1. No category

International University Bremen
Joachim Vogt
Space Physics and Global Geophysics
— Sample Solutions —
Sheet 6
Fall 2005
21. Earth’s moment of inertia and core density
The average density %E of the spherical Earth model is given by
4π
3
%E R E
= ME
3
⇒
%E =
3ME
= 5.515 g/cm3 .
3
4πRE
We use %E and RE for normalization purposes and define:
%ˆm = %m /%E ,
%ˆc = %c /%E ,
ˆ c = Rc /RE .
R
(a) The total mass of the spherical two-layer model is given by
M =
h
i
4π
4π
4π
3
3
ˆ 3 ) + %ˆc R
ˆ3 .
%m (RE
− Rc3 ) +
%c Rc3 =
%E R E
%ˆm (1 − R
c
c
3
3
|3 {z }
=ME
Identifying M = ME yields
ˆ 3 ) + %ˆc R
ˆ3 ,
1 = %ˆm (1 − R
c
c
and, finally,
h
i
ˆ c3 ) /R
ˆ c3 .
%ˆc = 1 − %ˆm (1 − R
ˆ c = 3480/6371 = 0.546 and %ˆm = 3/5.515 = 0.544 gives
Using R
%ˆc = 3.345
⇒
%c = %ˆc %E = 18.45 g/cm3 .
(b) The two-layer spherical model is composed of two homogeneous spheroids, namely, (1) a
spheroid with density %m and radius RE , and (2) a spheroid with (excess) density (%c −%m )
and radius Rc . The moment of intertia of a homogeneous spheroid of mass M and radius
R is given by (2/5)M R2 , thus we obtain
2 4π
2 4π
3
2
%m R E
RE
+
(%c − %m )Rc3 Rc2
5 3
5 3
C = C1 + C2 =
=
h
i
2 4π
5
ˆ5 .
%E R E
%ˆm + (ˆ
%c − %ˆm )R
c
|5 3 {z
}
=(2/5)ME R2
E
1
2 = C
ˆ (2/5)ME R2 with Cˆ = 0.827
Inserting the observed value C = CE = 0.331 ME RE
E
yields
ˆ c5 )ˆ
ˆ c5 %ˆc .
Cˆ = (1 − R
%m + R
With
ˆ c3 ) + %ˆc R
ˆ c3
1 = %ˆm (1 − R
from part (a) we have a set of two linear equations for the unknowns %ˆm and %ˆc which can
be solved by standard methods (e.g., Cramer’s rule). We finally obtain
%ˆm =
%ˆc =
ˆ c2
0.827 − 0.5462
Cˆ − R
= 0.754 ,
=
ˆ2
1 − 0.5462
1−R
c
ˆ c5 ) − C(1
ˆ −R
ˆ c3 )
(1 − R
= 2.27 ,
ˆ c3 (1 − R
ˆ c2 )
R
or, in dimensional variables,
%m = %ˆm %E = 4.16 g/cm3 ,
%c = %ˆc %E = 12.5 g/cm3 .
Since the mantle density is much closer to the average value than the proxy obtained from
surface rocks, this estimate of the core density is much better than the value obtained in
part (a).
22. Spinning satellite with wire booms
As given in class, the inertia tensor of a homogeneous spheroid of mass M and radius R is given
by




Ixx Ixy Ixz
1 0 0
2
2




I =  Iyx Iyy Iyz  = M R2  0 1 0  = M R2 U
5
5
0 0 1
Izx Izy Izz
where U denotes the identity tensor. The inertia tensor of an ensemble of point particles
{m1 , m2 , . . .} located at positions {r1 , r2 , . . .} is




Ixx Ixy Ixz
yn2 + zn2 −xn yn −xn zn
X




I =  Iyx Iyy Iyz  =
mn  −xn yn x2n + zn2 −yn zn  .
n
Izx Izy Izz
−xn zn −yn zn x2n + yn2
Here we have mn = m, (x1 , y1 , z1 ) = (r, 0, 0), (x2 , y2 , z2 ) = (0, r, 0), (x3 , y3 , z3 ) = (−r, 0, 0), and
(x4 , y4 , z4 ) = (0, −r, 0).
(a) Looking at the point mass contributions to the inertia tensor, we first notice that all offdiagonal vanish because zn = 0 for all spherical probes, and either xn = 0 or yn = 0. For
the diagonal elements we get
probes
Ixx
= m2 y22 + m4 y42 = 2mr2 ,
probes
= m1 x21 + m3 x23 = 2mr2 ,
Iyy
probes
Izz
= m1 x21 + m2 y22 + m3 x23 + m4 y42 = 4mr2 .
2
The core satellite contribution to the inertia tensor is (2/5)M R2 U. We obtain




