# After the big-bang, tiny black holes may have formed. If one with a mass of 1 x 10^11 kg (and a radius of only 1 x 10^-16 m) reached Earth, at what distance from your head would its gravitational pull on you match that of the Earth's?

Aug 31, 2016

#### Explanation:

The acceleration you would experience is:

$a = \frac{G M}{r} ^ 2$

Where $M = {10}^{11} k g$ is the mass of the black hole, $G = 6.674 {m}^{2} k {g}^{- 1} {s}^{- 2}$ is the gravitational constant and $r$ is the distance from the black hole.

The acceleration due to gravity on Earth is $a = 9.81 m {s}^{- 2}$.

So, to experience 1g of acceleration, thee distance is:

${r}^{2} = \frac{G M}{a} = \frac{6.67}{9.81}$

This gives $r = 0.82 m$.

Being so close to a black hole puts you in the region where tidal effects can occur. At $r = 0.6 m$, $a = 18.5 m {s}^{- 2}$. At $r = 0.4 m$, $a = 41.7 m {s}^{- 2}$.

Incidentally, the Schwarzschild radius for a black hole is given by:

$r = \frac{G M}{c} ^ 2$

Where $c$ is the speed of light. A black hole with a mass of ${10}^{11} k g$ has a radius of $7.4 \cdot {10}^{- 17} m$, which is slightly smaller than ${10}^{-} 16 m$.