Innermost stable circular orbit in Schwarzschild solution I've been reading about GR recently and I can follow the derivation of a Schwarzschild solution to it's final and well known form in Schwarzschild coordinates.
The orbit stability argument (for a massive test particle) is also clear - no stable circular orbit can exists for $r<6M$.
What usually follows after that is a calculation for the Earth:
$r = 6GM/c^2 = 0.03m$
radius of the Earth $= 6300km$.
So comparing them one notes that it is not a problem for the Earth because 0.03m is well below the surface.
My question is - how can we make such a comparision? Radius of a planet is measured in spherical coordinates but $r$ in $r=6M$ is in Schwartzschild coordinates - while deriving Schwartschild solution one starts with spherical coordinates but makes a lot of coordinates transformations so the resulting $r$ is really a very complicated function of a spherical radius and comparing their values seems wrong.
 A: The   Schwarzschild coordinate $r$ is defined so that the area of an $r=const.$ surface is $4\pi r^2$ with the area being evaluated using  the metric at fixed $t$. This means that one can regard  our (to a very good approximation) flat space-time  radius $r$ as coinciding with the Schwarzschild coordinate $r$ once we are outside the body of the earth. (The Schwarzschild metric does not apply inside he earth)
A: The comparison is still valid, although what it means is a little hidden. The Schwarzschild r coordinate is defined as the square root of the surface area of a sphere at that distance divided by $4 \pi$. In other words $A=4\pi r^2$. 
So saying that the radius of the earth is greater than $0.03 m$ is saying that the surface area of the earth is greater than $0.01 m^2$, which is clearly true. And saying that the radius of the Earth is $6300 km$ is saying that it’s surface area is $4\pi (6300 km)^2$ which if we are approximating the earth as spherically symmetric and non rotating is approximately true. 
