Question

Left over from the big-bang beginning of the universe, tiny black holes might still wander through the universe. If one with a mass of L x 1011 kg (and a radius of only 1 x L0-16 m) reached Earth, at what distance from your head would its gravitational pull on you match that of Earth's?

Answer 1

Answer:

The distance as calculated is 0.823 m

Solution:

As per the question:

Mass of black hole, M = $1\times 10^{- 11}\ kg$

Radius of blck hole, R = $1\times 10^{- 16}\ m$

Now,

At the time when the black hole's gravitational force exerts a pull on the body, the gravitational force due to Earth on the body and that due to black hole will be the same.

Gravitational force due to Earth is given by:

$F_{G} = \frac{GM'm}{R'^{2}}$

where

M' = Mass of earth

m = Mass of the body

R' = Separation distance between the body and the center of the Earth

Now,

Gravitational force of the black hole on the body is given by:

$F_{B} = \frac{GMm}{R^{2}}$

Now,

$F_{G} = F_{E}$

$\frac{GM'm}{R'^{2}} = \frac{GMm}{R^{2}}$                     (1)

where

$\frac{GM'}{R'^{2}} = g = 9.8$                                          (2)

From the eqn (1) and (2):

$R = \sqrt{\frac{GM}{R'^{2}}}$

$R = \sqrt{\frac{6.67\times 10^{11}\times 1\times 10^{11}}{(1\times 10^{- 16})^{2}}}$

R = 0.823 m