## Last Time

- Revisiting Kepler's Third Law
- Ellipses in terms of \( L \)
- Barycenter
- Kinetic and Potential Energy
- Orbit classification
- Orbital velocity
- Transfer orbits
- Stellar Parallax

TA James Ternullo

- Revisiting Kepler's Third Law
- Ellipses in terms of \( L \)
- Barycenter
- Kinetic and Potential Energy
- Orbit classification
- Orbital velocity
- Transfer orbits
- Stellar Parallax

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

- Exams 9/29 and 10/27.
- +1% EC for visiting office hours

- Like the apparent slant of falling rain when viewed from a moving vehicle, the Earth's motion introduces a slight difference in the apparent position of the stars, that varies as the Earth’s velocity changes direction over the year. $$ \tan \theta = v_{\mathrm{earth}}/v_{\mathrm{light}} \approx 10^{-4} $$ $$ \theta \approx 0.006^{\circ} \approx 10^{\prime\prime} $$
- This was explained by James Bradley in 1729

- Rotating or accelerating coordinate systems cause particular challenges
- Newton's law for a rotating coordinate system must be rewritten $$ \vec{a} = \vec{F}/m + \underbrace{2 ( \vec{v} \times \vec{\omega} )}_{\mathrm{Coriolis~force}} - \underbrace{\vec{\omega} \times ( \vec{\omega} \times \vec{r} )}_{\mathrm{centrifugal~force}} $$
- These are "pseudoforces", forces which arise only because of the choice of coordinate system

- The Coriolis force gives an apparent push to motion perpendicular to rotation
- Typically a small effect, but observed by Foucault in 1851
- Causes a pendulum to rotate slightly over the course of a day

- The centrifugal force is the counterpart of the centripetal force
- The centrifugal force from the Earth's orbit
around the sun is is 0.34 m/s
^{2} - But this effect is zero at the pole

- Most of our previous work assumed:
- planets were point masses
- planetary orbits share the same plane
- This causes several important effects:
- Precession of the pole
- Tides
- The frequency of eclipses

- The Earth's oblate shape is caused by the centrifugal force
- The centrifugal acceleration, \( 0.034~\mathrm{m}~\mathrm{s}^{-2} \), is 0.35% of the acceleration of Earth's gravity
- In turn, the Earth's diameter at the equator is 0.33% larger than that across the poles

- Precession is caused by the gravitational influence of the Sun and the Moon acting on Earth's equatorial bulge.
- Because the Earth’s Pole is inclined to the Ecliptic and the Earth is oblate, there is an unbalanced component of the Sun’s gravity on the near bulge and a smaller component on the far bulge.
- This unbalanced force on the rotating Earth, produces a torque that drives polar precession
- Newton explained this effect, observed in 150 BC

- The force of the Moon on a part of the Earth varies, both in magnitude and direction, with location on the Earth. $$ \vec{F}^{\mathrm{moon}}(r,\theta) $$
- If we factor out the force on the center of the Earth, what is left is the tidal force. $$ \vec{F}_{\mathrm{moon}}(r,\theta) = \vec{F}_{\mathrm{moon}}(r=0) + \vec{F}_{\mathrm{tidal}}(r,\theta) $$

- From an astronomical perspective, the lunar tides should strictly track the progress of the moon across the sky.
- Because the Moon is moving in its orbit in the same sense as the Earth is rotating, the tidal period $$ P_{\mathrm{tide}} = (1 + 1/P_{\mathrm{moon}})~\mathrm{days} = 24^{\mathrm{h}},50^{\mathrm{m}} $$
- For any terrestrial locality, the shape of the coastline and the undersea terrain can alter this timing. In some locations, the local effects are so dramatic as to render the tides diurnal (1 high & low tide each day) rather than the normal semi-diurnal (2 high & low tide each day)
- These same local features can also alter the magnitude of the tide, turning the typical 1 m displacement into as much as 12 m tides.

- The solar and lunar tides have similar sizes, thus the combinations become important.
- When sun and moon align, the most extreme tides result, the spring tides.
- When the solar tide is maximally out of phase with the lunar tide, weak tides result, the neap tides.

- The tidal force is $$ F_{\mathrm{tides}} \propto F_{\mathrm{grav}} \times R_{\mathrm{planet}} / R_{\mathrm{orbit}} $$ more precisely $$ \vec{\Delta} F = \frac{G M_{\mathrm{moon}} m R_{\oplus}}{r^3} ( 2 \cos \theta \hat{i} - \sin \theta \hat{j}) $$
- Tidal Force points toward the Moon on the near side, away from the Moon on the far side.
- At the top and bottom of the Earth, the Tidal Force points inward, at half the equatorial strength.