Astronomy 6570
Physics of the Planets
Tidal Evolution
Tidal Interactions
x = r sin z = r cos
Accel'n of point P due to m =
Gm
Gm
a = 3 d = 3 aẑ r cos ẑ r sin x̂
d d
Gm
3 a z ẑ xx̂ d
(
(
)
)
1
1
2
2z 2
d = a z + x2 a 1 a
(
And
)
a 1 2
z
if r << a
a Gm z
a 3 1 + 3 a z ẑ xx̂ a
a (
)
(
)
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 :
2
n ma
ẑ
M+m
Gm
(Kepler's 3rd law: G M + m = n2 a 3 )
= 2 ẑ
a
The Tidal Accel'n, aT , is the difference between a & ac :
x Gm z
aT = 2 2 ẑ x̂ a a a
âc =
(
)
If we write aT in terms of a tidal potential,
aT = VT x, z ,
Gm 2 x 2 then
VT = 3 z 2
a ( )
(
)
Gm 2 1
r
3cos 2 1
3
2
a
Gm
VT = 3 r 2 P2 cos
a
1
Compare the centrifugal potential, Vc = + 2 r 2 P2 cos
3
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 k2T :
(
)
R5
T Gm
VG = k2 3 i 3 P2 cos
r
a
For the Earth, satellite measurements indicate that
(R = planet's radius)
k2T 0.29
much less than the value derived previously for the rotational distortion:
k2 = 0.94.
(Recall that k2 =
3
for a uniform density fluid planet.)
2
The reduction of k2T 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
( )=h
r (
)
mR3 2
i
P cos
R
Ma 3
where the observed value of the Love number is
T
2
h2T = 0.59.
Compare this with the rotational Love number associated with the Earth's oblateness:
h2 = 1.94.
Again, the reduction of h2T is due to the Earth's elasticity.
The oceans, of course, do not respond elastically. If their response were purely hydrostatic we would have
(
)
h2T ocean = 1 + k2T
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:
2 basin dimension
~
P=
res
gd
d = mean ocean depth ~ 4 km
P~
109 cm
2 104 cm
~ 5 104 sec ~ 13 hrs.
sec
Summary of Tide heights and Love numbers:
At the surface of a body the total potential is:
GM GM
GM
Gm 2
+ 1 + k2T VT r, + 2 r 1 + k2T
R P2 cos
R
r
R
a3
If the surface is hydrostatic, the potential is a constant
(
) ( )
(
() (
)
(
)
(
)
)
mR 4
P cos .
r H 1+ k
Ma 3 2
In general, a solid surface does not deform hydrostatically, and we write for convenience
()
T
2
(
)
mR 4
r S = h
P cos .
Ma 3 2
For the Earth we find from measurements of solid earth tides
T
2
h2T 0.59,
While the ocean responds closer to the hydrostatic limit:
(
)
h2 ocean 1 + k2T = 1.29
relative to the land surface, the ocean response is given by
(
)
h2 ocean h2T 0.70
The tidal distortion is
3 mR3
1.1 107 h2
T = h2
3
2 Ma
Lunar-induced solid body tide amplitude = T R = 41 cm.
Lunar-induced ocean tide = 58 cm (rel. to solid ).
Solar induced tides are ~
1
of these values.
3
()
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:
(
GmR5 3
VG = k
r P2 cos
a3
T
2
)
Where is measured from the long axis of the bulge. The torque on m is
1 VG T = m a evaluated at r = a, = r
3 T GmR5
= k2
sin 2
2
a6
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).
(
GmR 2
Tidal potential, VT = P2 cos
3
a
1
P2 μ 3cos 2 1
2
Potential due to distorted planet,
( )
(
)
R
VD r, = k2VT R, r
( )
( )
)
( )
(
3
quadrupole field ~r 3
)
GmR5
VD a, = k2
P2 cos including the lag angle
6
a
Torque on satellite,
1 VD
T = mai
a 3 Gm2 R5
= k2
sin 2
6
2
a
The lag angle, , is related to the energy dissipation rate in the planet due to the periodic tidal forcing. Approx.,
1
where the quantity Q is given by
tan(2 ) =
Q
E dt
1
Q =
2 Emax
E is the rate of energy dissipation, E is the peak potential energy stored in the bulge, & the integral extends
max
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 =
where
C 0
T
C = MR 2 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
mM
R r, satellite radius
Q
and T now refers to tides in the satellite.
k2
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.
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%&)
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&-/&%.
,+/&%&%
))/&%&'
Satellites by their planets.
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"
(/&%,
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,(/&%)
'/&%*
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+/&%)
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)/&%(
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+&/&%)
"
(
T 1 = 5T1 4 T 1 = 243.16 d
(
or n = 4 n n
)
)
1
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 mM
d
horb = na 2 dt M + m
dt
( )
(
)
1
= m GM
2
)
m GM
(
d 12 a
dt 1
2
a = 3
k
T
2
Q
i
1
2
a
1
2
1
2
G m R5
1
2
11
2
if we assume that m << M
a
a
11
2
since sin2 Q 1.
M a
The solution to this equation is straightforward
13
t t0
2
where a0 = initial semi-major axis.
