Spindown time The amount of time, t, for a moonless planet to spin-decelerate from an initial rotational period P (in seconds) to a 1:1 spin-orbit tidal lock with the star it orbits. t = ωa⁶IQ / (3GM²k₂R⁵) ω = 2π/P a = semimajor axis of the orbit, in meters (for Earth, a=1.495978707e+11 meters) I = planet's moment of inertia (for Earth, I=8.038e+37 kg m²) Q = planet's tidal dissipation factor (typically Q=50, for Earth Q=13) G = 6.6743e-11 m³ kg⁻¹ sec⁻² (CODATA 2018) M = star's mass in kilograms (for the sun, M=1.9885e+30 kg) k₂ = planet's Love number (for Earth k₂=0.308 minimum, 0.353 maximum) R = planet's equatorial radius in meters (for Earth, R=6.378e+6 meters) Earth's spindown time from an initial 8-hour rotational period to a 1:1 tidal lock is about 10¹⁸ sec or about 31 billion years, according to the equation. However, there's a problem. When I calculate the time to go from an initial 24-hour rotational period to tidal lock, I get only 10.5 billion years, implying that Earth took more than 20 billion years to brake from an 8-hour period to a 24-hour period, and Earth's age is only a quarter of that. So there's something wrong with. . . oh. It's the moon.
@deepastronomy Жыл бұрын
I recognize the equation guy! You used to have the handle David Simms if I recall. Welcome back!
@Jenab7 Жыл бұрын
@@deepastronomy Yep. David Sims. Thanks!
@areamusicale Жыл бұрын
1:47 LOL what? 🤣 Perhaps you mean: depending how close is the planet to its star, the surface may be covered in volcanos. I don't think a planet can surface can reach 2500ºC just be the light it receives from the sun. No! That's too hot for an atmosphere on a small rocky planet to exist, and without atmosphere the heat dissipates.
@deepastronomy Жыл бұрын
We're talking about tidally locked exoplanets here close to red dwarf stars. Of course a planet can get high temps by the radiation from a star with plenty of volcanic activity on the day side. LOL