# Black holes and extended horizons

1. Jun 24, 2010

### Kokitsunemaru

I'm currently writing a sci-fi/fantasy novel and although I know everything in this genre doesn't have to be technically accurate I still like to keep one foot firmly in the realm of reality and combine it with aspects of the fantastic.

At the moment I'm working on a plotline that involves a planet being swallowed by a black hole but on researching it have found that there's no way a planet that supports life would be able to exist near enough to a star to be effected by a black hole.

I understand that the range of a black hole's horizon depends on the mass of the star that existed before the final stages of it dying. I was wondering if it was possible for a rotating black hole's horizon to expand in some way?

If I can work some real life mechanics into this it'd be great but if its not scientifically possible is there any highly plausible if innaccurate theories that I could use the explain this away?

Some of you may say who cares its sci fi! Unfortunately I'm one of those people who likes to pick holes in things which inadvertently mean I'm writing for an audience that I expect to have similar bad habits. :uhh:

2. Jul 2, 2010

### The riddler

If you added mass to a blackhole its event hozion and overall gravitational attraction would increase in radius.

I don't really know what you are talking about when you say "on researching it have found that there's no way a planet that supports life would be able to exist near enough to a star to be effected by a black hole", blackholes do move around so it is possible for a star with an orbiting planet to be absorbed by one. Just to point out, if the sun turned into a blackhole this very second the orbit of the earth would still be exactly the same because the sun would have the same mass and pose the same gravitaitional force on the Earth

One way in which a blackhole could absorb your life baring star is through the colliding of Galaxies. (ex: The Milkyway and Adromeda galaxies are predicted to one day collide).

Hope this helps :)

Last edited: Jul 2, 2010
3. Jul 3, 2010

### stevebd1

You also need to take into account tidal forces that an object will experience as it crosses the event horizon which is the sum of radial stretching and sideways compression-

$$\Delta g= \frac{2Gm}{r^3} l_1 +\frac{Gm}{r^3} l_2$$

for a static black hole where l1 is the radial length of the object perpendicular to the BH and l2 is the width of the object. If l1 and l2 are equal then the equation can simply be written $\Delta g=3Gm/r^3$. For arguments sake, lets say that l is equivalent to the diameter of Earth and the largest change in gravity the Earth can experience from one side to the other is 1 earth g, if we substitute r for the Schwarzschild radius $(r_s=2Gm/c^2)$ then the equation to find the minimum size of a black hole that an Earth sized planet would survive crossing the EH would be

$$\Delta g=\frac{3Gm}{(2Gm/c^2)^3}l$$

$$\Delta g=\frac{3Gmc^6}{8G^3m^3}l=\frac{3c^6}{8G^2m^2}l$$

solve for m

$$m=\sqrt{\frac{3c^6}{8G^2\Delta g}l}$$

if Δg=9.81 (Earths gravity) and l=2 x 6.731e+6 (Earth's radius) then the minimum mass for the BH would be ~141.672 million sol, this is just to reach the event horizon. Maybe an Earth like planet can withstand greater tidal forces but this provides an approximation and Δg can change to suit. Note that this would get a bit more complex for a rotating black hole.

$$A=4\pi(r_+^2+a^2)$$
where $r_+=M+\sqrt{(M^2-a^2)}$, r+ being the outer event horizon, M and a representing the mass and spin of the black hole in geometric units where $M=Gm/c^2$ and $a=J/mc$.