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

I had to get through A lEvel Physics without really understanding about frames of reference (and the teacher wasn't much help, because he didn't understand it either).

I finally understand it, thanks to your article!

Thanks!

You're most welcome! Glad I could help.

Jules

I've often wondered what would be the effect of testing this in the presence of a concentrated gravitational source. Imagine parking a little distance from a neutron star, dropped objects would fall in a line toward the star rather than directly 'down', this deviation should be measureable and thus indicate a gravitational mass rather than an inertial mass. If you come back to the solar system and accelerate under free fall to provide the same acceleration and test again dropped objects will fall directly 'down'.

Is this relevant or am I missing a point somewhere?

I think you have missed the point. Or maybe (and this is more likely) I haven't explained it well enough!

Objects will naturally fall towards a gravitational centre, as you pointed out. So, carefully releasing two stationary objects inside a laboratory freely falling towards a planet will result in the two objects appearing to drift towards each other; the further apart the objects and the further the laboratory drops, the more noticable this will be. Doing this in a laboratory nowhere near any gravitational bodies, the objects will stay put.

This effect would be quite clearly measurable (one only has to consider two objects falling towards opposite ends of the Earth - of course they're drifting towards each other!), but this does not alter the fact that objects fall at the same rate. That's all the P of E states: Gravitational Mass = Inertial Mass. There is nothing about direction of fall here, or about not being able to tell the difference between gravity and (say) the effect of booster rockets. I simply used that (oft used) example to illustrate the point.

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MATHS BIT
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I have absolutely no idea how comfortable you are with Physics and algebra, but if you are (and apologies in advance if I have over- or under-simplified), here's the maths of it:

Newton's laws of motion give us the following equation:
F = m a (Force = Mass times Acceleration)

The "m" in this equation refers to "inertial mass" - the resistance to acceleration. If you increase the mass, you either have to put up with less acceleration, or increase the force.

Newton's laws of gravitation give us the following equation for the gravitational force between two objects:
F = G m1 m2 / (d^2)

("G" is the gravitational constant and is always the same)
(Force = Gravitational constant times the mass of object 1 times the mass of object 2 all divided by the square of the distance (centre-to-centre) between them).

This equation works for objects of any size, but, if we're talking about dropping stuff onto the Earth's surface, object 1 can be the Earth, and object 2 can represent our test object(s). We shall replace "m1" with "M" to represent the mass of the Earth, and "d" with "C" to represent the distance from the centre of the Earth.

When we drop the object, the force it experiences will be:
F = G M m2 / (C^2)

This can be simplified further, because "G" is a constant throughout the Universe (as far as we know), and "M" and "C^2" are effectively constants because we're testing this from the same distance above the same planet. We can combine all these into a single constant and call it "g". It just makes this posting a lot easier to read and type, if nothing else!

So now we have:
F = m2 g

But, from Newton's laws of motion, F = m a, so:
m a = m2 g

and "a" is now our acceleration due to gravity.

The argument from here can now split into two possible routes. The easy way out is to say that, using the P of E the inertial mass "m" is equal to the gravitational mass "m2". Therefore:
a = g.

As "g" is a constant, so "a" is a constant, in other words, acceleration due to gravity is a constant independent of the mass of the test object "m2".

The other route (and perhaps more relevant to my article) is as follows. The most simple relationship the two types of mass (gravitational and inertial) can have is a ratio:
m/m2 = R, or
m = R m2

From "m a = m2 g" we now have:
R m2 a = m2 g
R a = g

Now perform the same experiment with a different object. This object will have a different inertial mass (m') and gravitational mass (m2'), but will still be related by the same ratio, R. However, we won't assume the object falls at the same rate, but at a different rate (a'). So, following exactly the same arguments as before but for this object, we get:
F = m2' g
F = m' a'

m' a' = m2' g
R m2' a' = m2' g
R a' = g

But "R a = g" so:
R a' = R a
a' = a

This is saying, that for a simple relationship between inertial and gravitational masses, all objects will still fall at the same rate. This neither confirms nor denies the P of E. For the P of E to be true R must be exactly 1, i.e. "m = m2" and "m' = m2'". The experiment (R.H. Dicke, et al) I mentioned in the article imposes a limit of R to somewhere between 0.99999999999 and 1.00000000001.

It is, of course, possible to imagine more complex relationships between inertial and gravitational mass, but the point is, the P of E is the simplest solution and all the evidence to date supports it.
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END OF MATHS BIT
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You can tell, from the experiment you outlined, whether your acceleration is due to gravity or not, but for each object, however, the differences in acceleration due to gravity and acceleration due to (say) rocket boosters is, according to the P of E, exactly the same. That is to say, 1 Kg of inertial mass will have exactly 1 Kg of gravitational mass and vice versa.

I hope this answers your question.


Jules

Ah, if I can paraphrase you correctly the P of E relates to the effect not the cause. My mistake - sorry

Yes, I think that sums it up. The P of E does simply state the effect, and merely implies a fundamental relationship between gravitational and inertial masses - i.e. it could be a coincidence or some as yet undiscovered physics at work.