The novel The Hitchhiker’s Guide to the Galaxy has a scene where Arthur Dent goes to a hyperspatial factory where planets are constructed. It made me wonder: What if I were in a spaceship inside that enormous sphere — and I fell out of the ship?
In other words, what would gravity be like inside an enormous, hollow, planet-sized sphere?
Would you:
a) get pulled to the precise centre of the sphere, and stay there?
b) get pulled toward the nearest section of wall, and go splat?
c) just float around wherever you are?
I was very pleased to find that last week’s Straight Dope column answered that very question.
Make your prediction, and I’ll see you in comments.
1 October 2011 at 1:29 pm
Isn't that counter-intuitive? It's not what I thought.
So it seems that, yes, you're close to some of the wall (and it's pulling you), but there's a lot more wall that's far away from you. It's pulling you, too, but it's farther away, so all in all, it balances out, and you're weightless.
1 October 2011 at 4:21 pm
This is elementary and I read about it in Asimov decades ago. The interior of a hollow sphere experiences no gravitational acceleration from the surrounding mass so everything is weightless.
Of course tidal forces from large masses orbiting outside the sphere will still be felt, as will the tidal forces of a star it orbits. Matter does not act like towards gravity like a "Faraday cage" does with electrical forces.
2 October 2011 at 3:40 pm
I and wife watched 'Journey to the Centre of the Earth' (newer version) last week, and I diatribed tediously about how there'd be no gravity in the centre of the earth (for the same reasons as for interior of sphere). Of course, in movie land, it was all earth-surface gravity pulling down, even tho there could be no down. a^{13}rgh!
17 June 2014 at 5:14 pm
The force on your mass is being pulled equally in all directions when you are in the center of the plant; hence, you experience weightlessness. If the size of the hollow center is sufficiently great, then as you approached a wall (e.g., by pulling on a rope tied to that wall0, you would experience a small attraction toward that wall as you approached it because the gravitational pull is inversely proportional to the square of the distance. Thus, as the distance from the wall decreases, the force of gravitation goes up while the force on your back diminishes. F=gmM/d2 @Newton
17 June 2014 at 5:35 pm
According to Newton, F=gmM/d2. Hence at the center of the earth the F of gravitation is exactly equal in all directions and you would experience this as "weightlessness." If the hollowness was sufficiently large, however, and you could approach a desired wall (perhaps by pulling on a rope attached to that point), then you would experience a small attraction toward the wall that would grow as your distance to the wall decreased. This force would cause you to accelerate toward that wall. Ft=mv If the wall was very close, you'd hit the wall with a light bump. If the wall was far away, you could develop a sufficient velocity to do real damage upon impact (since the v=velocity is directly proportional to the average F times the t=time). Since you are inside the sphere, the F (increasing) toward the wall is always countered by the F (decreasing) upon your back. Hence, the resulting F accelerating your body toward the wall is always far short of what it would be if you were outside the sphere.