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

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

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 $8.18 \times 10^{30} k g$ $8.18×10^30 kg$ such that the string connecting them lies along a radius.

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Answer 1 · 2016-07-11T10:47:15

$3380 N$ $3380" N"$ . 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:

$F_{O A} = \frac{G M_{O} M_{A}}{r_{O A}^{2}}$ $F_{OA}={GM_OM_A}/{r_{OA}^2}$

$F_{O B} = \frac{G M_{O} M_{B}}{r_{O B}^{2}}$ $F_{OB}={GM_OM_B}/{r_{OB}^2}$

As the A and B masses are equal we substitute $M_{A}$ $M_A$ for $M_{B}$ $M_B$ and subtract to get the tension:

$T = F_{O A} - F_{O B} =$ $T=F_{OA}-F_{OB}=$
$G M_{O} M_{A} (\frac{1}{r_{O A}^{2}} - \frac{1}{r_{O B}^{2}})$ $GM_OM_A(1/{r_{OA}^2}-1/{r_{OB}^2})$

Now put in numbers and calculate:

$G = 6.67 \times 10^{- 11} \frac{{Nm}^{2}}{{kg}^{2}}$ $G=6.67xx10^{-11}" Nm"^2/"kg"^2$ (same units in SI system as those given)

$M_{A} (= M_{B}) = 1 kg$ $M_A(=M_B)=1" kg"$ , taken as exact values

$r_{A} = 686, 000 - 0.5 = 685, 999.5 m$ $r_A=686,000-0.5=685,999.5" m"$

$r_{B} = 686, 000 + 0.5 = 686, 000.5 m$ $r_B=686,000+0.5=686,000.5" m"$

Then

$T = 3380 N$ $T=3380" N"$ . 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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Given the following, what is the tension in the string when the nearest mass is a distance of 686 km from the Black hole? $G = 6.673 \times 10^{- 11} m^{3} k g - 1 s - 2$ $G=6.673×10^-11 m^3kg-1s-2$

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 $8.18 \times 10^{30} k g$ $8.18×10^30 kg$ such that the string connecting them lies along a radius.

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

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=6.673×10−11m3kg−1s−2G=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 8.18×1030kg8.18×10^30 kg such that the string connecting them lies along a radius.

1 Answer

Explanation:

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

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 $8.18 \times 10^{30} k g$ $8.18×10^30 kg$ such that the string connecting them lies along a radius.