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As mentioned earlier, the escape velocity can not go to zero at the center of the planet. It has to always decrease with distance. Here is a free-hand drawing of what I would expect to see. [ Invalid Attachment ] The change in escape velocity may be a curve somewhat like what you have, in which case you would integrate it from distance d from the center of the planet to infinity (area under the curve).

Ve in my graph above could really be considered to represent a gradient with respect to mass cancelled moving out from the centre of gravity which is different altogether. It represents freedom of vertical movement away from the COG.

Well I am using Ve = SQRT(2GM/r) which is the calculation for escape velocity.

Quote from: jeffreyH on 28/02/2014 11:48:51Well I am using Ve = SQRT(2GM/r) which is the calculation for escape velocity.That equation is giving you an answer that doesn't make any sense. Thus, one must conclude that it is not valid for subsurface acceleration due to gravity.If A(x) is the acceleration due to gravity at distance (x) from the center of the planet, then the escape velocity would be:v = for d is distance from the center of Earth.Above the surface, A(x) = GM/r^{2}Substituting it into the integral above, one gets:v = = = 0 - = Oops, it looks like velocity is in time, and I calculated it with respect to distance... See the ...sorry, you cannot view external links. To see them, please REGISTER or LOGINSo it should be: (1/2)v^{2} = , and v = Now, we know that the acceleration due to gravity function below the surface of earth is NOT A(x) = GM/r^{2}, so we should expect the escape velocity function to fail.We can break the function down into two parts.A_{b}(x) = Acceleration due to gravity below the surface of Earth.A_{s}(x) = Acceleration due to gravity in space (above the surface of Earth)Set the radius of Earth = 6,371 km.And one gets the equation: (1/2) v^{2} = + We've already solved the second half of the equation above, but you need a good approximation for A_{b}(x) for the acceleration due to gravity below the surface.Consider: A_{b}(x) = 9.8*x/(6,371 km)That would give you the case for the constant density of Earth (straight line on the Wikipedia graph above).Simply integrate it, and one gets:(1/2) v^{2} = + = = So, your escape velocity (subsurface) isv_{e} = Ok, so I graphed the above equation, giving a nice curve, of the general shape of what I had scribbled by hand above. [ Invalid Attachment ] Note, this uses the constant density of Earth. You'll probably want to substitute in your preferred subsurface density/gravitational acceleration function.

Your graph now looks like the one I made. It never hurts to double check my calculations as my calculus is getting a bit rusty by now.Your graph will vary a little if you change the density/acceleration function subsurface.You can't really have a gravity function that pulls towards the surface. I think I read that the gravity within a hollow shell is zero. And, potentially if you had a hollow shell with a non-uniform density, then the gravity could pull towards one side or the other, but that is quite a special case that is not reflected on Earth.As far as falling in a hole. Most of these escape velocity calculations ignore wind resistance. In a planet, since the gravity is always positive towards the center, falling into the hole, you will always experience positive acceleration as you approach the center... and thus speeding up (or zero acceleration at the center). Then once you pass the center and approach the opposite side, you would experience negative acceleration. In an "ideal system", falling through the sphere, you would accelerate towards the center, slow down going away from the center and barely reach the opposite surface.

How did you get such a steep asymptote at the center of the body?It looks like you used the equation: That would be find for a point source mass. However, for a body with volume, the gravity goes to zero as one goes subsurface towards the center of mass.