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对话内容[edit | edit source]

Moon parking and day length[edit | edit source]

Generated: 2025-11-08 22:17:41

Prompt: Would parking a moon in earths orbit make the day longer?

ChatGPT: Yes — adding another moon (or “parking” one) in Earth’s orbit could make Earth’s day longer over time, but not immediately.

Here’s why:

The tidal interaction between Earth and its moon transfers rotational energy from Earth to the moon’s orbit. This slows Earth’s rotation slightly while pushing the moon farther away.
If you add another sizable moon, it would generate additional tidal forces, further enhancing that braking effect.
The magnitude depends on the new moon’s mass and distance — a large, close moon would noticeably increase the rate of tidal braking; a small or distant one would barely matter.

So:

Short-term (immediate): The day wouldn’t suddenly lengthen.
Long-term (millions of years): Yes, the day would gradually become longer, just as Earth’s current 24-hour day used to be about 18 hours long when the Moon was much closer.

Would you like a rough estimate of how much longer the day could get for a second moon of a given size?

Prompt: Yes, I would like to make the day an hour longer.

ChatGPT: Nice — let’s make the day exactly 1 hour longer (24 → 25 h). I’ll show the physics, a simple calculation, and what it means in plain terms.

What has to change

Earth’s rotation has to lose angular momentum. To slow the 24-hour spin to 25 hours Earth must lose

ΔL≈2.33×1032 kg\cdotpm2/s.\Delta L \approx 2.33\times10^{32}\ \text{kg·m}^2/\text{s}.ΔL≈2.33×1032 kg\cdotpm2/s. How to store that angular momentum in a moon

The orbital angular momentum of a circular satellite is Lorb=mGM⊕aL_{\rm orb}=m\sqrt{G M_\oplus a}Lorb=mGM⊕a.
If you give the entire ΔL\Delta LΔL to a single parked moon at orbital radius aaa, its mass would need to be

m≈∣ΔL∣GM⊕a.m \approx \frac{|\Delta L|}{\sqrt{G M_\oplus a}}.m≈GM⊕a∣ΔL∣. Using that formula for three representative parking radii:

Near current Moon distance (a ≈ 384,400 km): → m≈6.0×1020m \approx 6.0\times10^{20}m≈6.0×1020 kg (≈ 0.8% of the Moon’s mass) If made of rock (~3000 kg/m³) that’s a spherical body ≈ 720 km diameter (roughly Ceres-sized).
Geostationary radius (a ≈ 42,164 km): → m≈1.8×1021m \approx 1.8\times10^{21}m≈1.8×1021 kg (≈ 2.4% of Moon’s mass) — bigger, because closer orbits carry less angular momentum per kg.
Low Earth orbit (7000 km): → m≈4.4×1021m \approx 4.4\times10^{21}m≈4.4×1021 kg (≈ 6% of Moon’s mass) — impractically large and also unstable / destructive for LEO.

Interpretation & important caveats

Those numbers show how much orbital angular momentum is required to change the day by one hour. A single object roughly the mass/size of a large dwarf planet (hundreds of km radius) placed at lunar distance contains about the right orbital angular momentum.
But placing such an object in orbit does not automatically slow Earth instantly unless the angular momentum is transferred from Earth into that object. How the transfer happens matters: * If you could magically assemble the moon already carrying that orbital angular momentum (i.e., bring it in from far away with the required L and park it), Earth’s rotation would change by conservation of total angular momentum — the numbers above apply directly. * If instead you “park” a body and expect tides to do the work, tidal exchange is usually very slow — it could take millions of years unless the moon is massive and/or very close (which brings huge stability and safety problems).
Real-world complications: three-body dynamics with the existing Moon, orbital resonances, tidal heating, long-term orbital stability, collision risk, and changes to tides/climate. A body hundreds of km across in near-Earth space would be a massive hazard.

Short takeaway: to gain one extra hour you need to remove ≈ 2.33×10322.33\times10^{32}2.33×1032 kg·m²/s from Earth’s spin. That is comparable to giving that angular momentum to a moon roughly 6×10²0 kg if parked at the current Moon’s distance — roughly a ~700 km diameter rocky body. Whether you achieve that by magically placing such a moon (instant effect) or by slow tidal braking (very slow) changes the feasibility and consequences dramatically.

Want me to:

convert those masses to familiar objects (Ceres / big asteroids), or
estimate how long tidal transfer would take for a given moon mass and distance (I can do a rough timescale estimate next)?