# 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.

• Much more simply: within Earth spacetime geometry is not Schwarzschild. It is Schwarzschild outside Earth if three simplifying assumptions are done: 1) Earth's mass distribution is spherically symmetric 2) Earth's rotation is neglected 3) Earth is alone into space (no Sun, no Moon, etc.). The value 0.03 m << Earth radius only means that spacetime geometry around Earth is very slightly affected by its mass - it departs very little from being Lorentzian. – Elio Fabri Sep 23 '18 at 13:07
• while deriving Schwartschild solution one starts with spherical coordinates but makes a lot of coordinates transformations This sounds to me like a misunderstanding, although it's hard to be certain without seeing what source you're reading. There is no underlying, flat-space system of spherical coordinates that later gets transformed to Schwarzschild coordinates. – Ben Crowell Sep 23 '18 at 14:46

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)
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.