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**Flat Earth Theory / Re: NASA claim on mass doesn't matter in a vacuum.**

« **on:**January 27, 2018, 08:39:19 PM »

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

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