Back to Home

Blackholes Forum Message

Forums: Atm · Astrophotography · Blackholes · Blackholes2 · CCD · Celestron · Domes · Education
Eyepieces · Meade · Misc. · God and Science · SETI · Software · UFO · XEphem
RSS Button

Home | Discussion Forums | Blackholes I | Post
Login

Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...
The Space and Astronomy Agora
Re: Hypothetical Gravity Questions

Forum List | Follow Ups | Post Message | Back to Thread Topics
Posted by A Broderick on May 19, 1997 09:21:42 UTC

First, the destructive force associated with a black hole is not due to the magnitude of the gravitational field, but its variance in space. For the same reason you feel weightless when dropped off a building, you would be weightless freefalling into a black hole. It is the "tidal" forces that cause the destruction of matter in a black hole. These tidal forces are due to the difference in magnitude of the forces acting at different radii. Because of the comparably low density of the earth, we don't notice this effect here. The real question is at what point does this difference exceed the limits of human endurance. Clearly we can make indirect use of the facts previously presented, such as blackouts at 9g's. Now we must make a few approixamations, namley, average hieght~2 meters, spherical symetry in the black hole (this is not a perfect assumption, rotation effects things a little but that complicates the matter greatly), and that the average person is rendered sensless at 100 Newtons (about 10 g's). Then from General Relativity, we get dG/dr=-rs(7rs/r^2-4rs^2/r^3-2/r)/2r^2 where G is the gravitational force, rs is the schwarzchild radius (event horizon), and r is the radial position of our unfortunate fellow who is at rest relative to the black hole. dG/dr is the rate of change of force with respect to radius, thus when dG/dr>50 our friend is no longer with us. Because solving this for r is not simple, merely graphing it yields that approixamatly the minimum radius which our person can survive at is the square root of rs. This is not exact, but fairly accurate at low rs. In your question you stated that you believed that a person would die long before the event horizon, but here we see that this is not true. You misconception might follow from a misunderstanding of what the event horizon is. The event horizon is not!! the surface of the black hole, black holes are pointlike. The event horizon is the radius at which even light cannot escape. It is important to understand that one could pass through the event horizon and notice nothing different, that is until they try to escape. Objects inside the event horizon are doomed to fall into the black hole, but they are not neccessarily destroyed yet. It will appear to observers outside the blackhole that the person would take an infinite amount of time to reach the event horizon, but, the traveler himself will note no change as he happily passes the point of no return. My calculations are valid only when the velocity is zero with respect to the black hole, there is ofcourse corections for non-zero velocities, but I think that this should at least get you started. If you are really interested you should read something on General Relativity, this is vital to the understanding of black holes.

Follow Ups:

    Login to Post
    Additional Information
    Google
     
    Web www.astronomy.net
    DayNightLine
    About Astronomy Net | Advertise on Astronomy Net | Contact & Comments | Privacy Policy
    Unless otherwise specified, web site content Copyright 1994-2024 John Huggins All Rights Reserved
    Forum posts are Copyright their authors as specified in the heading above the post.
    "dbHTML," "AstroGuide," "ASTRONOMY.NET" & "VA.NET"
    are trademarks of John Huggins