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.

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" 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.

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