## Do black holes exist in Newtonian physics?

Posted by Diego Assencio on 2013.11.10 under Physics (Gravitation)

According to the theory of general relativity, if you compress an object of mass $M$ to a sphere with radius equal to or smaller than the Schwarzschild radius ($R_s$) for that object: $$R_s = \displaystyle\frac{2GM}{c^2}$$ then the escape velocity for a particle at a distance $R_s$ from the center of mass of $M$ will be equal to $c$, the speed of light. Since no object can exceed the speed of light, you have then created a black hole. The surface of this sphere of radius $R_s$ is called the event horizon of the created black hole. Any particle located at or below this surface will never be able to escape the gravitational field of the (compressed) mass $M$.

One very curious fact is that this same radius appears in the Newtonian theory of gravitation. In fact, let's compute the escape velocity for a particle of mass $m$ when in the presence of the gravitational field of a spherical body of mass $M$. Since energy is conserved as $m$ escapes the gravitational field of $M$ (the gravitational force is conservative), we have that: $$E_i = E_f \Longrightarrow \displaystyle\frac{1}{2}mv_i^2 - \displaystyle\frac{GMm}{r_i} = \displaystyle\frac{1}{2}mv_f^2 - \displaystyle\frac{GMm}{r_f}$$ where $v_i$ and $v_f$ are the initial and final velocities of the particle $m$ and $r_i$ and $r_f$ are the initial and final distances between $m$ and the center of mass of $M$ respectively.

The escape velocity ($v_i = v_e$) which $m$ needs to initially have to escape the gravitational field of $M$ is such that, when it reaches "infinity", its velocity $v_f$ is zero ("infinity" here means an $r_f$ large enough so that the gravitational force of $M$ on $m$ becomes negligible). If $v_f \neq 0$ then $m$ had initially more kinetic energy than necessary to escape from $M$. From these facts we have that: $$\frac{1}{2}mv_e^2 - \displaystyle\frac{GMm}{r_i} = \frac{1}{2}m(0)^2 - \displaystyle\frac{GMm}{\infty} = 0 \Longrightarrow v_e = \sqrt{\displaystyle\frac{2GM}{r_i}}$$

If $m$ is initially at a distance $r_i = 2GM/c^2$ from $M$, then its escape velocity is: $$v_e = \sqrt{\displaystyle\frac{2GM}{r_i}} = \sqrt{2GM \left(\displaystyle\frac{c^2}{2GM}\right)} = c$$ In other words, Newtonian gravity predicts that an object of mass $m$ situated at a distance $r_i = R_s = 2GM/c^2$ from an object of mass $M$ must have an initial velocity equal to or larger than to the speed of light ($c$) to escape the gravitational field of $M$. Assuming that a photon has mass (this must be true according to Newtonian gravity as it says only particles with mass interact gravitationally and we know for a fact that light is bent by gravitational fields), then light itself would not escape such a gravitational field.

Before you jump to the conclusion that there are things such as black holes in Newtonian gravity, notice that given enough fuel and a sufficiently powerful engine, any object (say, with mass $m$) can escape any gravitational field in a Newtonian universe. For that to happen, the object needs not exceed its escape velocity at any point of its escaping trajectory; one could simply attach the mentioned engine to it and have this engine exert a force larger than the gravitational force due to $M$ at all times in the escaping process. This will get $m$ to escape the gravitational field of $M$.