# Approaching to black hole?

1. Aug 21, 2006

### hasman

Hi,
I am a newbie.I wonder how close to a hole a star(like sun) could approach before being torn apart by tidal forces?

this black hole has a mass 100 times of sun's mass.

How can we solve that?

From kepler's third law?

or anything else...

2. Aug 21, 2006

### DaveC426913

Roche limit

3. Aug 21, 2006

### hasman

but how?
I really can not figure it out

4. Aug 21, 2006

### Labguy

You can get a decent explanation at http://en.wikipedia.org/wiki/Roche_limit" [Broken].

Last edited by a moderator: May 2, 2017
5. Aug 22, 2006

### Theor

im sorry everyone this is totally mindless and unintelligent but it just crossed my mind.. wouldnt it be great if they made a movie about our galaxy/solar system falling into a black hole? like a "war of the worlds" type approach to the movie.. like show local reactions to the events through a single character, and then have sweet 3d visuals of the earth being sucked from existence!!! i would see it.. for sure.. but anyways, continue on, pardon my interruption :D

6. Aug 22, 2006

### hasman

from

d = r (2M/m) 1/3

we know that Schwarzschild radius of 100 Msun is 300km

d = 300 (2 x 100Msun / Msun)1/3

= 1754 km

is this true?

7. Aug 22, 2006

### DavidBektas

I think you have to insert the Radius of the sun for r (look at the link). Thats why the formula is often given in the form d=R*(2D´/D)^1/3 so that you don´t need to know the radius of the satellite. Other than that it should be correct.

8. Aug 22, 2006

### hasman

but how can we find the density of black hole?

9. Aug 22, 2006

### Zeit

Hello,

May be I'm wrong, but I try it anyway

First, you have to find Schwarzschil radius :

R_s = 2GM/c²

So, the density of the black hole is :

rho = M/V = M/(4*pi*R_s³/3) = 3*c^6/(32*pi*G³*M²)

Last edited: Aug 22, 2006
10. Aug 22, 2006

### hasman

nope not again I think

Because I tried it with the samples I found on internet
it says for 8 solar masses ,400 km
6 solar masses, 5300 km
but when I calculated from the formulas it does not match (both of them)

(but I am not sure if they are ok or not :grumpy: 400 km seems not ok 5300 km is more sense.)

P.s: by the way are black holes sphere?(4/3 pi r³) for volume is sense?

Last edited: Aug 22, 2006
11. Aug 22, 2006

### Zeit

Hello,

Schwarzschild black hole is the simpler model of a black hole : no rotation, no electric charge. There is no many chance to find this kind of black hole I thing...

12. Aug 22, 2006

### DavidBektas

Hey Hasman, you don´t necessarily need the density. Just insert the radius of the sun for the radius of the black hole in the above calculation and you are there!

And I believe the calculation for the density is correct.

Sorry for my english...

13. Aug 23, 2006

### hasman

so the formula I should use is d = r (2M/m) 1/3 ?

so my calculation above is ok?

This question should not be hard :(

14. Sep 4, 2006

### navneet1990

if thw mass od a black hole is m
then what will be its density/????

15. Dec 1, 2006

### Chris Hillman

Relativistic Roche limit? How about Newtonian Roche limit?

Hi, hasman,

DWR (Danger Will Robinson)! Wikipedia is unstable and lacks any effective quality control. My impression from hanging around Physics Forums for a few weeks, compared with my experience last year at Wikipedia (see http://en.wikipedia.org/wiki/User:Hillman/Archive) is that the general relativity board at PF tends to have better quality control, at least right now.

This generic warning aside, the version of this Wikipedia article which I am looking at, http://en.wikipedia.org/w/index.php?title=Roche_limit&oldid=91178742, does seem to give the right idea: set the magnitude of the tidal acceleration pulling apart antipodal bits of star, at a given distance from the massive object (call it a black hole if you like), equal to the magnitude of the acceleration on a bit of star at the surface of the star, due to the self-gravitation at the surface of the star. It seems appropriate to me to use Newtonian approximations for the tidal force and acceleration due to self-gravitation of the star, for reasons I'll mention in a moment.

Then we obtain $$m/r^2 = 4 r \, M/R^3$$ whence
$R = \left( 4M/m \right)^{1/3} r$
Here, on the left, I used the magnitude of the radial outward acceleration of a static fluid particle at radius r in a mass m Schwarzschild solution
$\frac{m/r^2}{\sqrt{1-2m/r}} \approx m/r^2$
and on the right I used the magnitude of the radially pulling tidal tensor component as measured by a static observer at Scharzschild radius R for a mass M Schwarzschild solution, namely $$2M/R^3$$, times the diameter of the star.

Since gtr is a nonlinear theory, this is clearly only a rough approximation! If you wanted a truly relativistic computation, you'd need to solve the vacuum EFE for two approaching massive bodies. (There is a solution, the so-called double Kerr solution, which claims to answer to this description, but there are good reasons to be cautious about accepting this interpretation!)

Since in ignoring nonlinearities and "$$v^2$$ effects", I secretly made pretty much the same assumptions as those used to obtain the Newtonian limit of gtr, it is not surprising that I obtained the same value for the critical distance R as given in the article on the basis of Newtonian theory (allowing for the fact that even in Newtonian theory, it seems to me, one should really use the radially pulling Coulomb component of the tidal tensor, not the orthogonally compressing ones).

Plugging in values for M = 100 solar masses, m=1 solar mass, r= very roughly 1 solar diameter, or $$M = 150 \, \rm{km}, \; m = 1.5 \, \rm{km}, \; r = 10^6 \, \rm{km}$$, I find $$R \approx 7.4 r$$, or about 2200 km, which essentially agrees with your computation.

Found WHERE precisely? If you found a figure at Kip Thorne's website (say) which differs drastically from our computation here, you should worry (although you might simply be misinterpreting something if so, so give a link, please). If you found a figure at some website put up my some anonymous nonprofessional physicist, you need not worry.

BTW, you can find a truly relativistic computation in this paper: http://www.arxiv.org/abs/gr-qc/0501084; [Broken] a quick glance suggests that their computation agrees approximately with the back-of-the-envelope estimate above.

Danger! Danger! DavidBektas and Zeit, hasman was right to be suspicious of this: one should really should avoid speaking of the "density" of a black hole. In this case, I guess this is a self-cancelling error, since we wound up using a Newtonian analysis to get a ball park figure.

Chris Hillman

Last edited by a moderator: May 2, 2017