Question

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.

Answer 1

`3380" 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_{OA}={GM_OM_A}/{r_{OA}^2}`

`F_{OB}={GM_OM_B}/{r_{OB}^2}`

As the A and B masses are equal we substitute `M_A` for `M_B` and subtract to get the tension:

`T=F_{OA}-F_{OB}=`
`GM_OM_A(1/{r_{OA}^2}-1/{r_{OB}^2})`

Now put in numbers and calculate:

`G=6.67xx10^{-11}" Nm"^2/"kg"^2` (same units in SI system as those given)

`M_A(=M_B)=1" kg"`, taken as exact values

`r_A=686,000-0.5=685,999.5" m"`

`r_B=686,000+0.5=686,000.5" m"`

Then

`T=3380" 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.