 Blackholes Forum Message Forums: Atm · Astrophotography · Blackholes · Blackholes2 · CCD · Celestron · Domes · Education Eyepieces · Meade · Misc. · God and Science · SETI · Software · UFO · XEphem Be the first pioneers to continue the Astronomy Discussions at our new Astronomy meeting place...The Space and Astronomy Agora Explanations : Orbitting A Black Hole And The Event Horizon Forum List | Follow Ups | Post Message | Back to Thread Topics | In Response ToPosted by Mark Chapman/">Mark Chapman on September 24, 1997 05:47:35 UTC

: the equation g=v squared/r is wrong. : traveling at a speed of 0, there would be no gravitational pull.

Wrong, here is an explanation.

The equation g=v^2/r is the simplification of the net force on a satellite. Two forces are involved: 1) gravitaional 2) centripetal

Gravitational Fg = G (m[satellite]*m[black hole])/(r^2) simplify g = G*m[black hole]/(r^2) assumes r is constant thus Fg = m[satellite]g

This force is pulling the satellite towards the black hole, and the countering force, centripetal, is what keeps the satellite from falling into the black hole.

Centripetal Fc = m[satellite]*(v^2)/r

So, if the satellite is in steady orbit --

Fg = Fc m[sat.]g = m[sat.](v^2)/r

divide out the m[sat.]

g = (v^2)/r

The conclusion of this is that the satellite must continue at the velocity v in order to maintain it's current orbit. A higher velocity would increase it's force away from the black hole and the satellite would accelerate away, and a smaller velocity, 0 would be smaller, would mean the centripetal force is no longer large enough to hold the satellite in it's orbit. So, the satellite would begin falling into the black hole. Too bad, heh!

------------------------------------------------------------

Using this to explain the event horizon:

At a certain distance from the black hole, a velocity equal to the speed of light would be necessary in order to continue an orbit of the black hole. Since our current theories predict that the speed of light is an impassable barrier, no particle could escape the event horizon. Equations:

Fg strong enough so that:

g = (c^2)/r

Since we defined g as G*m[black hole]/(r^2) the distance r from the center of the black hole that solves the equation:

r = G*m[black hole]/(c^2)

would be the distance of the event horizon from the center of the black hole.

Don't die!

Mark Chapman

Astronomy-Physics Undergraduate student at the University of Wisconsin-Madison

e-mail : machapma@students.wisc.edu

Follow Ups:

Login to Post
Additional Information Web www.astronomy.net About Astronomy Net | Advertise on Astronomy Net | Contact & Comments | Privacy Policy
Unless otherwise specified, web site content Copyright 1994-2019 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