Classical Physics to Calculate Rotation Periods of Planets and the Sun

Abstract

The rotation period of the Earth was calculated from the fundamental quantities, mass, distance, and radius, of the Earth-Moon system by almost an exact number 24^h3^m5^s. The rotation periods of Mars, Jupiter, Saturn, Uranus, Neptune, and the Sun were also calculated by the same equation. The Earth spin axis which is inclined by 23.45° with respect to the Earth orbit was derived from the gravitation of the Sun acting on the Earth and calculated by almost an exact number, 23.487°. An optical experiment to measure the reaction torque on the Earth acted by the Moon is proposed and discussed.

Appendix

9.1.1 Validity of the Initial Velocities Being Zero

The assumption in Eq. 9.4 that initial velocities v ₁₀, v ₂₀ which had not been driven by the forces f ₁ and f ₂ will be damped out may be validated by estimating the time to take for an initial velocity to decrease to a hundredth of its initial value. The increased radius, ΔR _e, of the Earth by the tidal force of the Moon can be inferred by an equation:

$$ -\frac{G{m}_1}{\left({R}_{\mathrm{e}}+\Delta {R}_{\mathrm{e}}\right)}\approx -\frac{G{m}_1}{R_{\mathrm{e}}}+\left(\frac{Gm{}_2}{d-{R}_{\mathrm{e}}}-\frac{G{m}_2}{d}\right), $$

(9.35)

where the gravitational potential at the increased radius is equal to the gravitational potential increased by the tidal force of the Moon. The increased radius is calculated ΔR _e = 21.6 m. The volume of the increased radius is given by ΔV ≈ (4πR _e ²ΔR _e) (1/3) (1/2), where the factor 1/3 comes from an increase in one direction among three perpendicular directions and the factor 1/2 from a rough average of the increased radius. The mass is given by Δm ≈ ΔVρ _e with the average density ρ _e of the Earth. For more accuracy, the radius of the Earth can be replaced by a mass-averaged radius R ^′ _e  = [∫rρ _e(4πr ² dr)]/m ≈ (3/4)R _e to use the average density ρ _e of the Earth instead of the surface density of the Earth. As the Earth rotates, gravitational potential of the mass Δm increases and decreases twice a rotation by height ΔR _e ′ ≈ ζ(3/4)ΔR _e = ζ × 16.2 m, where ζ is a constant representing the difference between the real increased radius and the calculated one here. The whole volume of increased radius per rotation may be approximated by 2 × ΔV, because there are two perpendicular directions on the equator plane. The potential energy loss per rotation is, approximately, ΔW ≈ η2 ⋅ 2 ⋅ Δm ⋅ g ⋅ ΔR _e′, which must be a loss of the kinetic energy of the Earth per rotation. The constant η is a nonconservative potential energy loss rate as the mass Δm moves by the increased radius. The number of rotations to lose all the kinetic energy of spin is E _k/ΔW ≈ (1/2)Iω ²/ΔW. If the initial velocity v ₁₀ makes 100 rotations per year, then the number is 5.86 × 10⁶ for ζ = 1 and η = 1. It takes 2.7 × 10⁵ years, many orders less than the Earth age, for the initial spin to be damped to one rotation per year, where R = R ₀exp(−Ct), N ₀–100 = N ₀exp(–Ct ₀), R ₀ = 100 rotations per year, and R = 1 rotation per year to calculate the time, N ₀ = 5.86 × 10⁶ and t ₀ = 1 year to calculate the constant C, are used. Even for ζ = 0.1 and η = 0.1, it takes 2.7 × 10⁸ years, still an order less than the Earth age.

As the same calculations are made with the reduced two-body systems of planets, for ζ = 1 and η = 1, it takes 1.77 × 10⁷ years for Jupiter, 4.55 × 10⁷ years for Saturn, 6.98 × 10⁷ years for Uranus, and 1.93 × 10⁷ years for Neptune, for 100 rotation per year of a initial spin to be damped to one rotation per year. For the Sun, it takes 1.93 × 10¹³ years, but if the imaginary planet were in a distance of Mercury in early stage of solar system, it would take 5.63 × 10⁷ years. For Mars, it takes 1.46 × 10¹² years calculated from its satellites. If the tidal force of the Sun was taken into account in much closer distance at the early stage, it may take much shorter time by many orders, which may apply to other planets. Although it is a very rough calculation, the assumption that if they had not been driven by torques, initial velocities were damped out to negligibly small values may be validated.

One further validity may be inferred from the spins of satellites in Table 9.3 where all major satellites but Hyperion are rotating synchronously to their orbital periods. It proves that if their initial spins were much faster than the synchronous spins, kinetic energy losses of satellites due to gravitations of mother planets were large enough for the spins of the satellites to decrease to the synchronous spins. If it is the same for the planets, the remaining initial spins of the planets are, at most, smaller than the synchronous spins.

9.1.2 Influence of the Fast Rotating Planets on the Spin of the Sun

If the Sun acts torques on the fast rotating planets, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, in the direction to decrease the speeds of spins, the reaction torques would make the Sun rotate faster than the planets in the opposite direction to the current spin of the Sun. A possible reason why it does not happen may be explained by a small or negligible magnitude of the torques being acted by the Sun. With reference to Fig. 9.2, for the gravitation of the Sun to make a torque on the Earth, the volume of increased radius must be in a certain angle between 0° and 90° or −90° with respect to the line connecting two centers of the Sun and the Earth. As the Earth rotates so fast, the increased radius may appear in an angle of ineffective direction or the volume of increased radius may spread over all 360°, due to the retarded response of the Earth materials. In this case, the Sun would not make a realistic torque on the Earth, which would be the same with the other fast rotating planets.

To explain the reaction torque on the Sun to hold the current spin of the Sun, the fast rotating planets would have some elliptic shape of orbital trajectories with their own satellites due to the tidal forces of the Sun. As the planets rotate around the Sun, the long axis of the elliptic orbit would always direct toward the Sun just as the synchronous rotation of the Moon. If the long axis of elliptic orbit deflects from the line connecting two centers of the Sun and the planet, then the gravitational force of the Sun would act a torque to make the elliptic orbit rotate synchronously with the orbital motion of the planet around the Sun. From this torque, there should be a reaction torque on the Sun to hold the current spin.