Gravity differences on asteroid with an O'Neill cylinder

Question

In my story there is something akin to an Island Three O'Neill cylinder buried vertically into the side of the asteroid Vesta. It is 5 miles across and 20 miles long. The station is buried but is spun up inside of a cylindrical shell carved out of the asteroid. The whole of Vesta does not need to be spun. The station is buried for two purposes: to protect from cosmic radiation and to give better access to the miners.

Vesta's gravity is about 0.25 m/s^2 which (I think) is around 0.03G. For things to feel like 1G inside, the cylinder should produce 0.9995G artificially.

Now, (if I'm doing the math right) using arctan(0.03/0.9995) the inside of the cylinder would feel like it has a 1.72° slope to it, which doesn't seem like much, but...

1. Would this feel like a hill? Would walking far surface-ward wear you out faster than? Could you throw a ball further "downhill" at this angle? Or would it not even feel any different?

2. Would it make sense to build terraces periodically to level things out?

3. Would this noticeably affect air pressure or water flow across the gradient of the cylinder?

I'm trying to feel out if this would even be worth mentioning my story, or if the effect is so negligible that it wouldn't.

Answer 1

So, a force toward the more-or-less center of Vesta of 0.03g and a perpendicular force of 0.9995g. Earth experiences magnitude variations of up to 0.7% and Vesta's gravity represents 3%. I'm not sure people would feel this that all, but it is "sideways," which would be odd. Let's run with it.

OK (it's been a while since I did inclined plane physics, so I might be wrong)...

Potential Energy = mgd sin(θ)

• m = mass
• g = gravity constant
• θ = angle of the hill
• d = distance travelled (we're going to assume "1")

At 0° the contribution due to the hill is 0 J. We want a contribution equal to the affect of Vesta's gravity. F= ma.

m(0.03g) = mg sin(θ)

0.03 = sin(θ)

θ = 1.72° (which you calculated! I'm on the right track.)

So, living in your environment means feeling like you're constantly walking up and down a 1.72° hill. What's that like?

Well... that was a long and fancy way of saying you're stepping up 0.03 meters (30 cm) for every meter walked or just over an inch for every 3.28 feet walked.

It's almost nothing. I doubt people would even notice it. According to this bicycling site a 3% grade (which this represents, rise/run*100 = 0.03/1*100 = 3%) is like riding your bike into the wind (of course, they don't tell you what wind...) but not considered much of a challenge to cyclists.

OK, whether or not a 3% grade is an issue depends on what's happeneing. For a person walking, it's likely not noticable. For a car moving at 70mph, it represents a risk if a sharp turn occurs. For a train, it's a big deal. It really depends on how much mass is being moved. As mass increases, the grade of the hill becomes more important (especially downhill) because the energy needed to overcome the grade increases with it. A 3% grade won't cause my Toyota Prius to recharge. I'm just sayin'

Oooh. You had more questions. You can't "level things out" with terracing. It may feel like you're living on the side of a 3% hill, but you actually aren't. You can't change an angle to make the potential energy due to Vesta's gravity go away. You can thank the need to spin your cyclinder for that.

It does mean that if you spill a glass of water, it's going to want to dribble in the direction of Vesta's core. Remember, 3% isn't much. If you spilled it on a big sheet of glass you'd see it move, but if you spilled it on concrete you probably wouldn't.

It will mean a slight increase in air pressure toward the center of Vesta. But, once again, I doubt it would be noticable.