Astronomy 6570

Physics of the Planets

Tidal Evolution

Tidal Interactions

x = r sin

z = r cos

Accel'n of point P due to m =

a =Gm

d3

d =Gm

d3

az r cos z r sin x( )

Gm

d3

a z( ) z xx

And d = a z( )2

+ x2

1

2a 1

2z

a

1

2

a 1z

a if r << a

aGm

a3

1+ 3z

aa z( ) z xx

This accel'n is responsible for the centripetal accel'n of P in its orbit about the system's center of mass, at a distance of ma M + m( ) from 0 :

ac=

n2ma

M + mz

=Gm

a2

z (Kepler's 3rd law: G M + m( ) = n2a

3)

The Tidal Accel'n, aT, is the difference between a & a

c:

aT=

Gm

a2

2z

az

x

ax

If we write aT

in terms of a tidal potential,

aT= V

Tx, z( ),

then VT=

Gm

a3

z2 x

2

2

=Gm

a3

r2 1

23cos2 1( )

VT=

Gm

a3

r2

P2

cos( )

Compare the centrifugal potential, Vc= +

1

3

2r

2P

2cos( )

Vc

&VT

have the same radial & angular dependence, but opposite signs.

tidal bulge is a prolate spheroid.

So we can apply the same theory as for centrifugal distortion. In particular, the tidal distortion

of the planet's interior results in a P2 component in the planet's gravitational potential, with an

amplitude given by the Love number k2

T :

VG= k

2

TGm

a3

iR

5

r3

P2

cos( ) (R = planet's radius)

For the Earth, satellite measurements indicate that

k2

T 0.29

much less than the value derived previously for the rotational distortion:

k2= 0.94.

(Recall that k2=

3

2 for a uniform density fluid planet.)

The reduction of k2

T in comparison with k2 reflects the elastic component of the Earth's response to the time-varying

tidal potential.

Similarly, the tidal distortion of the Earth's solid surface is given by

r ( )R

= h2

Ti

mR3

Ma3P2 cos( )

where the observed value of the Love number is

h2

T= 0.59.

Compare this with the rotational Love number associated with the Earth's oblateness:

h2= 1.94.

Again, the reduction of h2

T is due to the Earth's elasticity.

The oceans, of course, do not respond elastically. If their response were purely hydrostatic we would have

h2

T ocean( ) = 1+ k2

T

1.29

so that the amplitude of the ocean tide would be approximately twice that of the solid body tide. In reality, the ocean's

response is not hydrostatic, because the tidal time scale is comparable to the natural resonant frequencies of the large

ocean basins:

P =2

res

~basin dimension

gd

d = mean ocean depth ~ 4 km

P ~ 109 cm

2 104 cmsec

~ 5 104 sec ~ 13 hrs.

Summary of Tide heights and Love numbers:

At the surface of a body the total potential is:

GM

r+ 1+ k

2

T( )VTr,( )

GM

R+

GM

R2

r 1+ k2

T( )Gm

a3

R2

P2

cos( )

If the surface is hydrostatic, the potential is a constant

r ( )H

1+ k2

T( )mR

4

Ma3

P2

cos( ).

In general, a solid surface does not deform hydrostatically, and we write for convenience

r ( )S= h

2

TmR

4

Ma3

P2

cos( ).

For the Earth we find from measurements of solid earth tides

h2

T 0.59,

While the ocean responds closer to the hydrostatic limit:

h2

ocean( ) 1+ k2

T= 1.29

relative to the land surface, the ocean response is given by

h2

ocean( ) h2

T 0.70

The tidal distortion is

T=

3

2h

2

mR3

Ma3

1.1 10 7h

2 ( )

Lunar-induced solid body tide amplitude = T

R = 41cm.

Lunar-induced ocean tide = 58 cm (rel. to solid ).

Solar induced tides are ~ 1

3 of these values.

Tidal Torques & Their Effects:

If the tidal bulge were perfectly aligned with the tide-raising body, m, then symmetry requires that

there be no net tidal torque between the two bodies. For real planets, however, the elastic

response of the planet lags behind (in time) the periodic tidal potential, resulting in an angular

offset of the maximum tide height from the direction to m. If the planet’s spin rate, , is

greater than the satellite’s orbital motion, n, then the bulge will lead the satellite, as shown in

the figure on the next slide. If < n, the bulge lags the satellite, and the torque changes sign.

To calculate the torque exerted by the bulge on m, we return to the induced gravitational potential:

Where is measured from the long axis of the bulge. The torque on m is

The torque acts to increase the orbital angular momentum of m, and the reaction torque of m on

the bulge acts to decrease the spin angular momentum of the planet (if > 0).

