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GR9677 #67
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Alternate Solutions |
elgo
2017-10-18 03:15:42 | To get the answer exactly you need to know the Schwarzchild radius is at R=2GM. This means for there to be a blackhole, all of the mass has to be inside that radius. To convert G to general relativity units you need to divide it by  . When you put in the values for the constants, you\\\\\\\'ll get the right answer of 1cm. |  | 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 | |
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Comments |
enterprise
2018-04-01 16:17:05 | It is just an order of magnitude calculation. escape velocity=speed of light. You can write this as 2*c=G+M-R. Every term is an order of magnitude. Take the equation and apply the logarithm on both sides. For example , M=9*10^24 so just set M=25 in the above equation. |  | elgo
2017-10-18 03:15:42 | To get the answer exactly you need to know the Schwarzchild radius is at R=2GM. This means for there to be a blackhole, all of the mass has to be inside that radius. To convert G to general relativity units you need to divide it by . When you put in the values for the constants, you\\\\\\\'ll get the right answer of 1cm. | | keradeek
2011-10-01 21:32:38 | well, I think the handwaving shown in the official solution is a bunch of crap, cuz energy is gamma*m*v^2, not 1/2. That's like saying the Bohr model of the atom is really informative as to what's going on, cuz it gets the right answer.
keradeek
2011-10-01 21:33:16 |
gamma*m*c^2, sorry
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| | pam d
2011-09-17 19:06:43 | I think down below evanb was the only one who gave a solution that involves general relativity. The Schwarzchild radius is R = 2M (unfortunately knowing this is a combination of what courses you have taken and/or a bit of luck) in c = 1, G = 1 units. You then figure out in SI you need a factor of  , simplify the algebra by using G = 7E-11 and M = 6E24, and arrive at about 1 cm.
That being said, I really like the plucky handwaving solutions. They demonstrate the type of thinking that will get you some extra points on this test. My favorite is the grape one. | | aqme28
2010-11-12 14:35:17 | This is a pretty poor solution, but in this case it works:
I realized the units of G are m^3/(kg*s^2)
Multiply G*M/R and you get units of (m/s)^2.
So by dimensional analysis I figured v is about the square root of GM/R
I got 4 millimeters, which is close enough to a centimeter. | | 1234
2010-09-03 11:34:19 | Set escape velocity = c:
Escape velocity = [(2*G*M)^(1/2)]/r = c. Solve for r.
x3818919209145
2010-11-11 06:30:12 |
Doesn't the r have to go inside the square root?
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| | 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.
Baharmajorana
2014-09-16 01:46:35 |
I like your solution, but we have to memorize the radius and mass of black hole!! I think the GREPHYSICS's solution is better.
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| | 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.
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|  | 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. ;)
Nebula
2015-09-16 22:29:35 |
I heard marble, but still works!
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| | 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...
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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. |
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