I have a colonized asteroid that is roughly spherical and has been hollowed out. To create artificial gravity I spin the asteroid creating a strip of 1g gravity around the "equator". What would happen as I walk towards the poles? At the poles is there any effect at all?

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    $\begingroup$ The gravity would lower until it reached 0 at the poles. If you stood at the pole, its just going to spin you in a circle and if someone was messing with you, they could just leave you in line with the pole, but out of reach and you would just float there until someone saved you. $\endgroup$
    – Shadowzee
    Commented Dec 13, 2018 at 0:37
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    $\begingroup$ Gravitational force should be same everywhere on the surface of ur spherical asteroid, it is the centrifugal force that matters... $\endgroup$
    – user6760
    Commented Dec 13, 2018 at 4:00
  • $\begingroup$ @Shadowzee, please don't put partial answers in the comments $\endgroup$
    – Separatrix
    Commented Dec 13, 2018 at 8:47
  • $\begingroup$ How are you going to stabilize the sphere? If you spin a large hollow sphere, the tension of its shell will be huge. Just imagine a simple rope bridge such that it's curvature matches that of your sphere's equator. Now put everything on that bridge that's supposed to be inside your sphere. For each kilogram you add, many kilograms of tensile stress are put on the bridge's material. The mass of the bridge itself just adds to this total tension. You cannot just hollow out an asteroid and spin it up, hoping that it'll stay one piece. Cause it won't stay one piece. $\endgroup$ Commented Dec 15, 2018 at 20:44

3 Answers 3


What would happen as I walk towards the poles?

As soon as you leave the equator you are walking uphill. The slope gets steeper and steeper until it becomes a vertical cliff at the poles.

At the same time, the higher you climb, the weaker effect of the spin-induced gravity.

The artificial gravity effect is proportional to the radius times the square of the rotation rate. Whatever the size and spin of your world, as you climb the gravity effect falls off evenly with altitude. Note however that altitude increases more slowly near the beginning of your climb.

The angle of the slope you are climbing is proportional to the arc-cosine of your radial distance from the axis, and is directly proportional to the distance you have traveled along the inside surface of the sphere.

At the halfway point on your track, the slope is already one-in-one (or 45 deg) but the gravity is still 70% of equatorial. When you have reached half your maximum altitude, the perceived gravity is 50% of equatorial, and the slope is two-in-one (or 60 deg).

This diagram charts the changes in slope, altitude, weight, and climbing difficulty as functions of distance along your journey.

necessarily dimensionless I assume you had the foresight to install a funicular rail, or at least a ladder, to the pole, so I have defined climbing difficulty as the work required to move a unit distance along the track of your climb.

As you would expect, you will encounter maximum climbing effort at the middle of your climb, while the beginning and end will be dead easy. The interesting fact is the symmetry of the difficulty curve. Before I did the math, I thought the effort would be skewed toward the end of the climb.

At the poles is there any effect at all?

The spin-induced gravity effect is zero at the poles.

  • $\begingroup$ Thanks for the graph, that really helps me visualise it. $\endgroup$ Commented Dec 15, 2018 at 23:42

You would be hard pressed to get to poles, you would need to use equipment for that.

As you go towards poles you would feel how centripetal acceleration gets weaker as you get closer to rotation axis, at the same time floor would "transform into" slope >>> steep mountain >>> cliff.

Lower gravity will make it physically easier to climb all that, but at the same time you would loose traction and make it super easy to launch your self off that slope. You can use climbing equipment, suction cups, rope tracks or other help, dependent on suitability for your case.

If you get to the pole, you would slowly spin in zero-G, don't forget a safety rope or means to land safely(parachute, glider), as should you shift your center of mass off axis you will start your drop into that direction.

Make note, be careful with spinning asteroids. Most of them are rocky, uneven, with a lot of defects. Hollowed asteroid will tend to fly apart. So You would need to make a lot of work to strengthen it, deal with pockets of volatile substances, especially, around equator.

Metallic asteroids are a thing and metals are way superior in tensile strength.


You would have zero gravity at the poles if you chose to only spin it around one access. one way you could potentially fix this is by having the asteroid spin to create a band of 1g along the prime meridian as well as the equator. it might be a little chaotic, but it would solve the gravity problem at the poles

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    $\begingroup$ Wouldn't that just be spinning along a diagonal line? (Just creating a new "equator" with higher gravity than 1 g?) $\endgroup$ Commented Dec 13, 2018 at 18:53

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