Question

A black hole is an object whose gravitational field is so strong that even light cannot escape from it . To what approximate radius would earth (mass = 5.98 x 10 {power-24} kg) have to be compressed to be a black hole ?

Answer 1

according the formula r= (2GM)/(c^2) the earth will have to be 8.7 millimeters in radius .

Answer 2

Schwarzschild Radius

Black Holes are stellar objects with so strong gravitational fields that even light cannot escape.

In other words, if the escape velocity of any object became greater than or equal to the speed of light in vacuum, it would become a black hole.

To have such great gravitational fields, an object must have lots of mass packed in a tiny region. Theoretically, compressing any object beyond a certain radius makes it a black hole.

The critical radius for a given body beyond which if it is compressed, the body becomes a black hole, is called the Schwarzschild Radius.

We can obtain the formula for Schwarzschild Radius from the formula of Escape Velocity, by putting Escape Velocity as speed of light.

$\sf \displaystyle v_{esc} = \sqrt{\frac{2GM}{R}} \\ \\ \\ \textsf{Put }v_{esc}=c \\ \\ \\ \implies c = \sqrt{\frac{2GM}{R_S}} \\ \\ \\ \implies c^2=\frac{2GM}{R_S} \\ \\ \\ \implies \boxed{R_S = \frac{2GM}{c^2}}$

We have our values:

$G = 6.67 \times 10^{-11}\ Nm^2/kg^2 \\ \\ M = 5.98 \times 10^{24}\ kg \\ \\ c = 3\times 10^8\ m/s$

Let's put in the values and find the Schwarzschild Radius of Earth.

$\sf\displaystyle R_S = \frac{2GM}{c^2} \\ \\ \\ \implies R_S = \frac{2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{(3\times 10^8)^2} \\ \\ \\ \implies R_S = \frac{79.7732 \times 10^{13}}{9\times 10^{16}} \\ \\ \\ \implies R_S \approx 8.86 \times 10^{-3}\ m \\\\\\ \implies \huge\boxed{\boxed{\sf R_S \approx 8.86\ mm}}$

Thus, the earth must be compressed to a radius smaller than approximately 8.86 mm to make it a black hole.