# Given the following, what is the tension in the string when the nearest mass is a distance of 686 km from the Black hole? G=6.673×10^-11 m^3kg-1s-2

## Two masses of 1 kg each are connected by a light, inextensible string of one metre in length. The two masses are now made to approach a Black Hole of mass 8.18×10^30 kg such that the string connecting them lies along a radius.

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Jul 12, 2016

#### Answer:

$3380 \text{ N}$. Don't try this at home: you might get torn apart (see the explanation).

#### Explanation:

The amount of tension is given by the difference in gravitational force between the two masses. Let A and B be the masses, with A being closer to the black hole O. Then:

${F}_{O A} = \frac{G {M}_{O} {M}_{A}}{{r}_{O A}^{2}}$

${F}_{O B} = \frac{G {M}_{O} {M}_{B}}{{r}_{O B}^{2}}$

As the A and B masses are equal we substitute ${M}_{A}$ for ${M}_{B}$ and subtract to get the tension:

$T = {F}_{O A} - {F}_{O B} =$
$G {M}_{O} {M}_{A} \left(\frac{1}{{r}_{O A}^{2}} - \frac{1}{{r}_{O B}^{2}}\right)$

Now put in numbers and calculate:

$G = 6.67 \times {10}^{- 11} {\text{ Nm"^2/"kg}}^{2}$ (same units in SI system as those given)

${M}_{A} \left(= {M}_{B}\right) = 1 \text{ kg}$, taken as exact values

${r}_{A} = 686 , 000 - 0.5 = 685 , 999.5 \text{ m}$

${r}_{B} = 686 , 000 + 0.5 = 686 , 000.5 \text{ m}$

Then

$T = 3380 \text{ N}$. This is over 170 times the gravity of Earth on the same amount of mass. A similar 100-g+ force acting on your mass would pull you apart, even though you are still relatively far from the hole's horizon.

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