Why does Schwarzschild radius increase linearly with mass?

Question

Suppose there are $2$ black holes with masses $M_{earth}$ and $M_{sun}$, shouldn't the $M_{sun}$ crush the matter within it more intensely compared to $M_{earth}$? Based on this, shouldn't the Schwarzschild radius of $M_{sun}$ be lesser than predicted by the equation:

$$r_s = \frac{2GM}{c^2}$$

More specifically, why does the radius have linear relationship w.r.t. the mass considering that every bit of addition of mass crushes the matter inside the blackhole more intensely?

If there are $3$ blackholes $a$, $b$ and $c$ with masses $M_a$, $M_b$ and $M_c$ ($M_a$> $M_b$>$M_c$) respectively, then wouldn't it make more sense if their Schwarzschild radii were given by:

$$r_i = \frac{2GM_i^{\frac{1}{k_i}}}{c^2}, \quad i\in\{a,b,c\} \quad \text{where} \quad k_a>k_b>k_c$$

where $k_i$ is a scaling factor (or a shrinking factor) that varies depending upon the blackhole mass.

Answer 1

Consider a non-black hole mass i.e. a star or planet or indeed any other body. The escape velocity is given by:

$$ v^2 = \frac{2GM}{r} \tag{1} $$

If we set the escape velocity to be the speed of light and rearrange to get the radius then we get:

$$ r_c = \frac{2GM}{c^2} \tag{2} $$

which as it happens is exactly the same as the equation for the Schwarzschild radius:

$$ r_s = \frac{2GM}{c^2} \tag{3} $$

though note that the meaning of $r$ is different in the two equations so don't be deceived by the apparent similarity. Equation (2) is not a valid calculation since the equation for Newtonian gravity does not apply for any object massive enough to make the escape velocity even close to the speed of light. My point is that even using everyday Newtonian gravity we find this radius $r_c$ is proportional to $M$ not to any power $M^{1/k}$ as you suggest.

What might be confusing you is that for any spherically symmetric body the escape velocity at a distance $r$ from the centre does not depend on how compressed the object is. Suppose we are the Earth-Sun distance from a massive object, then it doesn't matter if that object is massively dense with a radius of a micron or a fluffy cloud with a radius of a million kilometres. The gravitational field at the distance $r$ depends only on $M$ and $r$ and not on the density of the body. Or to put this another way, if the Sun were to become a black hole of the same mass it would not affect the orbits of the planets.