[Macarios]

I've honestly no idea how this got onto black holes. My physics is a bit rusty but I think this is correct. The formula for gravitational attraction is:

f = G (M1 x M2) / r2

G being a constant, M1 and M2 being the masses of the two objects. r being the distance between the two objects' centre of gravitys
So if M1 is my mass, M2 is the earth's mass then the above gives you the force the earth exerts on me (which is my weight, that's what weight is).

But we also know that

f = ma

F = force, m = mass, a = acceleration.

This can be arranged as

a = f/m

This, by the way, is what makes super sonic travel so expensive, the more "m" there is, the more "f" you have to provide to produce "a".

So the acceleration on me because of gravity is the force of gravity on me divided by my mass. Using the above two formulas that is:

(G (M1 x M2) / r2) / M1.

The two M1s cancel themselves out so it's:

(G M2 / r2)

Point being, this force is independent of MY mass, it only relies on the mass of the earth which pretty much remains constant and my distance from the earth's centre of gravity - which does vary slightly because the earth is not a perfect sphere. But the headline is that objects of different mass will fall at the same rate.

That is what Galileo proved by dropping cannonballs of different sizes out of high buildings and observing that they hit the ground at the same time. The reason this doesn't work with feathers and hammers on earth is air resistance which slows the feather's fall (and the hammer's, but not enough so's you'd notice because of the mass of the hammer). On the moon there is no atmosphere and so no air resistance so they fall at the same rate, hence the astronaut's exclamation "what do you know, Mr Galileo was right!"

The effect was recreated in a vaccuum chamber for a BBC series

https://www.youtube.com/watch?v=frZ9dN_ATew

That explanation is ignoring that under the theory of gravity the falling object is also pulling the earth towards it. "Mass doesn't matter" is clearly wrong.

I would agree that it is wrong, and that mass matter.
Faling objects get pulled all to center of mass of the whole system.

But if one object is for several orders of magnitude more massive than all other objects, then the pull by those smaller objects can be neglected.
Center of mass of the whole system will be very close to center of mass of the massive object, and apparently they will all just fall to the massive object.

If its mass is great enough, then the pull will be below our measuring abilities, and acceleration of small objects won't depend on their masses.
Weight will depend, but it is another story.

[douglips]

Two black holes colliding would happen faster than a feather falling into a black hole, because in the case of two black holes, BOTH black holes are accelerating towards each other. In the case of the feather, the black hole's acceleration towards the feather is negligible.

The "different mass objects fall at the same rate" only refers to the acceleration of the single object under test. While a very very massive object would accelerate at the same rate, it would also cause the earth (or other surface) to accelerate towards it, making the apparent acceleration larger. When one object is several orders of magnitude more massive than the other, the more massive object doesn't accelerate much compared to the lighter one.

Great. So you admit that greater masses would appear to fall faster and that the OP is correct.

"Greater" is relative. Do you know a way to drop an object that has a mass near that of the earth? Anything with a mass 6 orders of magnitude smaller than the earth is going to fall at the same apparent rate, because the difference according to newton would be so tiny as to be unmeasurable.

So no, they won't "appear" to fall faster unless you have an oil tanker full of neutronium or something.

[JohnAdams1145]

Haha douglips. You speak as if Tom were talking about something 6 orders magnitude less than the mass of the Earth; he was probably thinking of a metric ton vs a feather. Anything 4 magnitudes smaller than the mass of the Earth will have a negligible, but measurable effect (with precision instruments; this is incidentally about the same size as ratios involved in the curvature of the Earth).

I should help some FE believers with the mass of the Earth (of course, their hypothesis could also deny the mass of the Earth! RE should start thinking about the rebuttal to that... so many ad hoc hypotheses...) -- it's on the order of 10^24 kg. A person is around 100 kg. A feather is about 1 gram. The formula for the magnitude of closing acceleration (defined as the time derivative of the closing velocity) is G(m_1+m_2)/d^2. Let's try this calculation in a vacuum at the Earth's surface for a feather, a person, and the entire Moon compressed into a chicken nugget.

Feather:
G(5.972 x 10^24 kg + 1 g) / (6371 km)^2 = 9.79811147 m / s^2

Person:
G(5.972 × 10^24 kg + 100 kg) / (6371 km)^2 = 9.79811147 m / s^2

USS Nimitz (the aircraft carrier):
G(5.972 × 10^24 kg + 10^8 kg) / (6371 km)^2 = 9.79811147 m / s^2

The Moon in a nugget:
G(5.972 × 10^24 kg + 7.346 × 10^22 kg) / (6371 km)^2 = 9.94043866 m / s^2

So, mass doesn't matter. It really just doesn't. Even the Moon nugget would hit the ground at about the same time as a feather. Also, you can see that the precision of my calculator isn't enough to even register a difference between the USS Nimitz and a feather.