Astronomy 217


Prof. Andrew W. Steiner

Sep. 17, 2021

TA James Ternullo

Last Time

  • Aberration of Starlight
  • Rotating Coordinate Systems
  • Corrections to Solar System Dynamics
  • Tidal Forces


  • Tides and Offsets
  • Impact on Earth and Moon's Angular Momentum
  • Tidal Locking
  • Librations
  • Extreme Tides
  • Tidal Disruption
  • Roche Limit
  • Hill Radius
  • Exams 9/29 and 10/27.
  • +1% EC for visiting office hours

Tidal Force

  • For the case of the Moon orbiting the Earth, the tidal force on the Earth $$ \vec{\Delta} F_{\mathrm{tidal}}(r) = \vec{F}_{\mathrm{moon}} - \vec{F}_{\mathrm{moon}}(r_0) $$ for a point at latitude \( \theta \) and longitude \( \lambda \) is $$ \vec{\Delta} F = \frac{G M_{\mathrm{moon}} m R_{\oplus}} {r^3} \left[ 2 \cos \theta \hat{i} - \sin \theta \hat{j} - \sin ( \lambda - \lambda_{\mathrm{moon}} ) \hat{k} \right] $$
  • Where R⊕ is the Earth’s radius, r is the distance from the center of the Earth to the Moon and \( R_{\oplus}/r \ll 1 \)

Tidal Comparison

  • Compare relative strengths of solar and lunar tides to the forces on the Earth $$ \frac{\Delta F_{\mathrm{moon}}}{F_{\mathrm{moon}}} \approx \frac{2 R_\oplus}{r_0} \approx \frac{1}{30} \qquad \frac{\Delta F_{\odot}}{F_{\odot}} \approx \frac{2 R_{\oplus}}{a_{\oplus}} \approx \frac{1}{12000} $$
  • Compare the relative strength of the tidal forces $$ \frac{\Delta F_{\mathrm{moon}}} {\Delta F_{\odot}} = \frac{M_{\mathrm{moon}}}{M_{\odot}} \left( \frac{a_{\oplus}}{r_0} \right)^3 = 2.27 $$
  • Compare the tidal force to the Earth's gravity $$ \frac{\Delta F_{\mathrm{moon}}}{F_{g,\oplus}} = \frac{2 R_{\oplus} G M_{\mathrm{moon}} m}{r_0^3} \frac{R_{\oplus}^2}{G M_{\oplus} m} = 2 \frac{M_{\mathrm{moon}}}{r_0^3} \frac{R_{\oplus}^3}{M_{\oplus}} $$ which is \( \approx 10^{-7} \)

Tidal Offset

  • The meter high mountain of water (and 20 cm high bulge in the Earth’s surface) generated by the tides does not move freely across the Earth, resulting in friction (costing roughly 2.5 terawatts for the lunar tide).
  • This friction is the result of Earth rotating (once per sidereal day) more quickly than the tidal bulge, which matches the Moon’s orbit (P = 29.5 sidereal days).
  • Friction drags the tidal bulge forward, resulting in a \( 10^{\circ} \) offset between the bulge and the Moon’s crossing of an observer’s zenith.
  • This results in a average delay of 40 minutes between the Moon’s upper (or lower) transit and the high tide.

Tidal Braking

  • The tidal offset introduces an asymmetry along the Earth- Moon line, similar to the misalignment of the Earth’s equatorial bulge with the Ecliptic.
  • Since the resulting force on the near side is stronger than on the far side, the resulting torque attempts to slow the Earth’s rotation.
  • The corresponding forces from the tidal bulges pull the Moon forward in it’s orbit. This transfers angular momentum from the Earth’s rotation to the Moon’s orbit.

