Question

Miniature black holes.Left over from the big-bang beginningof the universe, tiny black holes might still wander through the universe. If one with a mass of$1\times {10}^{11}\text{kg}$(and a radius of only$1\times {10}^{-16}\text{m}$) reached Earth, at what distance from your headwould its gravitational pull on you match that of Earth’s?

Short answer

The distance at which the gravitational pull on you matches that of the Earth is$0.8\text{\hspace{0.33em}m}$ .

Step-by-step solution

The given data

i) Mass of a black hole,$m_b=1\times {10}^{11}\text{\hspace{0.33em}kg}$

ii) Radius of black hole$r_{bp}=1\times {10}^{-16}\text{\hspace{0.33em}m}$

iii) Radius of Earth,$r_{ep}=6.37\times {10}^6\text{\hspace{0.33em}m}$

iv) Mass of Earth,$m_e=5.98\times {10}^{24}\text{\hspace{0.33em}kg}$

Understanding the concept of Newton’s law of gravitation 

This problem is based on Newton’s law of gravitation. According to Newton’s law of gravitation, the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. By balancing both forces, we can find the unknown distance between the black hole and the person.

Formula:

Force of gravitation, $F=\frac{G{m}_1{m}_2}{r^2}$ …(i)

Calculating the required gravitational pull to match that of Earth’s

Using equation (i), Force between Earth & Person can be given as:

$F_{ep}=\frac{G\times m_e\times m_p}{r_{ep}^2}$

Using equation (i), Force between Black hole& Person can be given as:

$F_{bp}=\frac{G\times m_b\times m_p}{r_{bp}^2}$

If these two forces are equal, then

$\begin{array}{c}{F}_{ep}=F_{bp}\\ \frac{G\times m_e\times m_p}{r_{ep}^2}=\frac{G\times m_b\times m_p}{r_{bp}^2}\\ \end{array}$

Hence,

$\begin{array}{c}\frac{r_{bp}^2}{r_{ep}^2}=\frac{m_b}{m_e}\\ \frac{r_{bp}^2}{{\left(6.37\times {10}^6\text{m}\right)}^2}=\frac{1\times {10}^{11}\text{kg}}{5.98\times {10}^{24}\text{kg}}\\ r_{bp}=\sqrt{\left(\frac{{\left(6.37\times {10}^6\text{\hspace{0.33em}m}\right)}^2\times 1\times {10}^{11}\text{kg}}{5.98\times {10}^{24}\text{\hspace{0.33em}kg}}\right)}\\ =0.824\text{m}\\ \approx 0.8\text{\hspace{0.33em}m}\end{array}$

The distance at which the gravitational pull match that to the Earth’s is .$0.8\text{\hspace{0.33em}m}$