Question

A rogue black hole with a mass 12 times the mass of the sun drifts into the solar system on a collision course with earth. How far is the black hole from the center of the earth when objects on the earth’s surface begin to lift into the air and ""fall"" up into the black hole? Give your answer as a multiple of the earth’s radius.

Answer 1

Answer:

Distance of blackhole from center of earth 2000 Re

Explanation:

we have mass of earth $= Me = 5.972 \times 10^{24} kg$

Mass of black hole $Mb = 12 \times M_{sun} = 12 \times 1.989 \times 10^{30} kg = 23.868 \times 10^{30} kg$

Object start to lift in air when net force on them becomes zero

$\frac{ G Mb m}{x^2} = \frac{GM_{sun} m}{Re^2}$

$\frac{MB}{x^2} = \frac{M_E}{R_E^2}$

$x = \sqrt{\frac{Mb}{M_E}} R_E$

$x = \sqrt{\frac{23.868 \times 10^{30}}{5.972 \times 10^{24}}} R_E$

$x = 1.99 \times 10^{3} Re$

x = 1999.16 Re

So, distance of blackhole from center of earth will be = x+ Re

     = 1999.16 Re + Re = 2000.16 Re