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 × 10²⁴ kg) have to be compressed to be a black hole?
1. 10^-6 m
2. 10^-2 m
3. 100 m
4. 10^-9 m

Answer 1

Correct Answer - Option 2 : 10^-2 m

CONCEPT:

• Escape velocity: It is referred to as the minimum velocity needed by anybody or object to be projected to overcome the gravitational pull of the planet earth.
• In other words, the minimum velocity that one requires to escape the gravitational field is escape velocity.
• The formula for escape velocity comprises of a constant G, which we refer to as the universal gravitational constant. the value of it is = 6.673 × 10^-11. N.m²/ kg²
• The unit for escape velocity is meters per second (m/s).

\(Escape Velocity = \sqrt{\frac{2(gravitational constant)(mass of the planet of moon)}{radius of the planet or moon}}\)

\(v escape = \sqrt{\frac{2GM}{R}}\)

•  Here, v escape refers to the escape velocity (m/s)
• G is the universal gravitational constant ( 6.673 × 10^-11 N.m² / kg²)
• M refers to the mass of the planet or moon (m)
• R is the radius of the planet or moon (m)

EXPLANATION:

• For the earth to be a black hole the escape velocity should be at least equal to the speed of light.

∴ escape velocity = speed of light

\(\sqrt{\frac{2GM}{R}} = C\)

\(R = \frac{2GM}{C^2}\)

\(R = \frac{2 × 6.67 × 10^{-11} × 5.98 × 10^{24}}{9 × 10^{-16}}\)

= 8.86 × 10^-3m 

= 10^-2 m Ans

option 2 is the answer.