Spherical shell enclosing the Earth's surface 
What is the minimum thickness of a spherical shell that encloses the Earth's surface?

The two spheres are concentric, the outer sphere encloses all the surface (to the highest mountain), and the inner sphere excludes all the surface (down to the deepest ocean trench).
          
Yes, I know this is not strictly a mathematics question, but it is mathematical in some extended sense.  Of course, just subtracting the highest mountain peak from the deepest ocean
trench ignores the oblate spheroid shape of the Earth. I am curious to learn how close the Earth is to a sphere, and the shell thickness divided by, say, the inner radius, is one measure of that closeness.  Thanks!
 A: Three notes:


*

*The size of the equatorial bulge is 42.72 km.

*It is a generally-accepted fact that the farthest point from the Earth's center is the peak of Chimborazo, in Ecuador. This is very close to the equator, and 6310 m above sea level.

*The deepest point in the Arctic Ocean is 5450 m below sea level. I'm guessing this is probably the closest point to the Earth's center -- obviously it's closer than anything Antarctic, and the bulge is so large relative to the size of anything geographical that I suspect the greater depth of any trenches would be beaten out by their greater distance from the poles. If I'm wrong about this, the obvious candidate to check would be the Aleutian Trench (which has a maximum depth of 7679 m).


Adding those first three numbers would give 54.48 km. Assuming I'm right that these are the sites we should be measuring between, this is definitely an overestimate (since Chimborazo isn't on the equator and the deepest point isn't on the pole), but it seems like a reasonable starting point. Unfortunately Wikipedia doesn't give the latitude of any of the oceanic sites we care about, which means you'd have to dig a little deeper to get any kind of exact value...
