[graham.d]

Alan, the key point to note is that the field INSIDE a spherical shell of uniform density is everywhere equal to zero. This is not very intuitive but results from the field being an inverse square law. This is true for an electric field from a spherical shell of uniform charge too. It reqires a page or so of calculus to show this. If you accept this then you can see that it would also apply to a concentric set of spherical shells which can represent all the mass above you as you tunnel down. The gravitational effect of all these concentric shells can be ignored leaving just the mass of the sphere beneath you having any effect.

Looking on the web came up with a nunber of sites that show this calculation. Here is one that may explain:

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/GravField.htm

[Alan McDougall (OP)]

I see so weight decreases at a constant rate down to zero at the core.

Many thanks guys!!

What if your drilled a hole right through the earth and let a peace of lead fall, would it oscillate betwen the surface points at oppsite sides of the earths circumference (assuming there was a vacuum)

[lyner]

Alan McDougall

I also do not get your point balancing out as zero, this would only happen at the middle of the earths core.

Inside a hollow spherical shell, wherever you are, the gravitational potential is zero; well known and easy to show the sums. Inside a solid sphere, the bits outside you can be looked at as a set of shells so still zero.
The gravitational force is proportional to the distance from the centre - also very easy to show. This means that a body would oscillate with simple harmonic motion if it were dropped down a hole through the centre. (Only a thought experiment) The time period would be about 90minutes. This also happens to be the time for a low Earth orbit.
As far as the barycentre is concerned, measuring your weight there would give a constant value; at any other point there would be 'tidal' forces due to the Moon and Earth's mutual orbit.

[turnipsock]

The gravity at the surface will be 1g and it will be 0g at the centre.

All you need now is somebody that can remember their calculus and you'll have your answer.

[lyner]

No calculus needed, really.
The force is proportional to the mass of the sphere beneath your feet (which is 4πρr³/3)
ρ is density - we can assume it's fairly uniform.
It's also proportional to 1/r² because of the inverse square law.
Between the two that makes it proportional to r. QED.

[Andrew K Fletcher]

If the gravitational force was 0 gravity at the core and 1 g at the surface, there would be no reason for the denser materials to be attracted towards the core.

[graham.d]

Sophie, the reason for the calculus is only that it is not obvious that the field is zero everywhere inside a spherical shell. Knowing this means you don't need the calculus, but if you need to prove this, then I think you do need calculus.

[Alan McDougall (OP)]

sophiec,

No calculus needed, really.
The force is proportional to the mass of the sphere beneath your feet (which is 4πρr3/3)
ρ is density - we can assume it's fairly uniform.
It's also proportional to 1/r2 because of the inverse square law.
Between the two that makes it proportional to r. QED

.

You answer is the the best and easy to comprehend , but the others also valid

Alan

[lyner]

Sophie, the reason for the calculus is only that it is not obvious that the field is zero everywhere inside a spherical shell. Knowing this means you don't need the calculus, but if you need to prove this, then I think you do need calculus.

I agree.
There is a fairly good 'arm waving' argument. Rather than calculating the force on you (sitting in the middle, somewhere), you think of the force on the surface of the shell (equal and opposite - so it's valid) and equate the gravitational, inverse square law, to the effect of ISL on illumination from a light source.  Two cones with the same solid angle, pointing in opposite directions from you, will ' illuminate' the shell with an equal amount of energy. The light is spread out on the distant bits and less spread out in the nearer bits. The total light flux hitting each end of the 'double cone' will be the same, though, and, by analogy, the total gravitational forces on the two areas formed by the two cones of light will also be the equal and opposite. Forces on the shell and, therefore, also on you, will be balanced in any direction. It's easiest to visualise if you are on a diameter but it works for any position - you don't need to consider the actual shape which the light beams form on the  inside of the shell.
(I made this up on my own and I'm quite pleased with it!)

[graham.d]

Oh yes, Sophie. That's very good and why it works is because a spreading flux and the inverse square law are effectively the same in Euclidean space; hmm, may be the same in other spaces but I am not even going to think about that one. I wonder how it would do as a physics undergrad answer?

[Bored chemist]

If, as has been shown, the force is proportional to the radius then an object dropped into this "idealised" hole (no moon, full of vacuum- the earth being solid with equal density all the way down) would exhibit simple harmonic motion.
The maths for that is well known so I will leave it as an exercise for the reader.

And, Sophie, That's a seriously good proof.

[lyner]

Lovely when you don't have to do anything other than wave your arms about.
I gave it some further thought and , first of all, it seemed that the argument would apply for any shape of shell. Then I realised that the shell has to be spherical if the density is uniform - for
 F = Gm₁m₂/r²
to apply.

I have written this elsewhere (and there was a TNS Podcast, too) but the period would be the same (90 minutes) for small, spherical asteroid and a pea sized object moving through a hole through the centre as long as the average density was the same as the Earth's.  What an experiment to try, one day.

[Alan McDougall (OP)]

Go to my other thread on gravity.It relates to a hollow shell like earth etc