Can the Schwarzchild radius for black hole be obtained from Newton's law of universal gravitation?

Question

The universal law of gravitation of Newton calculate the intensity of gravitational force by the radius and mass of an body.. so in a black hole there is very much small radius and a large amount of mass... So I think newtonian gravity would predict large amount of force.... That is similar to the great pull of black holes due to gigantic space time curvature...

Answer 1

You can get an order-of-magnitude estimate for "the gravitational radius" from dimensional analysis. If you're interested in the gravitational escape velocity from an object with mass $M$ as a function of radius, the only parameters in your problem are going to be the mass $M$ itself, the gravitational coupling constant $G$, and the velocity and radius of interest, $v$ and $R$. The units are

\begin{align} [G] &= \frac{\rm N\ m^2}{\rm kg^2} = \rm m^3\ kg^{-1}\ s^{-2} \\ [M] &= \rm kg \\ [v] &= \rm m\ s^{-1} \\ [R] &= \rm m \end{align}

The only product $G^a M^b v^c$ with units of distance is $GM/v^2$. So if your speed of interest is the speed of light, the associated distance is going to be $R_\text{light} = f \cdot GM/c^2$, where $f$ will be a dimensionless number like $\frac13$ or $\sqrt 2$ that comes out of doing algebra.

A Newtonian-graviational sphere whose escape velocity is the speed of light has $f=2$. The Schwarzchild radius also has $f=2$. But the alignment is a coincidence, rather than anything fundamental. There are lots of ways to do a page of algebra and have everything cancel out to get a factor of two.

Answer 2

Actually, one can obtain the information about Blackhole by merely looking at the escape velocity. $$v^2 = \frac{2GM}{r}$$

$\bf{Explanation} :-$ Imagine you fire a cannon, from the top of a mountain. The projectile would travel in horizontal direction because of its kinetic energy, $$T = \dfrac{1}{2}mv^2$$

But, simultaneously, it will also have vertical direction because of the Gravitational Potential energy, $$V = -\dfrac{GMm}{r}$$

Thus, the total energy of the system will be $$E = \dfrac{1}{2}mv^2 - \dfrac{GMm}{r}$$

Now, if the velocity is low, then Energy of the system will be negative. If, a body has negative energy, it essentially means their is attraction and when it has positive energy, it means system would no longer a bound system and they tend to repel each other.

Now, if one keep raising the velocity of the projectile then there comes a state when total energy of the system becomes zero. Lets call it $v_0$, This is known as the escape velocity of the system.

Then, $$\dfrac{1}{2}mv^2_{0} - \dfrac{GMm}{r} = 0$$ $$v^2_{0} = \dfrac{2GM}{r}$$

Now, if we apply Einstein's second postulate, which states that, "Nothing can travel faster than speed of light". One can show that $$r_{s} = \frac{2GM}{c^2}$$

This is our expression for Schwarzschild radius. Now, for $r < r_{s}$ one must require to move greater than speed of light in order to escape from its gravitational pull. So, yes, one can obtain Schwarzschild radius using Newton's law.