Question

Calculate what should be the radius of the earth so that it becomes a black hole without change in its mass.

Answer 1

A black hole is a section of space where it’s found that gravity pulls everything that not even the light which has such high speed will pass through it. This is observed because a very huge mass is squeezed to fit into a very little space. This is often seen or observed when a star dies out. For this question, we’ll apply the formula of escape velocity and then compare this velocity with the escape velocity of the earth at black hole condition to find the radius of the earth.
Formula used:
${v_e} = \sqrt {\dfrac{{2GM}}{R}}$
Here, $G$ is the gravitational constant $G = 6.67 \times {10^{ - 11}}$,
$M$ is the body mass to be escaped from $M = 5.98 \times {10^{24}}$,
$R$ is the body radius from the center.

Complete step by step answer:
General relativity theory predicts that to form a black hole, a sufficiently compact mass will distort space-time. The event horizon is named the limit of the area from which no escape is possible. It’s predicted that black holes of stellar mass will develop when very large stars collapse at the end of their life cycle.
Escape Velocity is termed as the minimum velocity needed to transcend the gravitational force of the planet earth by someone or object to be projected. In another way, escape velocity is the minimum velocity that one has to escape the gravitational field.
The equation of escape velocity is given as:
${v_e} = \sqrt {\dfrac{{2GM}}{R}}$
For the earth to be a black hole the escape velocity must be at least equal to the speed of light.
$\therefore escape$ $velocity = speed$ $of$ $light$
$\sqrt {\dfrac{{2GM}}{R}} = c$, Where $c = 3 \times {10^8}$
$\sqrt {\dfrac{{2GM}}{c}} = R$
Squaring both the sides we get,
${c^2} = \dfrac{{2GM}}{R}$
Rearranging the above equation we get,
$R = \dfrac{{2GM}}{{{c^2}}}$
Substituting values in the above equation we get,
$R = \dfrac{{2 \times 6.67 \times {{10}^{ - 11}} \times 5.98 \times {{10}^{24}}}}{{{{\left( {3 \times {{10}^8}} \right)}^2}}}$
$R = \dfrac{{79.77 \times {{10}^{13}}}}{{9 \times {{10}^{16}}}}$
$R = 8.86 \times {10^{ - 3}}m$ $\left( {or} \right)$ $R = 0.886 \times {10^{ - 2}}m = {10^{ - 2}}m$
$R = 8.86mm \approx 9mm$
Hence, Earth must be compressed to a radius of ${10^{ - 2}}m$ to be a black hole.

Note: While solving this problem, most of the students take the escape velocity to be greater than the speed of the light itself. There’s a fact that till now, nothing has been found which travels greater than the speed of light. No matter can survive its strong pull and not even light can escape in a black hole. A hypothetical spaceship will survive if it can travel at more than the speed of light.