Quiz: cannibalism in close binary stars Laws of Gravity III Tides

star m1 bloats up, part (d m) of its envelope becomes dominated by

the gravity of m2 and is transferred from m1 to m2

-- does the binary unbind or spiral-in? How to estimate?

● Use mass conservation? Yes m˙ 2=−m˙ 1 Cataclysmic Variable ● No! Energy lost Use energy conservation? m from binary 1 ● Use angular momentum conservation? m m For a circular orbit, L= 1 2 √GMa M m2

m˙ 1 m˙ 2 1 a˙ Hence, L˙ =L + + =0 ( m1 m2 2 a ) Dana Berry

a˙ m˙ (m2−m1) With m˙ =−m˙ 1=m˙ 2 , one finds =2 a m1 m2

Tidal forces and the tidal bulge Force = Lunar gravity + centrifugal force*

Gravitational acceleration centrifugal acceleration

GM G M GM GM 2 2 2 GM G M 2 (r+R) r (r−R) 2 (r−R) (r+R) C r 2 M B A Moon/Sun R

Radius R M : lunar mass; ME : Earth mass; r : Earth to center of mass distance; Orbital separation r E r M : Moon to center of mass distance. r=r +r The tidal acceleration is due to E M 2 GM diferential Centrifugal accelation: Ω r E= Lunar gravity across the Earth. X r 2 *The centrifugal force is relevant because we are considering a rotating coordinate system fied to Earth. Net acceleration GM 2 Tidal periods --- lunar & solar tides At A: a = −Ω r A (r −R)2 E GM GM 2R 2 Solar tides and lunar tides have comparable heights ≈ + −Ω r r 2 r 2 r E GM 2R Rotation of the Earth causes: ≈ , r 2 r semi-diurnal tides -- earth rotates every 24 hours GM 2R At B: a ≈− , tide rises and falls every ~ 12.4 hours B r 2 r dominates in atlantic coasts At C: a =0 C diurnal tides -- dominates in some pacifc coasts C M resonance of tidal forcing with ocean basin

B A Moon/Sun Orbital phases of the Sun and the Moon: 'spring' tides and 'neap' tides Orbit of the Earth causes: fortnightly tides -- moon's distance from us varies, e=0.055 semi-annual tides -- earth's orbit around the Sun, e=0.017

Tidal bulges points at and away from the Moon

Tidal Height Tidal Evolution Tides = ocean tide + atmosphere tide + solid tide Over most of the world, ocean tide ~ 0.7 metres, balancing enhanced self-gravity (due to tidal bulge) with Locally (observer on Earth) the tidal acceleration (see extra note at end for an order-of-magnitude estimate) tidal sloshing energy dissipation (into heat) Bay of Fundy: tidal height ~ 9 m (highest in the world) Nearby PEI: ~ 2.5 m and a spin faster than Globally Ω Moon's orbit tidal bulges lead Earth-Moon line lunar torque on bulges - slows Earth's spin; - pushes away Moon

Total energy (orbit + spin) steadily decreases over Gyr timescale, while the total angular momentum (orbit + spin) is conserved. Final state: synchronised & circularised Tidal Evolution Earth-Moon system Tidal Evolution fnal state

As a result of tidal dissipation: Pluto & Charon Earth-Moon: Earth is spinning down Orbital period: 6.387 days orbital period: 27.32 days Pluto spin period: 6.387 days Earth spin period: 1 day --angular momentum transfer-- Charon spin period: 6.387 days Moon spin period: 27.32 days and the Moon is receding e=0 e = 0.05

x Observable Consequences: lengths of day & month are increasing number of days in a month is decreasing x Evidence from: laser ranging, historical eclipse records, coral & nautilus fossil, (F.Verbunt) mud deposit For an eicellent review (by F. Verbunt), see www.astro.utoronto.ca/~mhvk/AST221/verbunt.pdf

Extra Note: Taylor expansion Extra Note: Height of the Tidal Bulges -- used to derive tidal acceleration -- an order-of-magnitude estimate

See also http://en.wikipedia.org/wiki/Taylor_series 3 GM R M R Tidal acceleration ∼2 E ∼2 ( E ) g 2 ( r ) M r E r E Generally, any function f (x) can be Taylor-expanded around some x 0 1 h Tidal bulge generates additional gravity f (x)=f (x )+(x−x )f ' ( x )+ (x−x )f ' ' ( x )... 0 0 0 2 0 0 A to balances the tidal acceleration d f (x) 2 where f ' ( x)≡ , etc. G(ME +ρ hRE ) GME dx g'∼ 2 − 2 RE RE 1 E.g., expanding f (x)= around x =0 , GME h h 1+x 0 with ρ∼ρE , g'∼ 2 ∼ gE RE RE RE −1 f ( x)=1+x ...≃1−x to first order 3 2 h M RE [ (1+x) ]x =0 Equating the two, we obtain ∼2 R M ( r ) GM GM GM 1 E E Similarly, 2 = 2 2 = 2 2 factor of order unity correction (r−R) r (1−R/r) r 1−2 R/ r+(R/ r) -7 Moon raises tide on earth ~ 10 RE ~ 60 cm from using correct density, etc. Sun on Earth ~ 25 cm GM 2 GM ≃ (1+2 R/ r+(R/r) )≃ (1+2R/r) r 2 r 2 Earth on Moon ~ 1.7 m Earth on Sun ~ (you do it!) Extra Note: Tidal evolution muscle feiing --> heat generated, tidal sloshing --> heat

Total energy (orbit + spin) is steadily decreasing over Gyr timescale,

GM ME 1 2 1 2 E =E +E =− + I Ω + I Ω tot orb rot 2a 2 E E 2 M M 2 2 Moment of inertia: IE =r g ,E ME RE 2π Spin frequency: ΩE≡ PE while the total angular momentum (orbit + spin) is conserved, M M L = E G(M+M )a(1−e2)+I Ω +I Ω tot M+M √ E E E M M E ∂E ∂E ∂E Final minimum energy state: tot = tot = tot =0 ∂ΩM ∂ΩE ∂ e synchronised & Hence, Ω =Ω =ω , e=0 M E circularised

Currently: ΩM=ω , e≈0, ΩE≈27ω (free energy from Earth's spin)