91Ó°ÊÓ

i) Mass of a black hole,$m_b=1×{10}^{11}\text{ k²µ}$

ii) Radius of black hole$r_{bp}=1×{10}^{−16}\text{ m}$

iii) Radius of Earth,$r_{ep}=6.37×{10}^6\text{ m}$

iv) Mass of Earth,$m_e=5.98×{10}^{24}\text{ k²µ}$

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, $F=\frac{G{m}_1{m}_2}{r^2}$ …(¾±)

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

$F_{ep}=\frac{G×m_e×m_p}{r_{ep}^2}$

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

$F_{bp}=\frac{G×m_b×m_p}{r_{bp}^2}$

If these two forces are equal, then

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

Hence,

$\begin{array}{c}\frac{r_{bp}^2}{r_{ep}^2}=\frac{m_b}{m_e}\\ \frac{r_{bp}^2}{{\left(6.37×{10}^6\text{m}\right)}^2}=\frac{1×{10}^{11}\text{kg}}{5.98×{10}^{24}\text{kg}}\\ r_{bp}=\sqrt{\left(\frac{{\left(6.37×{10}^6\text{ m}\right)}^2×1×{10}^{11}\text{kg}}{5.98×{10}^{24}\text{ k²µ}}\right)}\\ =0.824\text{m}\\ ≈0.8\text{ m}\end{array}$

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