So I was in the shower today...

[SJ Zero]

Let's start with the basic gravitational formula.

Code:

F = G(m1m2)/R^2

Now let's set m1 and m2 to 1 for ease of use, and put aside G for now. This will make more sense later.

Code:

F = 1/R^2

F = R^(-2)

Now, we're going to integrate this between the two far reaches of the circle, -R and R, which, if we use b(the radius location of the particle we're looking at) as a reference point, become:

Code:

R + b
R - b

Now we can integrate.

SR^(-2) = 1/-R |(R + b to R-b)
Now we do the limits.

1/-(r + b)  -  1/-(r - b)

1/-(r + b) + 1/(r-b)

-(R + b) + (R - b)  /  -(R^2 + b^2)

-R - b + R - B  / -(R^2 + b^2)

-2b / -(R^2 + b^2)

2b/(r^2 + b^2)

And we now have an equation to determine the gravity in any point in a sphere. I'm using a sphere instead of an oblate spheroid because it's not neccessary to prove what I'm trying to show. I'm using a circle in this case, because the added complexity of a full sphere isn't needed either. To add the extra dimension would simply be an added integration at the end of this, and is largely unneccessary to show what I'm trying to show.



The above is the plot given by this equation. I have a good clue as to exactly how the drop-off occurs. The zero crossing represents the switch in direction of the force. Note, however, that the magnitude remains relatively significant for the majority of the planets radius.

The next part involves a fairly heavy calculus equation. I'd rather not do it that way because I'll be here all night, so I'm going to give R a value of 100, and use a program to manually sum the forces exerted upon the core by the 'spike' I've chosen.

Code:

defdbl a-z
screen 18
R = 100
print "Force and pressure characteristics"
for a = -100 to 100
    x = 5000 * ((2*a)/(r^2 + a^2))
    y += (x / 5000)
    pset (200 + a,100 + x)
    pset (200 + a,100 + 25 * y), 1
    pset (200 + a, 100)
    pset (200, 100 + a)
next a

for a = -100 to 0
    x = 5000 * ((2*a)/(r^2 + a^2))
    y -= (abs(x) / 5000)

next a
for a = 100 to 0 step -1
    x = 5000 * ((2*a)/(r^2 + a^2))
    y -= (abs(x) / 5000)
next a
    circle (200 + a,100 + 25 * y), 4,2

sleep

This code will show you something like this:



This graph is relevant because it shows that even taking into account the altered gravity as you head into the planet, the pressure build-up due to gravity is still substantial.

EDIT: there's an error in my equations, I'll adjust tomorrow when it's not 230am. Basically, I need to remember that mass is variable with radius as well. I hate very much to admit it, but it's possible that with this alteration we'll see results closer to what agamamnus predicts. We'll see then.