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.

Short answer

Distance of blackhole from center of earth$2000R_e$

Step-by-step solution

Given information

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

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

Explanation

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

$$\begin{gathered} \frac{G{M}_{b}m}{x^2}=\frac{G{M}_{sun}m}{{R^2}_e} \\ \frac{M_b}{x^2}=\frac{M_e}{{R^2}_e} \\ x=\sqrt{\frac{M_b}{M_e}}{R}_e \\ x=\sqrt{\frac{23.868\times {10}^{30}}{5.972\times {10}^{24}}}{R}_e \\ x=1999.16R_e \end{gathered}$$

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

$=1999.16R_e+R_e=2000.16R_e$