From: lazycat 9/06/99 21:53:31 Subject: journey to the centre of the earth post id: 17001 Here is a hairy one!!!!! IF you could core a whole through the earth from one side to the other, and then drop a lets say tennis ball down through it where would it stop???? Karl??? From: Chris W (Plebeian) 9/06/99 22:38:11 Subject: re: journey to the centre of the earth post id: 17008 Try reading here. If you still have questions please feel free to ask. From: Martin B 10/06/99 15:39:47 Subject: re: journey to the centre of the earth post id: 17263 Is this the question? So if you assume the sphere has a temperature of 0 degrees throughout, would it necessarily follow (as that rogue seemed to suggest) that the atmosphere inside the hole would therefore be at STP? I can't even get my mind around a "gravity map" in such a sphere, let alone work out how that would affect the characteristics of the atmosphere inside the hole. But I have this hunch that it couldn't be homogeneous. I'll have a bash at it... First the "gravity map" bit. I bet you can get your head around it - if you assume a sphere of constant density, then from the surface to the centre the gravitational field decreases linearly. Isn't that nice? (Yes I know the Earth is not a sphere of constant density, but let me have my little moment here) The second bit: What would the pressure look like? It certainly wouldn't be uniform. The air at the centre of the Earth would have acolumn of air above it pushing down on it - even though there would be no net force operating on these air molecules, there would be substantial atmospheric. As we go up the column of air, this pressure would decrease, much as it does in the atmosphere above the surface. STP is only STP at sea level - once you start going upwards in the atmosphere it's no longer very S. So what exactly would the pressure gradient look like? In a uniform gravitational field (which approximates the situation above the surface) the pressure falls of exponentially with height. (Before some smartypants jumps down my throat, if you are very close to the surface of the Earth and ignore the effects of differing concentration of air molecules with height then the pressure drops off linearly) In this case however, the gravitational field is not constant, but is increasing linearly. This means that it takes more energy to sustain a molecule higher in the column of air, so there will be relatively fewer there - the concentration at the centre will be greater, so the pressure will drop off even more quickly than exponentially ie it will still be exponential in form but with a factor of e^(-r*r) rather than e^(-r). This forum is un-moderated. The views and opinions expressed are those of the individual poster and not the ABC. The ABC reserves the right to remove offensive or inappropriate messages.