(2/5)M R2 + 2mr2
0
0
Ixx Ixy Ixz




0
(2/5)M R2 + 2mr2
0
I =  Iyx Iyy Iyz  = 
 .
2
2
0
0
(2/5)M R + 4mr
Izx Izy Izz
Using r = R and m = 10−4 M yields


0.4002
0
0

2
0
0.4002
0
I = = MR 
 .
0
0
0.4004
The angular momentum is
2π
2π
L = I |{z}
ω = Izz ˆ
z = 0.4004M R2 ˆ
z.
T
T
=ωˆ
z
(b) If after boom deployment the distance r of all spherical probes to the center has increased
to r = 20 R, then the moment of inertia
Izz = (2/5)M R2 + 4mr2 = 0.4M R2 + 4 · 10−4 202 R2 = 0.56M R2 .
The initial rotation period T ini = 4 seconds is related to the final rotation period T fin after
boom deployment through
ini
Izz
2π
fin 2π
= Lz = Izz
,
T ini
T fin
therefore,
T fin = T ini
fin
Izz
0.56
= T ini
' T ini 1.4 = 5.6 seconds .
ini
Izz
0.4004
The rotation period has increased by a factor of 1.4 .
23. Differential operators in spherical coordinates
If the function Φ is given by
Φ(r, ϑ) = Φ0
R
r
3
3 cos2 ϑ − 1
,
2
then
∂Φ
∂r
∂Φ
∂ϑ
∂Φ
∂ϕ
= Φ0 R3 (−3) r−4
3 cos2 ϑ − 1
,
2
= Φ0 R3 r−3 3(− sin ϑ) cos ϑ ,
= 0.
3
The gradient in spherical coordinates is given by
∇Φ =
∂Φ
1 ∂Φ
1 ∂Φ ˆ
ϑ+
ˆ
r+
ϕ
ˆ,
∂r
r ∂ϑ
r sin ϑ ∂ϕ
thus,
"
3 −4
B = −∇Φ = 3Φ0 R r
#
3 cos2 ϑ − 1
ˆ
ˆ
r + cos ϑ sin ϑ ϑ
2
which means that
Br = 3Φ0 R3 r−4
Bϑ = 3Φ0 R3 r−4
3 cos2 ϑ − 1
,
2
cos ϑ sin ϑ .
The divergence of B is given by
∇·B =
=
1
∂
1 ∂Bϕ
1 ∂ 2 r Br +
(sin ϑBϑ ) +
r2 ∂r
r sin ϑ ∂ϑ
r sin ϑ ∂ϕ
2
1
3 3 cos ϑ − 1 ∂
−2
3Φ
R
r
0
r2
2
|∂r {z }
=−2r −3
+
∂ 1
3Φ0 R3 r−4
sin2 ϑ cos ϑ
r sin ϑ
|∂ϑ
{z
}
=sin ϑ(3 cos2 ϑ−1)
h
i
= 3Φ0 R3 r−5 −(3 cos2 ϑ − 1) + (3 cos2 ϑ − 1) = 0 .
The components of the curl of B are
(∇ × B)r =
(∇ × B)ϑ =
(∇ × B)ϕ =
1 ∂Bϑ
1 ∂(sin ϑBϕ )
−
,
r sin ϑ
∂ϑ
r sin ϑ ∂ϕ
1 ∂(rBϕ )
1 ∂Br
−
,
r sin ϑ ∂ϕ
r ∂r
1 ∂(rBϑ ) 1 ∂Br
−
.
r ∂r
r ∂ϑ
Since Bϕ = 0 and ∂/∂ϕ = 0, we obtain
(∇ × B)r = 0 ,
(∇ × B)ϑ = 0 ,
and
(∇ × B)ϕ
1 ∂
1 ∂ 3 cos2 ϑ − 1
=
r 3Φ0 R3 r−4 cos ϑ sin ϑ −
3Φ0 R3 r−4
r ∂r
r ∂ϑ
2
1
1
=
3Φ0 R3 (−3)r−4 cos ϑ sin ϑ − 3Φ0 R3 r−4 (− sin ϑ)3 cos ϑ
r
r
3 −5
= −9Φ0 R r cos ϑ sin ϑ + 9Φ0 R3 r−5 cos ϑ sin ϑ = 0 .
4
!
24. Tidal friction in Earth-Moon system (∗)
The total angular momentum of the (model) Earth-Moon system is
CE ωE + CL ωL + ME a2E ωo + ML a2L ωo = const .
(a) The various terms on the left-hand side of the given equation have the following meaning:
(i) Earth’s (daily) rotation around the planetary axis: first term,
(ii) Earth’s orbital motion around the center-of-mass of the Earth-Moon system: third
