|
GR9677 #67 |
|
|
|
|
Alternate Solutions |
daverbeans
2008-11-06 12:49:13 | A really quick and dirty way to do this is to consider the radius of a "stellar mass" black hole, which is about 10 solar masses. These are around 30km in radius. The sun is approximately  earth mass. A stellar mass black hole would then be  earth masses. so 30km/ is on the order of .01m or 1cm. |  | scottopoly
2006-11-03 19:55:34 | I once heard that if the Earth were compressed into a black hole, it would be about the size of a grape. I think that is a much faster solution. ;) | | ben
2006-11-03 15:21:30 | i've generally seen the schwarzchild radius derived from the escape velocity as shown in
http://scienceworld.wolfram.com/physics/SchwarzschildRadius.html | |
|
|
Comments |
daverbeans
2008-11-06 12:49:13 | A really quick and dirty way to do this is to consider the radius of a "stellar mass" black hole, which is about 10 solar masses. These are around 30km in radius. The sun is approximately earth mass. A stellar mass black hole would then be earth masses. so 30km/ is on the order of .01m or 1cm. | | ivalmian
2008-03-25 19:42:17 | Um, well, the reason it's called the Schwarzschild radius is that it comes from Schwarzschild's solution of Einstein's equations. Take a look:
http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution |  | bat_pesso
2007-10-31 13:50:31 | i don't think relativity is such an issue here, the escape velocity depends only on the mass of the earth, not on the mass of the particle.
thus 0.5c^2 is still GM/R | | Jeremy
2007-10-18 13:17:00 | Not having seen such hardcore hand waving before, I would not feel comfortable with this solution even if I had thought of it. Looking in my books afterwards, I found something that's pretty cool: you can get  from dimensional analysis! (Of course you have to pick the right parameters.) And check this out,  , so unless the proportionality factor is several orders of magnitude, the closest answer is (C).
Moral: If thou needeth formulae, consulteth dimensional analysis. I think I gave up on this problem too quickly, when I could have at least submitted an educated guess. | | chri5tina
2006-11-28 05:24:22 | So in all this handwaving, did he explain the 1/2?
chri5tina
2006-11-28 05:32:11 |
well, I answered my own stupid question, however I'm hitting a bug when I try to edit my above post. I get this error when I try to edit:
omg are you trying to edit someone else's comment, you phisher?! (if you're innocent, then you ought to remember to login.)
when I am already logged in. and I'm not sure where to submit bugs.
|
|  | scottopoly
2006-11-03 19:55:34 | I once heard that if the Earth were compressed into a black hole, it would be about the size of a grape. I think that is a much faster solution. ;) | | ben
2006-11-03 15:21:30 | i've generally seen the schwarzchild radius derived from the escape velocity as shown in
http://scienceworld.wolfram.com/physics/SchwarzschildRadius.html | | pablojm
2006-10-28 15:30:36 | This is a good argument to convince someone who doesn't know physics, but how do you know it works...?
evanb
2008-07-01 17:22:24 |
Unfortunately, the way would know it works is by doing the general relativity. You can deduce that in the Schwarzschild metric for a point mass, any timelike or null (lightlike) trajectory (ie. any trajectory that a particle might take) that starts within R = 2M will ever wind up outside of R = 2M ( G = c = 1).
Un-unfortunately, GR reproduces lots of "Newtonian" results. You can even do most cosmology with Newtonian math. Why does that work? Because someone did it fully-general-relativistically and found that the math wound up the same...
|
| |
|
| Post A Comment! |
You are replying to:
Not having seen such hardcore hand waving before, I would not feel comfortable with this solution even if I had thought of it. Looking in my books afterwards, I found something that's pretty cool: you can get from dimensional analysis! (Of course you have to pick the right parameters.) And check this out, , so unless the proportionality factor is several orders of magnitude, the closest answer is (C).
Moral: If thou needeth formulae, consulteth dimensional analysis. I think I gave up on this problem too quickly, when I could have at least submitted an educated guess. |
Bare Basic LaTeX Rosetta Stone
|
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces  .
|
| type this... |
to get... |
| $\int_0^\infty$ |
 |
| $\partial$ |
 |
| $\Rightarrow$ |
 |
| $\ddot{x},\dot{x}$ |
 |
| $\sqrt{z}$ |
 |
| $\langle my \rangle$ |
 |
| $\left( abacadabra \right)_{me}$ |
_{me}) |
| $\vec{E}$ |
 |
| $\frac{a}{b}$ |
 |
|
|
|
|
The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it... |
|
|
|