VG= k

2

TGmR

5

a3

r3

P2

cos( )

T = m a1

r

VG evaluated at r = a, =

=3

2k

2

TGmR

5

a6

sin2

Tidal potential, VT=

GmR2

a3

P2

cos( )

P2μ( )

1

23cos2 1( )

Potential due to distorted planet,

VD

r,( ) = k2V

TR,( )

R

r

3

quadrupole field ~r3

VD

a,( ) = k2

GmR5

a6

P2

cos( ) including the lag angle

Torque on satellite,

T = m a i1

a

VD

=3

2k

2

Gm2R

5

a6

sin 2

The lag angle, , is related to the energy dissipation rate in the planet due to the periodic tidal forcing. Approx.,

tan(2 ) = 1

Q where the quantity Q is given by

Q 1 =E dt

2 Emax

E is the rate of energy dissipation, Emax

is the peak potential energy stored in the bulge, & the integral extends

over 1 tidal period.

For the Earth, ~ 2 .4, implying that Q 12. This value is NOT likely to be typical of other terrestrial planets, as

the dissipation occurs largely in the Earth's oceans. The solid Earth probably has Q ~ 100, or ~ 0 .3, based on

observations of seismic oscillations.

Despinning Timescales.

The time for a satellite (or the sun) to despin a planet is simply

despin

=C

00

T

where C = MR2 and 0= initial spin rate.

The same expression applies to tides raised by the planet on a satellite acting to despin the satellite, except that

m M

R r, satellite radius

and Q

k2

T now refers to tides in the satellite.

See the table of despinning times on the next slide. Note that the values of k2T used for the

satellites are much less than unity – this reflects the greater importance of mechanical strength

(i.e., elasticity) vs. gravity in smaller bodies. For a uniform density body of rigidity μ, kT2

Despite the reduced values of k2T , we see that satellite despinning times are quite short, except for

very distant satellites such as Iapetus, Hyperion, and the outer jovian satellites. In fact,

Iapetus is known to be despun to the synchronous state, but Hyperion is apparently not.

Among the planets, only Mercury and Venus have despinning timescales less than the age of the

solar system, 4.5 109 yrs. Mercury spins with a period of 58 days, in a spin-orbit resonance

[an alternative end-point to synchronism, for moderately eccentric orbits]. Venus spins

retrograde, with a period of 243 days, for reasons which remain mysterious.

Tidal Despinning Timescales

• Planets by the Sun.

• Satellites by their planets.

T1= 5T

14T

1= 243.16 d( )

1

or n = 4 n n( )

Orbital Evolution Timescales

The tidal torque on the satellite acts to increase the satellite/planet system's orbital angular momentum.

For simplicity, we consider a zero-inclination, circular orbit:

T =d

dth

orb( ) =d

dt

mM

M + mna2

m GM( )1

2d

dta

1

2 if we assume that m << M

=1

2m GM( )

1

2 a

1

2 a

a = 3k

2

T

Qi

G

1

2 m R5

M

1

2 a

11

2

a

11

2 since sin2 Q 1.

The solution to this equation is straightforward

a

13

2 = a0

13

2 +13

2t t

0( )

where a0 = initial semi-major axis.

At an given time, the timescalefor orbital expansion is given by

torbit

~a

a= a

13

2 1 ~ a

13

2 m 1,

where the very strong dependence on an almost guarantees that tidal orbital evolution is most important for

the innermost satellite(s) of a planet.

This dependence of a on m for a given value of t t0 is illlustrated onthe following slide.

Satellite Tides vs Planet Tides:

satellite tidal effects

planet tidal effects

Mp

ms

2

Rs

Rp

5

ks

kp

Qp

Qs

~p

s

Rp

Rs

ks

kp

Qp

Qs

… But, for synchronously rotating satellites the dominant semi-diurnal tide has zero frequency, since s= n,

and thus the associated lag angle is zero. The most important effect is due to energy dissipation associated

with orbital eccentricity, or spin axis obliquity, :

E =GM

p

2 Rs

5n

a6

ks

Qs

21

2e2

+3

2sin2

=GM

pm

s

2a2a

Since no net torque is exerted a synchronous satellite, the orbital angular momentum is conserved

e is damped:

Lz= GM

pa 1 e2( )

1

2 cos i

e =1 e2

2eaa if i

e=

e

e=

2

21

ms

Mp

a

Rs

5

Qs

ks

n 1

3 105Q years for Triton, as illustrated in the numerical integation.

