Question

Nothing can escape the event horizon of a black hole, not even light. You can think of the event horizon as being the distance from a black hole at which the escape speed is the speed of light, $3.00*{10}^{8}m/s,$making all escape impossible. What is the radius of the event horizon for a black hole with a mass $5.0$times the mass of the sun? This distance is called the Schwarzschild radius.

Short answer

The radius of the event horizon of the black hole is$14.74km.$

Step-by-step solution

Given information

Mass of black hole = $5$times of the mass of the sun, escape velocity =$3\times {10}^{8}m/s.$

Calculation

We know, escape velocity : $v=\sqrt{\frac{2G\times M}{R}}.$

Substituting the values of G, M, v in the above equation, and solving for R we get :

localid="1648487065595" $$\begin{gathered} 3\times {10}^{8}m/s=\sqrt{\frac{2(6.67\times {10}^{-11}N{m}^2/k{g}^2)(5\times 1.989\times {10}^{30}kg)}{R}} \\ 9\times {10}^{16}m/s=\frac{2(6.67\times {10}^{-11}N{m}^2/k{g}^2)(5\times 1.989\times {10}^{30}kg)}{R} \\ R=\frac{2(6.67\times {10}^{-11}N{m}^2/k{g}^2)(5\times 1.989\times {10}^{30}kg)}{(9\times {10}^{16}m/s)} \\ R=14740.7\mathrm{m} \\ R=14.740\mathrm{km}. \end{gathered}$$

Final answer

The radius of the event horizon for the black hole is $14.74\mathrm{km}.$