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I've created a Neptune sized shell world with properties that I find are not typical of other shell worlds. Its gravity does not come from a gas giant or a black it was constructed around, but from the sheer mass of the shell itself; which for 1G, is 12 billion kg/m2 of matter for every m2 of living space.

This was done so that I could utilize a combination of radioactive decay and tidal heating to keep a molten "mantle" of rock for geological activity. And since the inner area of the shell world is...odd. Though it still produces a magnetic field.

If we ignore the inner area of the world and the composition of the shell material:

Can orbital mechanics work the way they as they normally would? Or does the fact that gravity originates from the upper surface only affect how orbits work.

Clarifications:

  • World is hollow with no core. What fills the space in the world is irrelevant
  • Gravity is produced by a layer of rock on top of the shell that is heated by radioactive decay and tidal heating to produce tectonic activity
  • The surface area of the world is that of Uranus; 8.083 billion km²
  • I'm mostly curios on the orbital paths of objects around the world and of the Roche limit.
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    $\begingroup$ You may be aware already, but this shell will immediately collapse under its own weight, leaving you with a molten lava world surrounded by a ring system a few minutes after starting the physics clock, unless magic / Sufficinetly-Advanced-Science is holding it spherical. $\endgroup$
    – g s
    Commented Apr 29, 2023 at 22:46
  • $\begingroup$ It’s a bit of both magic and science really. $\endgroup$
    – Seraphim
    Commented Apr 29, 2023 at 23:07
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    $\begingroup$ "That also means you walk on the inside of the shell and experience similar gravity to the outside." About that... $\endgroup$
    – Unrelated String
    Commented May 1, 2023 at 0:30
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    $\begingroup$ @gs: "leaving you with a molten lava world". Lava so hot it wouldn't freeze the molten Tungsten. Setting aside this issue and the no-gravity-on-the-inside issue, the inside also would be "thermodynamically flat". Very little heat can escape, so in order to avoid cooking very little energy can be fed into the ecosystem. The entire inside, minus a few "hot spots" would be dark, windless, extremely barren and at constant temperature. $\endgroup$
    – Kevin Kostlan
    Commented May 1, 2023 at 4:30
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    $\begingroup$ As Unrelated String and Kevin Kostian alluded to, another part of the Shell Theorem is that gravity is zero inside the shell. You cannot walk on the inside. As a hard-sf reader I would accept some kind of inner shell material to hold up the mantle, but not outward-pointing gravity in the shell. But it's your story! $\endgroup$
    – Mark Foskey
    Commented May 1, 2023 at 4:54

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The thing you probably care about is the shell theorem. Newton proved that

A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center.

and if Gauss and Newton say you can have things orbiting your shell world in the usual way, then you can probably assume it'll be OK.

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The orbit around the sun should be similar/identical to a denser object of the same mass. However, the orbit around the shell planet would be very different. The strength of gravity is dependent on F = G(m1m2)/R^2, where G is a constant, m1 and m2 are the masses of the two bodies, and R^2 is the distance between them. The shell world gravity of a Neptune-sized planet that's hollow (say, a 5 km deep shell) would be:

Neptune Diameter (ND = 50,000 km). Neptune Mass (NM = 1.0241 × 10^26 kg). Earth Diameter (ED = 12,756 km) Earth Mass (EA = 5.97 x 10^24 kg)

Neptune's earth-gravity at the surface is: 11.15 m/s². Despite being 12 times more massive than earth, the gravity is less than double, because the diameter of Neptune is much larger, and gravity is proportional to the distance between masses squared.

On your shell world, the shell mass (SM) would be: [(4/24)* pi* ND^3 - (4/24)* pi* (ND - 5km)^3]* Density.

SM = 1.96×10^10 m^3 * Density.

To have the same gravity as earth at the surface, your shell mass needs to be:

FR^2/Gm2 = m1, F = 9.8 N, m2 = 1 kg m1 = 9.177×10^25 kg.

Therefore, the density of your shell building material must be: Density = SM/(1.96×10^10) = 4.67×10^15 kg/m^3. For reference, the density of steel is 7840 kg/m^3.

The thicker the shell, the less dense your material would need to be, but the less shell-like it becomes.

If you didn't care about the gravity on the surface and made your shell out of a steel-like material (in density), then your shell would only have a mass of SM = 1.96×10^10 m^3 * 7840 kg/m^3 = 1.54×10^14, translating to a gravity of 1.6437214738×10^(−11) Newtons on the surface. This ISS is 450,000 kg. The ISS, at the surface, would experience only 0.0000074 (7.410^(-6)) Netwons of force. That's such a small force that essentially nothing would be able to orbit the shell world. On the upside, it would also be very easy to land and take off from the surface, and the structure itself wouldn't collapse from its own gravity well. Orbiting around the shell world isn't even needed because many spaceships ship could generate 7.410^(-6) newtons of thrust for years on end without expending more than a hundred kg of ion fuel.

In conclusion, the shell either needs to be made of extremely dense material, or it shouldn't have a noticeable gravity on its surface, meaning nothing could orbit it.

This shell would have no Roche limit, as it produces essentially no gravity itself. However, it would be very susceptible to other celestial bodies' forces.

If any of my math is wrong, I apologize. Please let me know and I will do my best to correct it.

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  • $\begingroup$ You're assuming a thin shell. My earlier forays into shell world math say that you need a shell as thick as what you're mimicking--you want Earth gravity, you need a shell of about 6,300km (the radius of the Earth) thickness. $\endgroup$
    – Loren Pechtel
    Commented May 1, 2023 at 15:08

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