# Why does the Schwarzschild radius become excessively large after a certain point?

Here's something that I've found difficult to wrap my head around. The relationship between the Schwarzschild radius and mass is linear. It's generally known that if you take an object in the universe and squeeze it down to it's Schwarzschild radius, that radius will always be smaller than smaller than the original object's radius. E.x, the sun has 1 solar radius, which is much larger than it's Schwarzschild radius of 3 km.

But if you started calculating Schwarzschild radius for an object with really high mass, things seem to get a bit funky. Take an object that has a mass of $10^{100}$ kilograms. The corresponding Schwarzschild radius for that is $1.48513 \times 10^{73}$ meters. But if we took an object that has the density of the Sun (1410 $kg/m^{-3}$) and tried to find it's radius normally, we end up with a radius of $1.19244 \times 10^{32}$ meters which is smaller than the Schwarzchild radius. How is this possible? I understand that you won't be able to have an object that large enough to even consider this to be "possible", but this still is confusing to me. Are my calculations off?

-
The schwarzschild radius becoming larger than the (naive estimate of the actual) radius is why there are black-holes. –  zhermes Apr 28 '13 at 18:56
Awesome reasoning. If you want to go further, ask yourself this: How dense would the universe have to be in order for its "radius" of 14 billion light years to match its Schwarzschild radius? This is basically what is known as the "critical density," and it turns out the actual density of the universe is about this value. –  Chris White Apr 28 '13 at 18:57

In fact, if the average density of the universe were the same as the atmospheric density of Mercury (at the surface), the Schwarzchild radius of the visible universe would be about 5 trillion times its known radius. The atmospheric pressure of Mercury? $10^{-14}$ times that of Earth. Just fyi –  Jim May 8 '13 at 14:08