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?

Question

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?

Alice Andrews · Accepted Answer

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

The speed at which you would accelerate is: #a=(GM)/r^2# Where #M=10^11kg# is the mass of the black hole, #G=6.674m^2kg^(-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=9.81ms^(-2)#. Therefore, the distance needed to feel 1 g of acceleration is: #r^2=(GM)/a=6.67/9.81# This gives #r=0.82m#. Being so close to a black hole puts you in the region where tidal effects can occur. At #r=0.6m#, #a=18.5ms^(-2)#. At #r=0.4m#, #a=41.7ms^(-2)#. By the way, a black hole's Schwarzschild radius can be found using: #r=(GM)/c^2# Where #c# is the speed of light. A black hole with a mass of #10^11kg# has a radius of #7.4*10^(-17)m#, which is slightly smaller than #10^-16m#.

Michael Adkins · Answer

undefined To find the distance from your head where the gravitational pull of the tiny black hole would match that of Earth's, you can use the equation for gravitational force:

$F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}}$

Where:
$F$ is the gravitational force,
$G$ is the gravitational constant ($6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}$),
$m_1$ and $m_2$ are the masses of the two objects,
$r$ is the distance between the centers of the two objects.

For Earth, $m_1$ is the mass of the Earth ($5.972 \times 10^{24} \, \text{kg}$), and for the tiny black hole, $m_2$ is $1 \times 10^{11} \, \text{kg}$. Let $d$ be the distance from your head to the tiny black hole.

Setting the gravitational forces equal to each other:

\(\frac{{G \cdot m_{\text{Earth