# 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

Your head would need to be about 0.82 metres from the black hole to experience 1g.

#### 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 ${10}^{-} 16 m$.