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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= $5.98 \times {10}^{24}$ kg) have to be compressed to be a black hole?
A. ${10}^{-9}m$
B. ${10}^{-6}m$
C. ${10}^{-2}m$
D. 100m
Answer
470.1k+ views
Hint: For earth to be a black hole, its escape velocity should be greater than or equal to speed of light. But nothing can be greater than the speed of light. Thus, the escape velocity should be equal to the speed of light. Use the formula for escape velocity and equate it with the speed of light. Substitute the values in the obtained expression. Solve it and find the radius R. To this obtained radius earth would have to be compressed to be a black hole.
Complete step by step answer:
Given: Mass of earth (M)= $5.98 \times {10}^{24}$ kg
For earth to be a black hole, the escape velocity should be equal to the speed of light.
$\Rightarrow$ Escape velocity= Speed of light
$\Rightarrow v= c$ …(2)
We know escape velocity is given by,
$v= \sqrt{\dfrac {2GM}{R}}$ …(2)
Where, G is the universal gravitational constant
M is the mass of the body is to escaped from
r is the distance from the center of the mass
Substituting equation. (2) in equation. (1) we get,
$ \sqrt{\dfrac {2GM}{R}}= c$
Squaring both the sides we get,
$\dfrac {2GM}{R}= {c}^{2}$
Rearranging the above equation we get,
$R= \dfrac {2GM}{{c}^{2}}$
Substituting values in above equation we get,
$R= \dfrac {2 \times 6.67 \times {10}^{-11} \times 5.98 \times {10}^{24}}{({3 \times {10}^{8})}^{2}}$
$\Rightarrow R= \dfrac {79.77 \times {10}^{13}}{9 \times {10}^{16}}$
$\Rightarrow R= 8.86 \times {10}^{-3}m$
$\Rightarrow R= 0.886 \times {10}^{-2}$
Hence, Earth has to be compressed to a radius of ${10}^{-2}m$ to be a black hole.
So, the correct answer is “Option C”.
Note: Escape velocity is the speed at which the sum of an object’s kinetic energy and its gravitational potential energy is equal to zero. The escape velocity only depends on the mass and size of the object from which something is trying to escape. The escape velocity is inversely proportional to the volume of the object. The escape velocity from the Earth for a stone will be the same as it would be for a rocket or a space shuttle.
Complete step by step answer:
Given: Mass of earth (M)= $5.98 \times {10}^{24}$ kg
For earth to be a black hole, the escape velocity should be equal to the speed of light.
$\Rightarrow$ Escape velocity= Speed of light
$\Rightarrow v= c$ …(2)
We know escape velocity is given by,
$v= \sqrt{\dfrac {2GM}{R}}$ …(2)
Where, G is the universal gravitational constant
M is the mass of the body is to escaped from
r is the distance from the center of the mass
Substituting equation. (2) in equation. (1) we get,
$ \sqrt{\dfrac {2GM}{R}}= c$
Squaring both the sides we get,
$\dfrac {2GM}{R}= {c}^{2}$
Rearranging the above equation we get,
$R= \dfrac {2GM}{{c}^{2}}$
Substituting values in above equation we get,
$R= \dfrac {2 \times 6.67 \times {10}^{-11} \times 5.98 \times {10}^{24}}{({3 \times {10}^{8})}^{2}}$
$\Rightarrow R= \dfrac {79.77 \times {10}^{13}}{9 \times {10}^{16}}$
$\Rightarrow R= 8.86 \times {10}^{-3}m$
$\Rightarrow R= 0.886 \times {10}^{-2}$
Hence, Earth has to be compressed to a radius of ${10}^{-2}m$ to be a black hole.
So, the correct answer is “Option C”.
Note: Escape velocity is the speed at which the sum of an object’s kinetic energy and its gravitational potential energy is equal to zero. The escape velocity only depends on the mass and size of the object from which something is trying to escape. The escape velocity is inversely proportional to the volume of the object. The escape velocity from the Earth for a stone will be the same as it would be for a rocket or a space shuttle.
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