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Posted by Keith N. Tullis on May 29, 2004 20:38:35 UTC

Hi, sup?
Yea your right, at thoses distances an object that size would be like viewing a cell phone screen in New York from L.A.. Mabey more difficult. But that is not the point anyway. With that strong of a gravitational field, you couldn't see it if it were a million miles wide, right?; the gravity thing again.
Let's use an analogy. If a very large man dances with a very small girl and we view their dance from the balconey above it would be very easy to plot the circuler course of each dancer. It should be quite obvious that the little girl is influencing the circular path of the big man. His path is not perfectly round. It can't be as neither would the path of the little girl.
This same observation of the star we "can" see makes the same statement. By ploting it's course and monitoring the situation for a while it's easy to equate how much mass it would take to pull the visible star out of it's "Path Of Least Resistance". That value would represent the mass at the center of the black hole. And when visible object pass behind this region it is quite plain to see that they disappear when they should not. This being the region of the "Event Horizon", where no light will pass. this is not conjecture, it has been proven. I invite all members to check his equations. However I don't see what the orders of magnitude or the radius of the nutron relate to this topic.
As I've wrote before, I don't do equations, they hinder the concepual prosess. I would like to ask that you use your ability at equations to verify what I have stated. Find the data you need. Locate the position of the visible star. Theres help on the internet with that. If you can get the data on the variable path of this star you should be able to equate the mass of the black hole and determin it's diameter yourself and present us with some convincing numbers. I'm not quite that good.
Have to go. Be back soon.
--"After all is said and done, "Gravity Rules".--

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