Schwarzschild Radius

Classical Mechanics Level pending

If the mass of an object, specifically a sphere, were to be compressed to a sphere of a certain radius, the escape velocity on the surface of that sphere would be equal to the speed of light. This radius is called Schwarzschild radius.

Given that the escape velocity, $v_e = \sqrt{ \dfrac{2GM}R }$ , where $G$ is the Gravitiational constant, $M$ is the mass of the sphere and $R$ is the radius, deduce the expression for Schwarzschild radius, $R_\text{Sch}$ and find out the Schwarzschild radius of Hercules-Corona Broealis Great Wall (largest known structure in the observable universe).

Details and Assumptions :

$G = 6.67408 \times 10^{-11} \text{ m}^3 \text{kg}^{-1} \text{s}^{-2}$ .
$M = 3.9782 \times10^{49}\text{ kg}$ .

\approx 252,000 \text{ trillion km}

\approx2.6 \text{ million km}

\approx 60,000 \text{ trillion km}

\approx 515 \text{ trillion km}

1 solution

Akshay Sreenivasan
Mar 14, 2016

$We\quad know\quad that\quad { v }_{ e }=\sqrt { \frac { 2GM }{ R } } \\ At\quad Schwarzschild\quad radius,\quad R={ R }_{ Sch }\quad ,\quad { v }_{ e }=c\\ \therefore \quad c=\sqrt { \frac { 2GM }{ { R }_{ Sch } } } \\ \Rightarrow { c }^{ 2 }=\frac { 2GM }{ { R }_{ Sch } } \\ \Rightarrow { R }_{ Sch }=\frac { 2GM }{ { { c }^{ 2 } } } \\ \Rightarrow { R }_{ Sch }=\frac { 2\times 6.67408\times { 10 }^{ -11 }\times 3.9782\times { 10 }^{ 49 } }{ { 299792458 }^{ 2 } } \\ \cong 6\times { 10 }^{ 22 }\quad m\\ \cong \boxed { 60,000\quad trillion\quad km }$

Schwarzschild Radius

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