Earth's Rotational Angular Momentum

  • The angular momentum of a rotating solid is \( L = I \omega \), where \( I \) is the "moment of inertia". For a sphere, $$ I = \frac{2}{5} M_{\oplus} R_{\oplus}^2 $$ Thus the Earth's angular momentum is $$ L = \frac{2}{5} M_{\oplus} R_{\oplus}^2 \left( \frac{2 \pi}{P_{\mathrm{rot}}} \right) $$
  • Since the mass and radius are not changing, $$ \frac{d L}{d t} = \frac{4 \pi}{5} M_{\oplus} R_{\oplus}^2 \left( - \frac{1}{P_{\mathrm{rot}}} \frac{d P_{\mathrm{rot}}}{dt} \right) $$
  • The Earth's rotation has been observed to be slowing $$ \frac{d P_{\mathrm{rot}}}{dt} = 5.3 \times 10^{-13} \Rightarrow \frac{d L}{d t} = - 4.4 \times 10^{16}~\mathrm{kg}~ \mathrm{m}^2~\mathrm{s}^{-2} $$

Moon's Orbital Angular Momentum

  • The orbital \( L \) of the moon is $$ L_{\mathrm{orb}} = M v r = M_{\mathrm{moon}} r \sqrt{\frac{G M_{\oplus}}{r}} = M_{\mathrm{moon}} \sqrt{G M_{\oplus} r} $$ where the centripetal velocity is $$ \frac{m v^2}{r} = \vec{F}_c = \vec{F}_g = \frac{G M m}{r^2} $$
  • If the change is slow and the orbit is still Keplerian $$ \frac{dL_{\mathrm{orb}}}{dt} = M_{\mathrm{moon}} \sqrt{G M_{\oplus}} \frac{1}{2} r^{-1/2} \frac{dr}{dt} = 3.7 \times 10^{25} \frac{dr}{dt} $$

Angular Momentum Conservation

  • If the Earth-Moon system is closed, then the \( L \) lost by the Earth must be gained by the Moon $$ \begin{eqnarray} - \frac{d L_{\oplus,\mathrm{rot}}}{dt} &=& 4.4 \times 10^{16}~\mathrm{kg}~ \mathrm{m}^2~\mathrm{s}^{-2} \\ &=& \frac{d L_{\mathrm{moon,orb}}}{dt} = 3.7 \times 10^{25}~\mathrm{kg}~\mathrm{m} ~\mathrm{s}^{-1}~\frac{dr}{dt} \\ &\Rightarrow& \frac{dr}{dt} = 1.2 \times 10^{-9}~\mathrm{m}~\mathrm{s}^{-1} \\ &=& 4~\mathrm{cm}~\mathrm{yr}^{-1} \end{eqnarray} $$
  • Tidal braking slows the Earth's rotation by 1.6 ms per century and is increasing the radius of the Moon's orbit by 4 m per century

Tidal Locking

  • The Earth also exerts tidal forces on the Moon $$ \frac{\Delta F_{\oplus}}{\Delta F_{\mathrm{moon}}} = \frac{G M_{\oplus} m R_{\mathrm{moon}}}{r_0^3} \frac{r_0^3}{2 G M_{\mathrm{moon}} m R_{\oplus}} = \frac{M_{\oplus} R_{\mathrm{moon}}} {M_{\mathrm{moon}} R_{\oplus}} $$ which are roughly 20x stronger than those the Moon exerts on the Earth.
  • Over the history of the solar system, this tidal force has brought the Moon into synchronous rotation, where its rotational period and orbital period are locked together.
  • As a result, the same side of the Moon points toward the Earth and the far side is not seen from Earth.
  • Such tidal locking is common for moons in our solar system, though not all of these spin-orbit resonances are 1:1.

Lunar Librations

  • The Moon’s synchronous rotation makes half of the moon unviewable from Earth, but 3 “imperfections” in the Earth-Moon system bring slightly different halves into view.
  • Diurnal Libration: The finite radius of the Earth allows one to see \( \approx 1^{\circ} \) further east/west in lunar longitude at moonrise/set.
  • Libration in Longitude: The slight eccentricity of the Moon’s orbit results in small variations in the orbital velocity from the average, providing a \( 6^{\circ} \) variation in the face toward the Earth.
  • Libration in Latitude: The Moon’s rotational axis is tilted \( 6.5^{\circ} \) compared to the orbital plane, exposing the Moon’s North and South Polar regions to the Earth’s view over the course of a year.
  • All together, these librations leave only 41% of the Moon restricted from view.

