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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...

Qmechanic
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Rehan
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1 Answers1

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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.

rob
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