13
2
13
2
0
a =a +
(
)
At an given time, the timescalefor orbital expansion is given by
13
13
a
1
2
torbit ~ = a ~ a 2 m1 ,
a
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:
5
2
satellite tidal effects M p Rs ks Q p planet tidal effects
ms Rp k p Qs p Rp k Q p ~ s s Rs k p 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, :
GM p2 Rs5 n ks 21 2 3 2 E =
Q 2 e + 2 sin a6
s
=
GM p ms
a
2a 2
Since no net torque is exerted a synchronous satellite, the orbital angular momentum is conserved
e is damped:
(
)
1
Lz = GM p a 1 e2 2 cos i
1 e2
a if i e=
2ea
5
e
2 ms a Qs 1
e = =
n
e 21 M p Rs ks
3 105 Q 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, n 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 n and ,
depending on the particular date sets used. A commonly quoted value is from Spencer-Jones
(1939) who obtained
n = 22".4 century2 .
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
of
and
= m d2
n
=
n
n
dt 2
( )
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 " cy2 ,
corresponding to a rate of expansion of the Moon's orbit of
a = + 370 cm cy1.
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 ± 1.2 109 cy1 ,
(
)
corresponding to an increase in the length of day (LOD) of
d
msec
(LOD) = (+1.8 ± 0.1)
. [See diagram for data.]
dt
cy
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.
is significantly
Curiously, the predicted value of , based on tidal theory and the observed value of n,
different from that obtained above:
= 27 109 cy1.
tidal
The Earth therefore has a positive non-tidal acceleration in its spin rate of ~6.6 109 cy1. This appears
to be due to a steady decrease in C, the polar moment of inertia, which has been independently observed
from satellite tracking:
3 C
d
J2 2 MR 2
dt
= 2.6 ± 0.6 109 cy1.
( )
(
)
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.
we can calculate the Earth's Q from Kepler's 3rd law and the equation for a derived above:
From n,
3 a
n
=
2 a
n
1
T
9 k2 G 2 mR5
=
i
.
1 13
2 Q
M 2a 2
T
Using k2 = 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 :
(
)
22 ± 1 109 cy1 ,
fossils
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:
= 0 + n t
Solar longitude,
1 2
= 0 + n t + nt
2
= mod 2
Lunar longitude,
Eclipse occurs when
(
(
)
)
1 2
= 2 m
0 0 + n n t + nt
2
whereas the predicted eclipse time, T is given by
i.e.,
(
)
0 0 + n n T = 2 m
tT (
2
nT
2 n n
)
2
nT
T 2n
The rotational position of the Earth at time t is
1
= 0 + t + t 2
2
0
and
pred = + T
(
)
1
pred = t T + T 2
2
1
n 2
T
= 2
n The (east) longitude at which the eclipse is seen is
() ()
= t t
( )
= T = pred 1
n 2
T
pred 2
n 2n
2
n nT
n 2
1
n T
2 n n Substituting appropriate numerical values, we get
n = 25" cy1 = 9.1 1014 rad d 2
P = 1.8 ms cy1 = 3.6 1012 rad d 2
n = 0.0172 rad d 1
n = 0.211 rad d 1
= 6.28 rad d 1
t T 27 Tcy2 sec, where Tcy is the time before the present in centuries,
and pred 00.029 Tcy2
= 260 over 3000 yr.
Note that the rotational displacement of the Earth due to alone is
1
T 2 = 00.138 Tcy2 33 s cy2
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.
From the Rotational notes, the centrifugal potental is
1
VC = 2 r 2 sin 2 2
1
= 2 x2 + y2
2
(
The tidal potential is symmetric about the X axis, and is given by
GM p 3
1
VT = 3 cos 2 r 2
2
a 2
GM p 2 y 2 z 2 = 3 x 2
2
a For a synchronous satellite,
GM p
= n2 = 2
(Kepler III)
3
a
so the combined potential is
3
1 VC + VT = n2 x 2 z 2 VTOT
2 2
)
Like VC and VT , VTOT is a quadratic potential, so the distortion of the satellite leads to an additional
gravitational potential = k2 VTOT , so that the total potential at the surface of the satellite is
(1 + k 2 ) VTOT
(
) (
)
1
= n2 3sin 2 cos 2 cos 2 r 2 1 + k2
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 Gm
+ 2 r + (1 + k 2 ) VTOT = const.
r
r
r ,
n2 r 3
1 + k 2 3sin 2 cos 2 cos 2 r
2Gm
(
)
(
(
)
)
1
1 + k 2 q 3sin 2 cos 2 cos 2 2
This describes a Triaxial Ellipsoid, with principal axes a(x), b(y), and c(z). The axis differences are:
(
)
3
a b = (1 + k ) qr
2
a c = 2 1 + k 2 qr
2
(
)
1
1 + k 2 qr
2
since the value of along each axis is given by +3 a , b and 1 c .
b c =
() ()
If the satellite is uniform in its interior, then k2 =
ac
= 5q,
r
a b 15
= q,
r
4
b c 5
= q
r
4
3
and we have
2
()
Note that, for a synchronous satelllite, Kepler's 3rd law implies
3
3n2
n2 r3 M p r =
=
.
q=
m a 4 G Gm
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 1
=
ac 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 ) = 0.27 ± 0.06
( a c)
in close agreement with an equilibrium figure. However, the predicted distortion is slightly greater:
( a c)
( n = 7.72 10
5
pred
= 5q r = 18.7 km
rad / sec, = 1.137 g cm3 q = 0.01874
)
A theory of the shape accurate to 0(q 2 ) 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
0.353 ± 0.013.
Mr 2
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
bc Note that F = 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)(ac) 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.