Tides on the Moon

  • Because the tidal force on the Moon is stronger, and the Moon's gravity is weaker, the tidal effect is much larger
  • Compared to the Earth $$ \frac{\Delta F_{\mathrm{moon}}}{F_{g,\oplus}} = 2 \frac{M_{\mathrm{moon}} R_{\oplus}^3} {r_0^3 M_{\oplus}} \approx 10^{-7} $$ the tidal force on the moon is a larger fraction of the Moon's gravity $$ \frac{\Delta F_{\oplus}}{F_{g,\mathrm{moon}}} = 2 \frac{M_{\oplus} R_{\mathrm{moon}}^3} {r_0^3 M_{\mathrm{moon}}} \approx 10^{-5} $$
  • An interesting way to write this is in terms of the mass density $$ \rho = \frac{3 M}{4 \pi R^3} \Rightarrow \frac{\Delta F_{\oplus}}{F_{g,\mathrm{moon}}} = \frac{3 M_{\oplus}}{2 \pi r_0^3 \rho_{\mathrm{moon}}} = 2 \frac{\rho_{\oplus}}{\rho_{\mathrm{moon}}} \left( \frac{R_{\oplus}}{r_0} \right)^3 $$

Extreme Tides

  • The effects of tides on Earth and the Moon are small and gradual, but tidal effects can be much more extreme.
  • In our Solar System, the innermost Galilean Moon Io provides an excellent example of the power of tidal interactions. Similar to our Moon in size and orbital separation, it experiences much stronger tides because Jupiter is 300 times \( M_{\oplus} \) and the orbital radius \( r_{J-Io} = 6 R_J \)
  • Io’s orbit retains a small eccentricity due to a 1:2:4 orbital resonance with Europa and Ganymede, increasing the tidal effect. \( 10^{14} \) Watts of tidal energy results in tremendous tectonic action and volcanos.

Tidal Disruption

  • As the tidal force becomes comparable to the gravitational force holding an orbiting body together, the body becomes increasingly prolate
  • The body then experiences differential rotation and is smeared into a disk or ring. Thus there is a minimum orbital radius, called the Roche Limit, at which the tidal force becomes comparable to gravity.

Tidal Disruption, Part II

  • The tidal force on the moon relative to the moon's gravitational force $$ \frac{\Delta F_{M}}{F_m} \propto \frac{\rho_M}{\rho_m} \left( \frac{R_M}{r_R} \right)^3 $$ $$ \Rightarrow r_R = 2.44 R_M \left( \frac{\rho_M}{\rho_m} \right)^{1/3} $$
  • In the case of the Earth and the moon: $$ \rho_{\mathrm{moon}} \approx 3300~\mathrm{kg}/\mathrm{m}^3 $$ $$ \rho_{\oplus} \approx 5500~\mathrm{kg}/\mathrm{m}^3 $$ $$ r_{R,\mathrm{moon}} = 2.44 R_{\oplus} \left( \frac{5500}{3300} \right)^{1/3} \approx 2.9 R_{\oplus} $$

Black Hole Tides

  • Near a black hole, the tidal forces can be quite disruptive.
  • If a white dwarf star were to approach close enough to an intermediate mass BH, tidal forces would compress the WD, leading to thermonuclear reactions and a complete disruption of the WD.
From Rosswog et al. (2008, 2009)

Maximum Orbit

  • With increasing distance, a moon is more weakly bound to its planet and more subject to perturbations from passing bodies, including the parent star.
  • If we define the difference between the acceleration of the Moon and Earth to the Sun $$ \Delta g = g_{\mathrm{moon}} - g_{\oplus} = \frac{G M_{\odot}} {(a_{\oplus}-r)^2} - \frac{G M_{\odot}}{a_{\oplus}^2} $$ $$ \Delta g \approx \frac{G M_{\odot}}{a_{\oplus}^2} \frac{2 r}{a_{\oplus}} $$
  • Setting this equal to Earth's gravity $$ \frac{G M_{\odot}}{a_{\oplus}^2} \frac{2 r}{a_{\oplus}} = \frac{G M_{\oplus}}{r^2} $$ gives the Hill radius $$ r_H = \left( \frac{M_{\oplus}}{2 M_{\odot}} \right)^{1/3} a_{\oplus} \approx 0.01~\mathrm{AU} \approx 4.5~\mathrm{r_0} $$