Gravity on a hollow non-enclosed world

In general, the problem with hollow-Earth setups is that objects in hydrostatic equilibrium cannot be hollow - and planets must be in hydrostatic equilibrium. Planets form through collisions of smaller pieces of rock and dust, and eventually accrete enough matter to become substantially large. There is absolutely no way for a planet to form with a hollow interior; it should have been compacted during its formation.

In a typical hollow world that is fully enclosed, there should be virtually no net gravitational force on a person inside. For a spherical mass distribution, the shell theorem states that anyone inside a spherically symmetric shell should feel no net gravitational force from the shell. There will, of course, be variations in mass density, but they wouldn't add up to anything substantial.

A semi-open shell means that the shell theorem is not fully valid, but it does seem to imply that there will be very little gravity. Your image shows the planet about two-thirds enclosed, meaning that most of the field will be canceled out. Certainly all vertical component will be, assuming the two shells are identical.

Additionally, having the shell only partially enclosed means that the atmosphere has a chance of escaping to the outside of the planet through diffusion, where it will indeed feel a net force - centered on the center of mass of the planet, which is probably the center of the shell. There would still be some left inside, I would assume, but much could escape to the outside. I don't know how significant this could be, though.

Jason K suggested using rotation to keep things on the inside surface of the planet inside, via the fictitious centrifugal force. Here, the magnitude of the centrifugal force is the same as that of the centripetal force at the equator - the place where it is strongest - is $$|\mathbf{F}|=F=m\omega^2R$$ where $m$ is the mass of an object, $\omega$ is the angular speed of the planet, and $R$ is the radius of the inside of the shell. The magnitude of the angular acceleration is then $$|\mathbf{a}|=a=\omega^2R$$ Assuming that the planet rotates at the same speed as Earth, and its inner surface has the same radius as the Earth, then $$|\mathbf{a}|=\left(7.29\times10^{-5}\text{ rad s}^{-1}\right)^2\left(6,370,000\text{ m}\right)=3.39\times10^{-2}\text{ m/s}^2$$ That's at the equator, where the force is at its greatest.

The planet would need to spin about 17 times as fast or have a radius 289 times as large in order to have $|\mathbf{a}|=g$. It could spin at about 10 times Earth's speed and generate Martian-level surface gravity, though, which would be decently suitable for humans. Apparently, the fastest spinning exoplanet known, $\beta$ Pictoris b, spins at roughly three times Earth's speed.¹ That means that to generate Martian gravity, the planet would have to have a radius about 11 to 12 times that of Earth - making it the size of Jupiter! For what it's worth, measurements place $\beta Pictoris b at $\sim1.46$ Jupiter radii. No terrestrial planets will get anywhere near that large.

^{1 That said, we haven't measured the rotation periods of many exoplanets.}

posted almost 8 years ago

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