RE: [math-fun] A modest proposal for carbon sequestration in space
According to Newton, the weight near the center of the earth is near zero, because most of the mass is around you in equal & opposite directions. You raise an interesting question, which I think may have been asked in my physics class when I was an undergraduate: If one drilled a Verne-like well all the way to the center of the Earth, what would the air pressure there be (assuming that the insulated walls of the well withstood the pressure of the rock/magma/iron/etc.), assuming that the temperature at the bottom was reasonable -- room temperature. At 03:20 PM 8/1/2006, Cordwell, William R wrote:
However, I've been wondering about what happens when one calculates the pressure near the center of the earth. If one uses a column, one gets a nice, finite pressure, integrating the weight of the rock/magma/stuff upward. However, if one considers that the surface area of the sphere around the center of the earth becomes arbitrarily small, it seems that all of that weight is borne by a tiny area--the pressure should approach infinity. This is, I think, equivalent to using the frustrum instead of the column. So, is there a compensating force (the walls of the frustrum are radial, so there might be a net upward force from the pressure around it?), or is there something wrong with the model of using a frustrum, or is it something else?
Bill
--- Henry Baker <hbaker1@pipeline.com> wrote:
According to Newton, the weight near the center of the earth is near zero, because most of the mass is around you in equal & opposite directions.
You raise an interesting question, which I think may have been asked in my physics class when I was an undergraduate:
If one drilled a Verne-like well all the way to the center of the Earth, what would the air pressure there be (assuming that the insulated walls of the well withstood the pressure of the rock/magma/iron/etc.), assuming that the temperature at the bottom was reasonable -- room temperature.
The air would be mostly liquid, so let's just fill the hole with water. Assuming incompressibility, the pressure is P = rho g h / 2 = (1000 kg/m^3) (10 m/s^2) (6.4*10^6 m) / 2 = 32 GPa = 3,200,000 atm. Taking account of compressibility, there's more water in the hole, so the pressure is higher. The division by 2 accounts for the parabolic shape of the gravitational potential, assuming uniform Earth density. Gene __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
--- Eugene Salamin <gene_salamin@yahoo.com> wrote:
--- Henry Baker <hbaker1@pipeline.com> wrote:
According to Newton, the weight near the center of the earth is near zero, because most of the mass is around you in equal & opposite directions.
You raise an interesting question, which I think may have been asked in my physics class when I was an undergraduate:
If one drilled a Verne-like well all the way to the center of the Earth, what would the air pressure there be (assuming that the insulated walls of the well withstood the pressure of the rock/magma/iron/etc.), assuming that the temperature at the bottom was reasonable -- room temperature.
The air would be mostly liquid, so let's just fill the hole with water. Assuming incompressibility, the pressure is
P = rho g h / 2
= (1000 kg/m^3) (10 m/s^2) (6.4*10^6 m) / 2
= 32 GPa = 3,200,000 atm.
Whoops, that should be 320,000 atm.
Taking account of compressibility, there's more water in the hole, so the pressure is higher.
The division by 2 accounts for the parabolic shape of the gravitational potential, assuming uniform Earth density.
Gene
__________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com
participants (2)
-
Eugene Salamin -
Henry Baker