Question

The condition for a uniform spherical mass $m$ of radius $r$ to be a black hole is : $[G=$ gravitational constant and $g=$ acceleration due to gravity]

• A. $\left(\frac{2Gm}{r}\right)^{1/ 2} \le c$
• B. $\left(\frac{2gm}{r}\right)^{1/ 2} =c$
• C. $\left(\frac{2Gm}{r}\right)^{1/ 2} \ge c$
• D. $\left(\frac{gm}{r}\right)^{1/ 2} \ge c$

Answer 1

Answer: (C)

$\left(\frac{2Gm}{r}\right)^{1/ 2} \ge c$

Answer 2

A black hole is an object so massive that even light cannot escape from it. This requires the idea of a gravitational mass for a photon, which then allows the calculation of an escape energy for an object of that mass. When gravitational potential energy of the photon is exactly equal to the photon energy, then $h v _{0}=\frac{G M h}{r c ^{2}} v _{0}$ ...(1) where $G$ is gravitational constant, $M$ is mass, $r$ is radius, $c$ is speed of light, $h$ is Planck's constant. Then from E (1) we have $r=\frac{G M}{c^{2}}$ Note that this condition is independent of frequency v. Schwarzchild's calculated gravitational radius differs from this result by a factor of 2 and is coincidently equal to the non-relativistic escape velocity expression. $v_{\text {escape }}=\sqrt{\frac{2 G M}{r}} \geq c$