Satellite Tides Yoder (1979) Eccentricity tides

Large amplitude tide

Low amplitude tide

Summary of the known and probable effects of tidal interactions:

•

•

•

••

•

•

•

Other tides e.g., those raised on the sun by planets, can be shown to have negligible

effects, even over the age of the solar system. Tides may also have played a role in

the capture of satellites by planets or proto-planets (e.g., Triton).

Tidal Effects in the Earth-Moon System

Since ~1750 it has been realized that ancient ( 0 AD) observations of solar eclipses were

inconsistent with predictions based on modern observations and the theory of the moon’s

motion. In the 1930’s it was realized that the discrepancies in the predicted and actual eclipse

locations, and in other ancient and historical observations (such as lunar occultations of stars

and transits of Mercury across the sun) were attributable to a combination of

1. An orbital deceleration of the moon, and

2. A slowdown in the Earth’s spin rate,

Early ( 1975) attempts to separate these 2 effects lead to a wide range of values for and

depending on the particular date sets used. A commonly quoted value is from Spencer-Jones

(1939) who obtained

Note that solar or lunar eclipses really depend on their apparent longitude difference, relative to a

co-ordinate system fixed on Earth, i.e., on

n

, n

n = 22".4 century

2.

of =m

and d

2

dt2( ) = n n n

m

More recently, lunar-laser-ranging (LLR) experiments, and direct satellite determinations of the Earth’s tidal lag angle, ,

as well as lunar occultations of stars referred to Atomic Time have all converged on a direct determination of n alone:

n = 25 ±1( )" cy 2 ,

corresponding to a rate of expansion of the Moon's orbit of

a = + 370 cm cy 1.

The ancient eclipse observations (Chinese, Babylonian & Arab) may now be interpreted using this n

to derive an average value of . Over the last 3000 years, we find

20.4 ± 1.2( ) 10 9 cy 1,

corresponding to an increase in the length of day (LOD) of

d

dt(LOD) = (+1.8 ± 0.1)

msec

cy. [See diagram for data.]

Direct modern determination of is frustrated by decade and possibly century-scale random variations,

in the Earth's spin due to changes in atmospheric circulation and/or interactions between the Earth's core and mantle.

Curiously, the predicted value of , based on tidal theory and the observed value of n, is significantly

different from that obtained above:

tidal

= 27 10 9 cy 1.

The Earth therefore has a positive non-tidal acceleration in its spin rate of ~6.6 10 9 cy 1. This appears

to be due to a steady decrease in C, the polar moment of inertia, which has been independently observed

from satellite tracking:

d

dtJ

2( )3

2

C

MR2

= 2.6 ± 0.6( ) 10 9 cy 1.

This decrease in C is presumed to be a result of post-glacial uplift of polar regions since the last Ice Age ~10,000 years ago.

From n, we can calculate the Earth's Q from Kepler's 3rd law and the equation for a derived above:

n

n=

3

2

a

a

=9

2

k2

T

Qi

G

1

2 mR5

M

1

2 a

13

2

.

Using k2

T= 0.29, we get Q = 11.7.

This value of Q, if extrapolated into the past, implies a very close approach of the Moon to Earth < 2 109 years ago,

with unacceptable geological consequences to both bodies (e.g., melting the Moon!).

Observations of growth lines on fossil corals & bivalve shellfish up to 400 My old support the current value of n :

fossils

22 ± 1( ) 10 9 cy 1,

so we are left with postulating a considerably higher value of Q at times before this epoch (perhaps as high as 30), if we are to avoid

the above problem. It is likely that this variation in Q reflects the role of plate tectonics in rearrangaing the distribution of continents

and ocean basins.

Ancient Solar Eclipses & Tidal Evolution:

Solar longitude, =0+ n t

Lunar longitude, =0+ n t +

1

2nt2

Eclipse occurs when = mod 2( )

i.e., 0 0+ n n( )t +

1

2nt2

= 2 m

whereas the predicted eclipse time, T is given by

0 0+ n n( )T = 2 m

t TnT 2

2 n n( )T

nT 2

2 n

The rotational position of the Earth at time t is

=0+ t +

1

2t2

and pred

=0+ T

pred

= t T( ) +1

2T 2

=1

2

n

nT 2

The (east) longitude at which the eclipse is seen is

= t( ) t( )

= T( )n nT 2

2 n pred

1

2

n

nT 2

=pred

1

2

n

n nn T 2

Substituting appropriate numerical values, we get

n = 25" cy 1= 9.1 10 14 rad d 2

P = 1.8 ms cy 1= 3.6 10 12 rad d 2

n = 0.0172 rad d 1

n = 0.211 rad d 1

= 6.28 rad d 1

t T 27 Tcy

2 sec, where Tcy

is the time before the present in centuries,

and pred

00.029 Tcy

2

= 260 over 3000 yr.

