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How far away would light have to be from a black hole with the mass of the Earth (6 x 1024 kg) in order to be able to escape from its gravitational pull? (Note: use the formula for escape speed again: vesc = 1.15 × 10-5 MR−−√MR

Guest Jul 6, 2017
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I don't know what your formula is supposed to calculate! I know it is about calculating the "escape velocity", but I don't need it here! You should know that if the Earth was squeezed into a black hole, the black hole would have a radius of 9 millimeters! The "photon sphere", or the orbit of light would be: 1.5 x 9 =13.5 mm from the center of the black hole. Or, 13.5 - 9 =4.5mm from the "event horizon". That is the best I can do!.

Guest Jul 7, 2017
edited by Guest  Jul 7, 2017
edited by Guest  Jul 7, 2017
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How far away would light have to be from a black hole with the mass of the Earth (6 x 1024 kg)

in order to be able to escape from its gravitational pull?

(Note: use the formula for escape speed again: vesc = 1.15 × 10-5 MR−−√MR )

 

Let speed of light \(c = 3\cdot 10^8\ \frac{m}{s}\)

Let gravitational constant \( = G \) \( \)

Let Radius = R

 

\(\begin{array}{|rcll|} \hline v_{esc} &=& \sqrt{\frac{2 G M_{\text{Earth}} }{R} } \\ v_{esc} &=& \sqrt{2 G} \times \sqrt{\frac{ M_{\text{Earth}} }{R} } \quad & | \quad \sqrt{2 G} \approx 1.15 \cdot 10^{-5} \\ v_{esc} &=& 1.15 \cdot 10^{-5} \times \sqrt{\frac{ M_{\text{Earth}} }{R} } \quad & | \quad v_{esc} = c \\ c &=& 1.15 \cdot 10^{-5} \times \sqrt{\frac{ M_{\text{Earth}} }{R} } \\ c^2 &=& 1.15^2 \cdot (10^{-5})^2 \cdot \frac{ M_{\text{Earth}} }{R} \\ R &=& 1.15^2 \cdot (10^{-5})^2 \cdot \frac{ M_{\text{Earth}} }{c^2} \quad & | \quad M_{\text{Earth}} = 6 \cdot 10^{24}\ kg \qquad c = 3\cdot 10^8\ \frac{m}{s} \\ &=& 1.15^2 \cdot (10^{-5})^2 \cdot \frac{ 6 \cdot 10^{24} }{(3\cdot 10^8)^2} \\ &=& 1.15^2 \cdot 10^{-10} \cdot \frac{ 6 \cdot 10^{24} }{ 9\cdot 10^{16} } \\ &=& 1.15^2 \cdot 10^{-10} \cdot \frac{2}{3} \cdot 10^{(24-16)} \\ &=& 1.15^2 \cdot \frac{2}{3} \cdot 10^{(24-16-10)} \\ &=& 0.88167 \cdot 10^{-2} \ m \\ &=& 8.8167 \cdot 10^{-3} \ m \\ &=& 8.817 \ mm \\\\ \mathbf{R} & \mathbf{ \approx }& \mathbf{ 8.8 \ mm } \\ \hline \end{array} \)

 

The light would have to be more than \(\mathbf{8.8 \ mm}\) from the black hole with the mass of the Earth

in order to be able to escape from its gravitational pull

 

laugh

heureka  Jul 7, 2017
edited by heureka  Jul 7, 2017
 #3
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This is (as JB calls it),”slop at the top!” Of course, Mr. BB mixed this slop with blarney; I would expect nothing less from the Blarney Master.   Mr. BB’s presentations are always s slop, but the contrast is quite sharp when you compare it to Heureka’s masterful presentation.

 

Let’s see:  you “don’t know what the formula is supposed to calculate!”  But you know it is about calculating the ‘escape velocity’” (it’s impressive that you know that).  “But you don’t need it here!”  That’s impressive too, because it’s a monumental metric for measuring Batshit Stupid!  

 

The asker is supposed to ignore the formula that gives the Schwarzschild radius, but know “that if the Earth was squeezed into a black hole, the black hole would have a radius of 9 millimeters!

 

How is he supposed to know that?  Oh, of course, ask your computer: it’s connected to the internet, and everything read on there is true; that’s a fact; you can read about on the internet!    

 

After he finds out (via magic incantation) that it is9 millimeters!”  He’s supposed to know “The ‘photon sphere’ or the orbit of light would be: 1.5 x 9 =13.5 mm from the center of the black hole. Or, 13.5 - 9 =4.5mm from the ‘event horizon’. He knows this the same way he knows the first part . . .  He asks his computer. No need for formulas or derivations. That’s just a waste of time!   

 

None of this matters because this is not the question.  The photon sphere is irrelevant here. Only the escape velocity is relevant. 

 

That is the best I can do!.

 

Oh surely not! Ask your computer this question:  What would the minimum orbital radius be if the partial were an electron?  

 

I can hardly wait!  I’ll be biding my time playing Monopoly with my dog and cat. It’s the cat’s turn to be the banker so it should be an interesting game!

GingerAle  Jul 8, 2017

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