quote:
Originally posted by apache:
First off, yes, the 3520 was a typo, one I noticed and thought I fixed, but apparently did not.
Now, no no no, density has nothing to do with gravity. It is only distance and mass, and density is a function of mass and volume. At a distance of R from some mass M, the gravitational acceleration at that distance is constant, no matter how dense the mass is.
If you have a black hole with the mass of the sun, it will definitely be a few million times denser than the sun. However, the gravitational force on the earth from the black hole will be exactly the same as the force the sun has on the earth, if the earth orbited a black hole at the same distance it orbits the sun.
Nice try at changing the ground rules, but we're talking about the gravity at the surface of a celestial body. The gravity at the surface of the black hole (however you define that) is nearly infinitely greater than the gravity at the surface of the sun because the black hole is nearly infinitely denser than the sun and thus has an infinitesimal radius. In the equation using mass and radius, this shows up as a very small r resulting in a very large gravitational force.
I realize, of course, what you are trying to say, but describing the phenomenon using only mass and radius is awkward. You could say: "Well, Uranus is very big and has a very big mass, and gravity only depends on mass and size, but its gravity is less than Earth's because . . . uh . . . well, the ratio of the mass to the size is not all that big." Since the ratio of the mass to the size is the density, the explanation becomes "Uranus has low surface gravity because of its low density." Simple, understandable and absolutely accurate.
Re: Atmosphere
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