Same gravitational acceleration of unequal masses

[TurtleMeister]

Yes, if we use a reference frame attached to the surface of the earth: http://en.wikipedia.org/wiki/Reduced_mass

Yes, that's what I meant. Sorry, I should have included "as viewed by an Earth bound observer".

 

[atyy]

Yes, that's what I meant. Sorry, I should have included "as viewed by an Earth bound observer".

BTW, the wikipedia reduced mass article has the two minus signs you missed in the post which DH said the errors canceled out.

 

[TurtleMeister]

BTW, the wikipedia reduced mass article has the two minus signs you missed in the post which DH said the errors canceled out.

Yes, I noticed that. a - (-a) = a + a.

 

[Canticle]

So does the same apply to all objects in a gravitational field.
Light passing a massive object like our Sun gets deflected or bent but a large object side by side with the light beam would probably fall into the Sun.
Same would happen to a car or go-kart rushing past a planet though the light would probably not be so nowticebly bent and would not fall towards it.
If the differences balanced out why does it not apply to very light or massless objects.

Yes, the same applies to everything. But you don't quite have it yet- Still in plain english;

If the 'large object' was going the same speed as the light beam it would certainly NOT fall into the sun! (momentum / intertia) It's track would bend the same.

Gravity is curved space time, so curves everything, massive or not, including all zero mass disturbances from gamma rays to radio waves.

The mass of a body defines the 'slope inclination' of space time. And remember, when you drop metal and balsa wood balls out of your window, although they accelerate the same the metal ball attracts the Earth more. Too little to notice of course, but make it the size of the moon then you'd notice!

Can you get your head round that lot?

 

[Buckleymanor]

If the 'large object' was going the same speed as the light beam it would certainly NOT fall into the sun! (momentum / intertia) It's track would bend the same.

Would not that require rather a large amount of energy.

And remember, when you drop metal and balsa wood balls out of your window, although they accelerate the same the metal ball attracts the Earth more.

So the metal ball would hit the Earth first.

 

[TurtleMeister]

And remember, when you drop metal and balsa wood balls out of your window, although they accelerate the same the metal ball attracts the Earth more.

So the metal ball would hit the Earth first.

Yeah, that's pretty much been the theme of this thread. The metal ball would strike first but only by an extremely small undetectable amount. The difference in the accelerations would be a = G(M1-M2)/r² where a = difference in acceleration, G = gravitational constant, M1 = metal ball, M2 = wood ball, and r = the distance between M1, M2 and the center Earth. Notice that the mass of the Earth has noting to do with it, which was my earlier misconception.

 

[sylas]

You guys don't actually mean it would hit "first". That would imply being dropped at the same time, in which case they'd hit at the same time (ignoring friction etc)

What you really mean is that if the experiment was repeated on two occasions, then the metal ball would hit the ground in slightly less time.

The difference is all to do with the acceleration of the ground towards the balls -- and you can't use that to have one ball hitting "first".

 

[TurtleMeister]

You guys don't actually mean it would hit "first". That would imply being dropped at the same time, in which case they'd hit at the same time (ignoring friction etc)

What you really mean is that if the experiment was repeated on two occasions, then the metal ball would hit the ground in slightly less time.

The difference is all to do with the acceleration of the ground towards the balls -- and you can't use that to have one ball hitting "first".

I've been thinking about that myself. The only reason I haven't brought it up is that I wanted to come to a conclusion first. But since you've already brought it up I'll say that my intuition tells me that the metal ball would still strike first even if they were dropped at the same time. But hey, my intuition has been known to be wrong. :) Dropping both at the same time complicates things a bit.

 

[Buckleymanor]

The difference is all to do with the acceleration of the ground towards the balls -- and you can't use that to have one ball hitting "first".

Why not,
Both balls "perfectly" round Earth "perfectly" round.
Earth's round edge would touch metal ball first because the edge would be pulled to it via the centre of gravity.

 

[Buckleymanor]

Dropping both at the same time complicates things a bit.

Certainly does.
The metal ball would allso accelerate towards the wood ball and vice versa but the Earth would accelerate more towards the metal ball.

 

[worldrimroamr]

Why is everyone making such a complicate deal out of something that can be expressed so straightforwardly? Gravitational force is exactly proportional to mass. But inertial resistance to motion is also exactly proportional to mass. (That's Einstein's equivalence principle -- the founding insight of general relativity.) So for twice the mass, you have twice the force and twice the inertia. They exactly cancel each other out. Thus, all objects fall at equal rates in vacuum.