Note that the rotational displacement of the Earth due to alone is

1

2T 2

= 00.138 Tcy

2 33 s cy 2

which coresponds to an error in sidereal clocks relative to Atomic time of

33 s cy 2 or almost 1 hour in 1000 yrs.

Satellite Shapes

Most planetary satellites have been tidally despun to a state of synchronous rotation (i.e., = n),

with their spin axes approximately normal to their orbit planes. In this situation, satellites

experience a permanent tidal potential in addition to the rotational potential VC.

The tidal potential is symmetric about the X axis, and is given by

VT=

GMp

a3

3

2cos2 1

2r

2

=GM

p

a3

x2 y

2

2

z2

2

For a synchronous satellite,

GM

p

a3

= n2=

2 (Kepler III)

so the combined potential is

VC+V

T= n

2 3

2x

2 1

2z

2V

TOT

From the Rotational notes, the centrifugal potental is

VC=

1

2

2r

2 sin2

=1

2

2x

2+ y

2( )

Like VC

and VT, V

TOT is a quadratic potential, so the distortion of the satellite leads to an additional

gravitational potential = k2 V

TOT, so that the total potential at the surface of the satellite is

(1 + k2) V

TOT

=1

2n2 3sin2 cos2 cos2( )r 2 1+ k

2( )

where is a longitude measured in the XY plane from the X (planet) direction.

As before, the equilibrium shape is an equpotential surface:

Gm

r+

Gm

r 2r + (1 + k

2) V

TOT= const.

r ,( )

r

n2r3

2Gm1 + k

2( ) 3sin2 cos2 cos2

1

21 + k

2( )q 3sin2 cos2 cos2

This describes a Triaxial Ellipsoid, with principal axes a(x), b(y), and c(z). The axis differences are:

a c = 2 1 + k2( )qr

a b =3

21 + k

2( )qr

b c =1

21 + k

2( )qr

since the value of along each axis is given by +3 a( ), b( ) and 1 c( ).

If the satellite is uniform in its interior, then k2=

3

2 and we have

a c

r= 5q,

a b

r=

15

4q,

b c

r=

5

4q

Note that, for a synchronous satelllite, Kepler's 3rd law implies

q =n2r3

Gm=

Mp

m

r

a

3

=3n2

4 G.

Note also that, regardless of the value of q, the equilibrium shape of such a satellite satisfies

where is the satellite's mean density.

b c

a c=

1

4

which can be used to test the hypothesis that a particular stellite has an equilibrium shape.

EXAMPLE: Dermott & Thomas (1988) studied the shape of Saturn's satellite Mimas

(r 200 km) and found it to be triaxial, with

a c = 17.7 ±1.0 km

b c = 4.8 ±1.0 km

b c( )a c( )

= 0.27 ± 0.06

in close agreement with an equilibrium figure. However, the predicted distortion is slightly greater:

a c( )pred

= 5q r = 18.7 km

n = 7.72 10 5rad / sec, = 1.137 g cm 3 q = 0.01874( )A theory of the shape accurate to 0(q2 ) predicts an even greater value of

a c( ) 20.3 km,

so D & T concluded that Mimas is not homogeneous, but centrally condensed, with

C

Mr 20.353± 0.013.

This value was derived from a modified form of the Darwin-Radan relation for rotational oblateness.

The following Table and Figures give current values of the shapes of the Saturnian icy satellites, as

derived by P.C. Thomas, et al., Icarus (2007) doi:10.1016/j.icarus.2007.03.012

Note that F =b c

a c.

P.C. Thomas, et al., Icarus

(2007)

doi:10.1016/j.icarus.

2007.03.012

Figure 8: Roughness of limbs of

saturnian satellites. Plotted

quantities are the root mean square

residuals to the limb fits (a) as a

fraction of the object’s mean radius

and (b) as absolute amounts.

P.C. Thomas, et al., Icarus (2007)

doi:10.1016/j.icarus.2007.03.012

Fig. 9. (a) Approach to equilibrium shapes.

The observed factor F = (b c)(a c) plotted

against object mean radius. Equilibrium

values for F are 0.21 for Mimas, 0.23 for

Enceladus, 0.24 for Tethys and 0.25 for

Dione, and Rhea. Iapetus is not included in

this plot. (b) Observed and predicted

differences in long and short axes of

saturian satellites. The predicted values

assume that the object is homogeneous.