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

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? $G = 6.673 \times 10^{- 11} m^{3} k g - 1 s - 2$ $G=6.673×10^-11 m^3kg-1s-2$

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

Impact of this question

3718 views around the world

You can reuse this answer
Creative Commons License

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.

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? $G = 6.673 \times 10^{- 11} m^{3} k g - 1 s - 2$ $G=6.673×10^-11 m^3kg-1s-2$

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:

Related questions

Impact of this question

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^-11 m^3kg-1s-2G=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 8.18×10^30 kg8.18×1030kg8.18×10^30 kg such that the string connecting them lies along a radius.

1 Answer

Explanation:

Related questions

Impact of this 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? $G = 6.673 \times 10^{- 11} m^{3} k g - 1 s - 2$ $G=6.673×10^-11 m^3kg-1s-2$

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.