term,
(iii) Moon’s orbital motion around the center-of-mass of the Earth-Moon-system: fourth
term,
(iv) Moon’s rotation around the lunar axis: second term.
(b) The condition
ME · aE = ML · aL
means that the center-of-mass of the Earth-Moon system is taken as a reference point, and
a3L ωo2 = const
is Kepler’s third law.1
(c) It is useful to define the parameter
µ =
ML
aE
=
= 1.23 · 10−2
aL
ME
and to write the Earth-Moon distance as
D = aL + aE = aL (1 + µ) = aE (µ−1 + 1) .
Neglecting the term CL ωL yields
Ltotal = CE ωE + ME a2E ωL + ML a2L ωL = CE ωE + ML a2L ωL (1 + µ)
= CE ωE + ML D2 ωL (1 + µ)−1 = 5.92 · 1033 kg m2 s−1 + 28.58 · 1033 kg m2 s−1
= 34.5 · 1033 kg m2 s−1
using the values for the parameters in the equation given in class and on the problem
sheet.
(d) In order to eliminate the parameter ωL from the angular momentum balance we use a3L ωo2 =
a3L ωL2 = const which implies
aL* 3/2
ωL = ωL*
aL
1
Strictly speaking, this condition only holds if both the Moon and the Earth behave like point masses or
spherical rigid bodies. The temporal changes implied by tidal friction are much slower than the orbital motion,
therefore, we can still consider this as an invariant of motion.
5
and
3/2
1/2
3/2
a2L ωL = aL (ωL* aL* ) = D1/2
ωL* D∗
(1 + µ)2
where the subscript ∗ indicates present-day values. We obtain
3/2
Ltotal = CE ωE + ML a2L ωL (1 + µ) = CE ωE + ML D1/2 ωL* D∗ /(1 + µ)
which can be rearranged to yield
Ltotal − 2πCE /TE
D =
!2
.
3/2
ML ωL* D∗ /(1 + µ)
Another (equivalent) form of the relationship between D and TE can be obtained by
starting from Kepler’s third law in the form
a3L ωL2 = GMred = GME /(1 + µ)
where Mred = ME /(1 + µ) is the reduced mass of the two-body problem. Inserting D =
aL (1 + µ) gives
s
D3 ωL2 /(1 + µ)3 = GMred = GME /(1 + µ)
⇒
ωL =
GME
(1 + µ) ,
D3
and for the term ML a2L ωL (1 + µ) in the angular momentum equation
ML a2L ωL (1 + µ) =
√ p
ML D 2 ω L
= ML D GME .
1+µ
The angular momentum balance can thus be written as
CE ωE + ML a2L ωL (1 + µ) = Ltotal = CE ωE* + ML a2L* ωL* (1 + µ) .
|
{z
}
|
{z
}
√ √
√ √
=ML D
GME
=ML D∗
GME
With ωE = 2π/T E we find
2πCE
1
1
−
TE* TE
p
= ML GME
√
D−
p
D∗
and, finally,
D =
2πCE /TE∗
T
√
1 − E*
TE
ML GME
+
p
D∗
2
.
(e) Using one of the formulas derived in (d) with TE = 22 hours gives D ' 370 000 km.
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