

Subject:
Black Holes
Category: Science > Astronomy Asked by: periclesga List Price: $25.00 
Posted:
02 Dec 2003 11:44 PST
Expires: 01 Jan 2004 11:44 PST Question ID: 282710 
What is the relationship between black hole mass and circumference? Specifically, at what mass/circumference is the "surface gravity" (ie, acceleration)immediately beyond the event horizon 1 g? 

Subject:
Re: Black Holes
Answered By: haversianga on 25 Dec 2003 14:25 PST 
Good afternoon periclesga, Taking your questions one at a time, the primary issue becomes the definition of circumference. As hfshawga mentioned, a commonlyaccepted definition of the size of a black hole is the size of its event horizon (the boundary at which the escape velocity is the speed of light). Naturally, the actual size of physical matter inside the black hole can not be known (as no information can escape), and is likely immaterial. Various particles with velocities in some distribution will exist within the black hole. Those with greater velocities will come closer to the event horizon (on the inside, naturally), but there will be no solid object whose radius can be measured. The Schwarzschild Radius can be found by R = 2GM/c^2. G is the gravitational constant, M the mass of the black hole, and c the speed of light in a vacuum. See ( http://imagine.gsfc.nasa.gov/docs/science/know_l2/black_holes.html ) for more details. By definition, the gravity at the event horizon cannot be 1G  it must be sufficient that light cannot escape. In theory at least, the gravity may be 1G at an arbitrary distance outside the event horizon however. For example, a black hole exerting 1G on an object about 6.4megameters from its center would have the same mass as the earth. Google reports ( ://www.google.com/search?hl=en&ie=UTF8&oe=UTF8&q=2+*+G+*+mass+of+earth+%2F+c%5E2&btnG=Google+Search ) that the Schwarzchild radius of an earthmass object is just under 9mm, but we know that at the surface of the earth (about 6.4megameters from center of mass), the gravity we experience is what we call 1G. Haversian Search terms: Black hole radius Schwarzchild radius 

Subject:
Re: Black Holes
From: fstokensga on 02 Dec 2003 14:08 PST 
Black holes have the same gravitational effect as any other massive object. If the Earth turned into a black hole, the gravitational force would be unchanged. An "earth black hole" would have a gravitational acceleration of 1g at one Earth radius from the center. If you were in your house, and your house somehow stayed in place while the Earth turned into a black hole, you would feel exactly the same force of gravity as before. 
Subject:
Re: Black Holes
From: hfshawga on 03 Dec 2003 08:08 PST 
The "radius" of a black hole is usually taken as the Schwarzschild Radius, R_s, which is equal to: R_s = 2*M*G/c^2 where M is the mass of the object, c is the speed of light in a vacuum, and G is the universal gravitational constant. The Schwarzschild Radius marks the boundary of the sphere from which nothing can escape, i.e., the sphere with radius = R_s defines the "event horizon"). Any object, no matter what its mass would become a black hole if it were compressed down to a sphere with a radius less than the Schwarzschild Radius. As to the second part of your question, if you are asking what mass/size black hole would have a gravitational acceleration equal to 1g at its event horizon, then there is no such combination. Almost by definition, the acceleration at the event horizon is infinite. Note that you cannot simply use the Newton's law of gravitation (acceleration = G*M/R^2) to calculate the acceleration in this case; one must use a general relativistic formulation (acceleration = G*M/R^2*[1sqrt(12*G*M/R*c^2)]). 
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