 

[TurtleMeister]

Yes, worldrimroamr, you are correct. But what we are talking about is the way it would be viewed by an Earth bound observer, who is in a non-inertial reference frame. I hope I get it right this time. :)

A1 = G * M2 / r²
acceleration of M1 toward M2

A2 = G * M1 / r²
acceleration of M2 toward M1

A = A1 - A2 = G * (M1 + M2) / r²
relative acceleration between M1 and M2

If you were located on M1 and you measured the acceleration of M2 toward you, then A is what you would measure, not A2. And if you were located on M2 and you measured the acceleration of M1 toward you, then A is also what you would measure, not A1. So the "relative" acceleration of an object in Earth free-fall is not independent of the objects mass "as measured by an Earth bound observer".

 

[D H]

An Earth bound observer is not in an inertial frame, no matter how you cut it.

• Newtonian mechanics. The Earth is accelerating toward the object. An accelerating frame is not inertial in Newtonian mechanics.
• General relativity. A stream of falling apples accelerates with respect to the observer. A frame with a stream of falling apples is not inertial in general relativity.

 

[TurtleMeister]

Thanks DH, I changed it.

 

[Canticle]

Wrong. Any acceleration of the Earth reduces it's pull on the ball. Thought experiment needed;
Take a binary system of 2 planets, one 20% more mass than the other. Your'e standing on the big one. In your reference frame the small one is falling towards you at velocity v.
Now jet over and land on the small one. Standing there in its reference frame the big one is falling at exactly the same velocity v.

No matter what the 'share' of mass between them the total acceleration is the same, (only the distance changes that - 2nd law).

So both balls would hit at the same time whatever. That's equivalence!

Except for Godels incompleteness theorem etc. There are loads of other tiny factors, there IS an ether so relative atmospheric drag comes into it, your spaceship taking off and landing, the relative pull of the sun whichever its closest to, etc. Thats why Mr Planck invented Plancks, and we have Quantas, supposedly if it's smaller than that it's undetectable (thanks Turtle) and we can ignore it. (though one day we'll find we need it!) Not using Quantas is fine, BA fly to Aus anyway.

 

[sylas]

Wrong. Any acceleration of the Earth reduces it's pull on the ball. Thought experiment needed;
Take a binary system of 2 planets, one 20% more mass than the other. Your'e standing on the big one. In your reference frame the small one is falling towards you at velocity v.
Now jet over and land on the small one. Standing there in its reference frame the big one is falling at exactly the same velocity v.

Good point. If you repeat the experiment with different balls, you are effectively using the other ball as part of the mass of the Earth!

 

[Buckleymanor]

Wrong. Any acceleration of the Earth reduces it's pull on the ball. Thought experiment needed;
Take a binary system of 2 planets, one 20% more mass than the other. Your'e standing on the big one. In your reference frame the small one is falling towards you at velocity v.
Now jet over and land on the small one. Standing there in its reference frame the big one is falling at exactly the same velocity v.

No matter what the 'share' of mass between them the total acceleration is the same,

The total acceleration is the same but how can you recognise if the big one or the small one falls or moves towards you at all.

 

[TurtleMeister]

Canticle, your thought experiment is correct but incomplete in so far as what we're talking about. In order to make it complete you should identify your reference frame as non-inertial and then repeat the experiment using an inertial reference frame.

Edit1
I noticed something a little different in your thought experiment. You said: "No matter what the 'share' of mass between them the total acceleration is the same". Thus far we have not talked about redistribution of mass between the two objects. In this case the relative acceleration A = G * (M1 + M2) / r² in the non-inertial frame will always be the same, as you stated. But the individual accelerations of the two objects in the inertial frame will change.

If you repeat the experiment with different balls, you are effectively using the other ball as part of the mass of the Earth!

That is correct. It is the combined mass that determines the relative acceleration of the two bodies toward each other. And that's the reason for (M1+M2) in the equation A = G(M1+M2)/r² for the non-inertial frame. So regardless of which object (M1 or M2) you are observing from, the mass of the other object WILL affect the observed acceleration. However, if you are an observer in an inertial reference frame you will see M1 and M2 accelerating at different rates. M1 will accelerate at A1 = G * M2 / r² and M2 will accelerate at A2 = G * M1 / r². It's really a very simple thing. It's just that we're describing the same thing from two different points of view. If you are located on M1 or M2 then you are accelerating, and so your reference frame is non-inertial.

In short, for the acceleration of an object(M2) who's mass is variable, in Earth(M1) free-fall:

A = G * M1 / r²
This is the inertial reference frame. The mass of the falling object(M2) does NOT affect the acceleration because of the equivalence principle.

A = G * (M1 + M2) / r²
This is the non-inertial reference frame. The mass of the falling object(M2) DOES affect the relative acceleration because we are located on Earth(M1) which would include our own acceleration toward the object. This is not to say that the equivalence principle is being violated because the equivalence principle is only valid in an inertial reference frame.