h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

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According to this, Fgrav is not the force of gravity acting only on the apple etc...
It's the force of gravity acting on both masses (earth and apple) at the same time.

Anyway, 2 or no 2, does indeed look like heavy objects fall faster!
But not much faster.

Maybe we do need the 2

?

Still puzzling over this.

(1) I can simplify your proof:

F = 2F - F

therefore ma = (2Gmm/rr) - ma

Then divide by m1. (I've dropped the subscripts for ease of typing, but I hope that's clear.)

(2) Since F = m1.a1 and F = m2.a2, we can freely switch the ma in the above between m1.a1 and m2.a2.

(2) I can produce equally certain maths that say force due to gravity is directly proportional to m1 (F=Gmm/rr) and acceleration due to force is inversely proportional to m1 (a=F/m), ergo these cancel and m1 has no effect on acceleration (a=Gm/rr, where the m left is mass of Earth).

(3) We're not talking about the force when they 'hit' - it's the force attracting the 2 masses when they're not touching.

(4) Your assumption that F (from F=ma) and F (from F=Gmm/rr) are equal is correct; it's one of Newton's laws that 'every action has an equal and opposite reaction'.

Gif

In fact, we can make it even simpler.

F = F

therefore m1.a1 = m2.a2

therefore a1 = (m2.a2)/m1

HOWEVER, a2 also varies in direct proportion to m1, and should thus cancel out.

Yes, I think I've got it there. By the same method, we can calculate that a2 = (m1.a1)/m2. Substituting for a2 eliminates m1 as a term.

Galilleo was right after all!

Gif

The gravitational pull of an apple is neglegable, as is the gravitational pull of a 40 tonne lorry.

So if you drop both on earth they will both fall at the same rate (9.8 meters per second per second), the acceleration due to gravity on earth.
However, very large objects, such as the moon have a significant gravitational pull.

So, if you were to drop the moon off a tall building (unlikely perhaps), the moon would fall towards the earth at 9.8m/s2 like the lorry and the apple. However the difference would be, that this time the earth would also fall towards the moon at 1.6 m/s2.

So the moon would appear to fall slightly faster due to it's huge mass, but in realistic terms, all things on earth fall at the same rate in a vacuum in spite of mass.

Oh, by the way, I didn't read/understand any of the maths above, so I may have totally the wrong end of the stick, in which case

wolfticket

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No so sure that he was.
The problem is that a2 doesn't vary in direct proportion to m1.
If it did we'd have a direct linear equation eg. a2 = m1.G

Wait a minute. Gonna have to think about this.

acceleration = force / mass

Fgrav = m1.m2.G/(d.d)____where G is a constant and d is distance between the masses.

Combine these and we get:
acceleration = m1.m2.G/((m1+m2).d.d)

Hmm.

http://www.astronomynotes.com/gravappl/s6.htm

"The problem is that a2 doesn't vary in direct proportion to m1."

a2.m2 = a1.m1

Therefore

a2 = a1.m1/m2

a2 is directly proportional to m1.

Interestingly, though, this only works if we have a fixed frame of reference. If we're stood on the Earth trying to measure the acceleration of a falling object, we will measure a greater rate because we will be accelerating towards it also. Einstein said there are no fixed frames of reference - everything is relative - and thus (I assume) says that heavy objects do fall faster than light ones.

If anyone actually knows anything about relativity and could confirm that, it would be helpful.

Gif

As I understand it, a heavy object will be pulled towards the earth with a greter force. However a heavier object also requires more force to achieve the same speed as a lighter object. That in essence gives both the same speed.

Yes, but the heavier object will also accelerate the Earth more towards it?

And thus, to an observer standing on the Earth, the heavier object will appear to fall faster?

Gif

Wouldn't the observer, the earth and the two falling objects all be in the same frame?

(Admittedly stretching my knowledge of general relativity)

Yes, the whole universe in 'in the frame'.

But if the observer is measuring things relative to the Earth, the heavier object appears to accelerate more than the lighter one.

If the observer is measuring thing relative to - let us say - the Sun, then there would be no difference in the acceleration of light or heavy objects, but the Earth would appear to accelerate more when displaced by a heavy object than by a light one.

In Newtonian terms, it's clear that heavy objects and light objects fall at equal rates. In relativistic terms, the question is far more complex, and it depends what you are using as your 'immobile point' to measure acceleration compared to.

I think.

Gif

Thanks, Gif, now my brain hurts. I bought Sam Lilley's book, 'Discovering Relativity for Yourself' twenty five years ago and my bookmark is still somewhere in the second chapter.

<<"The problem is that a2 doesn't vary in direct proportion to m1."
a2.m2 = a1.m1
Therefore
a2 = a1.m1/m2
a2 is directly proportional to m1. >>


What I'm talking about is the relationship between a2 and m1.

If you could double the mass of the earth, a2 will not double.
What I mean is that a2 does not increase in direct proportion to the mass of the planet.

Eg. the earth is much more massive than the moon . Its mass is around 600e+22kg compared 7.349e+22 kg for the moon.

The earth is around 80 times more massive than the moon.

However the gravitational force of the earth is only 6 times greater.

There is not a directly proportional relationship between Fgrav and m1.
That's why you can't just say an increase in m1 is balanced out by an increase in a2.

Hi WG,

I think you're confusing a couple of things here.

a2 is proportional to m1, as I showed mathematically. As a2 doubles, so does m1 *assuming all other variables are kept the same*.

However, the Earth is also a different size to the Moon. F is affected not only by m, which is 80 times greater, but also the inverse of the square of the distance between the centres of gravity of two bodies. The Moon is significantly smaller than the Earth, so bodies on its surface are much closer to its centre of mass, thus increasing the force acting on them. Hence gravity on the Moon being a sixth (?) that of the Earth. If you flew a spacecraft in lunarstationary (*) orbit at the radius of the Earth, presumably gravity would be 1/80th of the Earth's.

I haven't done the maths to back this up, but it should work out.

Gif

(*) And how's that for a neologism?

In the equation a1.m1 = a2.m2, what you have to remember is that all the components (apart from m2) are effected by a change in m1. We are dealing with four variables, not constants.

As we increase m1, both a1 AND a2 increase, keeping both sides of the equation equal. If a1 were a constant then it would be a direct linear equation.

If the relationship between m1 (the earth or moon) and a2 (the acceleration of a ball dropped on the earth or moon) were directly proportional, then an earth ball would fall 80 times more "acceleratedly" than a moon ball, rather than just 6 times more "acceleratedly".


(I think it's something to do with it being a geometric, not arithmetic relationship).

Simulpost or disagreement?

Gif