the barycenter. In other words, these coordinates are not inertial, unlike the barycentric ones which are. So the starting coordinate acceleration alone does not determine the coordinate trajectory in this frame, as it does in the barycentric frame.

The qualitative logic above does not change when the two masses are unequal, as for the Earth and the Moon vs. the Earth and a 1 kg mass starting at the same distance as the Moon, but of course the numerical details will be different.

 

[PeroK]

ok then lets consider that we drop these objects at different times. so in theory the object with more mass will reach the ground first (this was agreed by you earlier).

I didn't agree that. I said the Moon would fall faster than a $1kg$ mass.

so doesn't contradict the idea that all masses reach the earth's surface at the same time when held at the same distance. how does this make sense

Every statement has a set of hypotheses under which it is valid. It does not apply if the hypotheses are not met. This often doesn't matter. But if you dig deep enough, the hypotheses on which the statement is based fail.

Either the theory is wrong

It's not wrong. It has what's called a domain of applicability. Also, almost all theories are approximations in some respect. For example, the Earth is not a perfect sphere, it's spinning and its s mass is not constant over time.

I'm trying to get you to think that the theory is approximate in ways that are more significant than the immeasurable motion of the Earth. Like considering the size and shape of objects and not just their mass.

 

[jeff einstein]

how about when i talk just about mass

 

[jeff einstein]

I didn't agree that. I said the Moon would fall faster than a $1kg$ mass.

But isn't that still two different masses Here we can see a large difference between this action but for masses with less weight such as 1.5kg and 1kg this should also be true
Shouldn't it?

 

[PeroK]

how about when i talk just about mass

I think we have exhausted this topic.

 

[jeff einstein]

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

 

[PeroK]

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

That's your decision.

 

[Dale]

so doesn't contradict the idea that all masses reach the earth's surface at the same time when held at the same distance

I have never seen that claim. All masses accelerate at the same rate when at the same distance from earth. That is a different claim than the claim that they reach the ground at the same time.

I think you are just misunderstanding the claim.

 

[jeff einstein]

so maybe I understood it wrong but in reality, different masses dropped at a certain height do reach the ground at different times?

 

[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.