What you have posted, with out attribution, here is Baumgardner's runaway plate tectonics theory. I can assure you, as a geologist and geophysicist, that it is entirely misguided and that if things happened as he suggests the whole world would have been turned into a molten ball.
Here is a nice simple explanation of why it is complete rubbish:
http://home.entouch.net/dmd/subduction.htm
I'm not a geologist, but even I was intrigued at the idea that the same slab that superheated the mantle just below the crust was still "cool ocean floor" by the time it had descended 2800 km to the outer core.
Then I started to think about buoyancy. Let's imagine the force required to force a rectangular slab of crust 1000 km wide and 30 km thick down to the outer core. I found a
What you have posted, with out attribution, here is Baumgardner's runaway plate tectonics theory. I can assure you, as a geologist and geophysicist, that it is entirely misguided and that if things happened as he suggests the whole world would have been turned into a molten ball."]USGS site that give approximate densities and thicknesses for the crust, mantle, etc. Let's try a back-of the envelope calculation.
Let's assume the density profile of the mantle is more or less linear with distance from the earth's center. Using the top and bottom of the mantle (at 6.371e6 and 3.48e6 meters from the center, respectively) as reference points, we end up with the linear relation
rho(r) = (-7.61e-4)R + 8.248e3
now we have to contend with gravity. Since a circular shell exerts no gravitational pull on the objects inside it, gravity at radius (r) will depend on the total mass of everything from radius r down to the center:
m(r) = 4*pi* Integ(R = 3.48e6 to r)[rho(R)*R^2dR] + K
Where K is the mass of the core (r < 3.48e6 meters). Doing the integration with r=6.371e6 meters and plugging in the total mass of earth gives us K = 9.79e23 kg.
So now we can complete the integration...
m(r) = (-2.39e-3)*r^4 + (3.455e4)*r^3 + 9.79e23
So now we can figure out the strength of gravity at radius r.
g(r) = G*m(r)*r^-2 = (-1.59e-2)r^2 +(2.31e5)*r+(6.53e24)*r^-2
Since the buyant force equals the weight of the displaced material, the total upwards force per unit volume of the slab is equal to:
dF(r)/d(v) = (rho(r)-rhocrust)*g(r)
let's be charitable and assume an unusually dense piece of crust, equal in density to the upper mantle at 3.4e3 kg/m^3.
dF(r)/d(v) = (rho(r) - 3.4e3)*g(r)
= (1.21e-4)*r^3-(2.53e2)*r^2+(1.12e9)*r-(4.97e21)*r^-1+3.17e29*r^-2
So if we imagine a 30km x 1000km cross-section slab of crust stretching all the way to the outer core, the total force is the integral
Integ(R = 3.48e6 to 6.371e6)[F(r)dv]
=3e10*Integ(R=3.48e6 to 6.371)[F(r)dr]
I'll spare you the huge polynomial, but I ended up with:
F=3.6e33 N
So what does that mean? Well, if the slab is 1000km * 30 km * (6371-3480) km, with a density of 3.4e3 kg/m^3, it would have a mass of
2.95e20 kg
Which would give us an upward acceleration of
1.22e13 m*s^-2
Wow.