What is the Correct Calculation for the Pressure at the Center of a Planet?

[Bill Wells]

Posted this on the homework page, got no replyies so thought I'd try here. It's no longer "homework", already sent in my work!

I have seen several references which give the pressure at the center of the Earth at about 3 x 10^11 Pa. (I've seen 3.0 - 3.6). An earlier post on this forum gives the equation for the pressure at the center of a planet as
2*(pi*G/3)*(R^2)*(rho)^2. When I crunch the numbers, I get 1.73 x 10^11, which is about half what most references give. I'm using rho = 5515 kg/m^3.
Where am I going wrong.
The above assumes constant density (rho). If you like the above problem, is there an equation which would give central pressure if the density increases with depth?

Thanks Much
Bill

 

[Chronos]

Integrate the gravitational constant given the mass of the earth.

 

[SpaceTiger]

I have seen several references which give the pressure at the center of the Earth at about 3 x 10^11 Pa. (I've seen 3.0 - 3.6). An earlier post on this forum gives the equation for the pressure at the center of a planet as
2*(pi*G/3)*(R^2)*(rho)^2. When I crunch the numbers, I get 1.73 x 10^11, which is about half what most references give. I'm using rho = 5515 kg/m^3.
Where am I going wrong.
The above assumes constant density (rho).

You're surprised that assuming a constant density gives you an answer that's off by a factor of 2? There is no simple formula to calculate the central pressure of the earth, but you can probably get it to a good approximation with an n \simeq 0.5 polytrope. If you don't know what polytropes are, I suggest looking it up.

 

[xArcturus]

Bill, here is an equation which will work for almost all celestial bodies:

Pressure = (Mass^2 * G) / (Volume * Radius)

Earth Mass is: 5.98E10^24 [kg]
Earth Radius is: 6.37E10^6 [m]
Newton's Constant G is: 6.67E10^-11 [m^3/kg*s^2]

V = (4 * pi * R^3) / 3 = 1.08E10^21 [m^3]

M^2 * G = (5.98E10^24 [kg])^2 * 6.67E10^-11 [m^3/kg*s^2) = 2.39E10^39 [kg*m^3/s^2]
V * R = 1.08E10^21 [m^3] * 6.37E10^6 [m] = 6.90E10^27 [m^4]

Earth Pressure = 2.39E10^39 [kg*m^3/s^2] / 6.90E10^27 [m^4] = 3.47E10^11 [kg/m*s^2]
which is: 347 giga Pascals of pressure at the center of the earth!

 

[qraal]

Hi xArcturus

Where did you get that one from? It's quite a good approximation, though more by chance than design. In terms of mass and radius here's the uniform density central pressure equation...

P= (3/8π)*(GM²/R⁴)

...so you can see the likeness.

Bill, here is an equation which will work for almost all celestial bodies:

Pressure = (Mass^2 * G) / (Volume * Radius)

Earth Mass is: 5.98E10^24 [kg]
Earth Radius is: 6.37E10^6 [m]
Newton's Constant G is: 6.67E10^-11 [m^3/kg*s^2]

V = (4 * pi * R^3) / 3 = 1.08E10^21 [m^3]

M^2 * G = (5.98E10^24 [kg])^2 * 6.67E10^-11 [m^3/kg*s^2) = 2.39E10^39 [kg*m^3/s^2]
V * R = 1.08E10^21 [m^3] * 6.37E10^6 [m] = 6.90E10^27 [m^4]

Earth Pressure = 2.39E10^39 [kg*m^3/s^2] / 6.90E10^27 [m^4] = 3.47E10^11 [kg/m*s^2]
which is: 347 giga Pascals of pressure at the center of the earth!

 

[travboat]

Since this forum has been so helpful to me, I'd like to give back! Here's my first post. Thanks for all the help, everyone.

@ qraal

Pascal's Principle states that P = ρgz where ρ is density, g is gravitational acceleration and z is depth.

If ρ=mass/volume, g=(G*mass)/(radius^2) then
P = [(G*mass^2)*z]/(volume*radius^2)

Then our equation transforms:
P = [(Gm^2)z]/(vr^2)
= (Gm^2r)/(vr^2)
= (G*mass^2)/(volume*radius)

Using the constants...
G = 6.674(10^-11) m^3*kg^-1*s^-2
m(earth) = 5.974(10^24) kg
v(earth) = 1.083(10^21) m^3
r(earth) = 6.371(10^6) m
z = depth = r, if we're thinking about the center.

P = 347(10^9) Pa at the center.

And there it is :)

 

[snorkack]

Since this forum has been so helpful to me, I'd like to give back! Here's my first post. Thanks for all the help, everyone.

@ qraal

Pascal's Principle states that P = ρgz where ρ is density, g is gravitational acceleration and z is depth.

If ρ=mass/volume, g=(G*mass)/(radius^2) then
P = [(G*mass^2)*z]/(volume*radius^2)

Then our equation transforms:
P = [(Gm^2)z]/(vr^2)
= (Gm^2r)/(vr^2)
= (G*mass^2)/(volume*radius)

Using the constants...
G = 6.674(10^-11) m^3*kg^-1*s^-2
m(earth) = 5.974(10^24) kg
v(earth) = 1.083(10^21) m^3
r(earth) = 6.371(10^6) m
z = depth = r, if we're thinking about the center.

P = 347(10^9) Pa at the center.

And there it is :)

Yes - but what you forget is integrating the g over z.

In the interior of planet, only part of the interior is gravitating - so unless the mass is concentrated enough in center, g decreases.

If the density is uniform through interior then g is simply proportional to r (distance from centre). Then the central pressure P=ρgR/2. Which is never exactly true, but there are 2 components of mass concentration. Compositional segregation and pressure compression. If both are negligible, e. g. in Moon, then the pressure should be close to what was stated above.

Also note that g itself is proportional to ρr: V=4/3πrˇ3; m=Vρ=4/3πρrˇ3; g=Gm/rˇ2=(4/3πG)ρr.
It follows that the central pressure is proportional only to the square of surface gravitational acceleration and independent of radius and density (because these cancel out). Earth and Saturn have the same core pressure - provided that the distribution of internal density were the same.

 

[mfb]

Bill, here is an equation which will work for almost all celestial bodies:

Pressure = (Mass^2 * G) / (Volume * Radius)

This has to be an effective formula which assumes something about the composition of the object. Consider earth: If you use the radius of the solid masses, you get a good approximation. But Earth also has another low-density layer outside: The atmosphere. It increases the radius (~1/60) and the volume (~1/20), but its mass and influence on pressure is negligible.
While the formula might be good with constant density (I think it is the same as the formula for that case), it cannot describe objects where the density varies a lot as function of radius.