5 ms·
Very fun thought experiment. It reminded me that we did a problem in calculus to compute the effect of gravity inside a hollow sphere. It turns out to be ZERO
by mckoss 15y ago
Very fun thought experiment. It reminded me that we did a problem in calculus to compute the effect of gravity inside a hollow sphere. It turns out to be ZERO.
So, if the Earth's mass, was all densely concentrated in a, say, 1 mile thick shell, you could drill a hole through the shell and experience total weightlessness when you popped out on the "inside" (assuming a total vacuum on the inside - if not - you'd experience a very small gravity toward the center based on the mass of the contained atmosphere).
- sjwright 15y agoIf the inside was a total vacuum, the moment you finished drilling through the 1 mile shell, the vast majority of the planet's atmosphere would be sucked through the open hole at an insanely fast rate.
- hammock 15y agoOK so let's say you install an airlock in between the inner and outer crust.
- ithkuil 15y ago> [..] insanely fast rate. Well, at the same rate as air at 1 atmosphere pressure will fill any vacuum container through a hole of that size. You seem to suggest that fact that that speed somehow depends on the size of the hollow earth and the quantity of the atmosphere. The pressure depends only on the height of the air column.
- liquids 15y agoSo on the surface of the shell planet, the effects of gravity would be the same as normal, but when you're on the inside there is no net gravitational effect? ie. you'd just float near the inside surface? That doesn't seem intuitive. Wouldn't the force from gravity be acting towards the centre of mass in both cases? Edit: Actually never mind, I think the density of the shell is what threw me off! On the outside the net gravitational effect is the local shell (RHS) plus the rest of the shell (RHS), but just on the inside - the net effect is the remaining shell surface (RHS) minus the local shell (LHS). Because the effect of gravity is 1/r^2 everything works out!