A theory for why Kerbin's gravity is disproportionate

[lajoswinkler]

Although a spacecraft theoretically could orbit a black hole, in reality only the exact center of mass of the vehicle would "feel" zero G. Any other area would feel a positive or negative gravity gradient due to being located elsewhere than the center of mass.

That's why NASA now uses the term "microgravity" for vehicles on orbit. First of all, the orbiting vehicle is perturbed away from a "perfect" orbit by solar wind, atmospheric drag, astronauts moving around, etc... but also only the exact center of mass feels "zero G".

Because gravity decreases with distance from the body, objects inside, say, the Space Shuttle, might feel +0.0001 G in one area of the spacecraft and -0.0001G somewhere else. That difference is the gravity gradient.

Now imagine orbiting a black hole. The gravity gradient might be something like +1000 G in one spot and -1000 G in a different spot. That gradient would certainly kill any living occupant orbiting a black hole, and would probably crush the spacecraft into a very small sphere (all points trying to get to the center of mass).

Crazy stuff....

Who says you've got to orbit it very close, where the tidal effects are very apparent? If you've got a 5x solar mass hole, you could orbit it at 20 AU without any problem. No tidal effects at metre level.

Problems with orbiting black holes come when you're relatively close to them. Other than that, they're perfectly normal bodies to orbit around and study with a spacecraft.

We're orbiting a black hole in the center of our galaxy. Do you see any tidal problems? I don't.

Ripping tidal forces at metre level appear way inside the event horizon for average black holes. You can pass the horizon without noticing anything except the visual psycho stuff with all the light distortions.

[SSR Kermit]

Gosh you are right. Need to finish my coffee before I post!

Thinking about it though, ANY place at the proper altitude for the orbital velocity the craft has would be the "zero G" point. So wouldn't the craft be PULLED into a thin "wire" (ultimately becoming a semi or complete RING) orbiting the black hole?

Doesn't need to be a black hole-- this is true for any massive body. The distance at which a second body will lose structural integrity due to tidal forces is known as the Roche Limit -- wikipedia has a sufficient article on it. Rings are usually the result of that kind of event. As I understand it a semi ring generally requires additional interactions such as clumping or jets.