Basically I'm wondering how big a rock in space would need to be to draw an average person to it with a gravitational force of a kilo or two. Say it was the average density of earth and it was spherical, what would its diameter be.
But you would als get closer to the center. Every time you halve the diameter, there will be 8 times less mass, the 1/r^2 in Newton's law. The end result is that an object with 1/4 times the diameter of Earth, will have 1/4 times the gravity, so you'll need 1/50 of the diameter.Well I'm pretty sure you would need 50 times less mass. Every time you half the diameter of a sphere you reduce it's volume by 8. So a sphere 1/4 the size of Earth would have 1/65th of the gravity.
Yes. He would see a horizon that's very near. Mars' horizon is something like a couple of kilometers, enough to be quite noticeable.basically I was wondering if a future astronaut landing on an asteroid that just had enough gravity to keep him on the surface would be able to tell he was on a small object by it's appearance.
The distance to the horizon on a perfect sphere is@willem2, ok so then that would be say a rock that was about 160 miles across. So an astronaut standing in the middle would see about 80 miles to his horizon.
See my last two posts on this thread.
Wouldn't it be the standard formula, but with the variable density replaced with just a number?There has got to be a formula that, for spherical objects of a fixed density, takes only mass and radius and returns the surface gravity.
No, not before your original comment in Post #11.Yes, 1. Had someone posted the formula? I did not see it.
And for a different density, you could complete the equation by plugging in the absolute density, or by ignoring the gravitational constant and dividing your final result by the relative density. (In other words, if your density is 0.8 that of earth, your final answer would be 128km/0.8 = 160km.In other words, a and r are proportional for a fixed density ρ.