PTYS 554 – Evolution of Planetary Surfaces
Solutions for Homework #1
1) Isostasy. On Venus plate tectonics is absent. Down-welling flows in Earth’s
mantle are usually associated with subduction whereas on Venus it’s thought
to cause shortening of the crust.
Think of a linear strip of the lithosphere that has a width wo. It gets
compressed and reduced in width to w. i.e. the compression factor is Cf =
wo/w. This compression builds mountains that are supported by Airy isostasy
i.e. they float like icebergs in the Venusian mantle. Show that the mountain
height is given by
h = TL
ρm − ρc
(C f − 1)
ρm
where the mantle and crust densities are ρm (~3300 kg m-3) ρc (~2750 kg m-3)
respectively and TL is the thickness of the crust. How tall do these mountains
get when crustal rocks get compressed by a factor of two?
There are two useful relations to derive here, conservation of volume and balance
between the weight and buoyancy forces. For conservation of volume, the lithosphere
strip originally has a volume (per unit length along the strip) of TLwo. After the
compression the lithosphere has mountains (height h) and a root sticking into the
mantle (height hr), giving a volume (per unit length along the strip) of (TL+h+hr)w, where
w is the new, shortened width.
wo (TL ) = w(TL + h + hr )
⎛ wo
⎞
− 1⎟TL = h + hr
⎜
⎝ w
⎠
⎛ h ⎞
h⎜1 + r ⎟ = TL (C f − 1)
h ⎠
⎝
−1
⎛ h ⎞
h = TL ⎜1 + r ⎟ (C f − 1)
h ⎠
⎝
Now think of the balance between buoyancy forces and weight. The lithospheric slab
usually has these forces in balance so we need only think about the extra masses i.e.
the mountains above the surface and the root below. The weight of the mountains and
root (again, per unit length along the lithospheric strip) are (whρcg) and (whr ρcg)
respectively, whereas the buoyancy force from the displaced mantle material is (whr ρm
g). Equating these gives:
w h ρ c g + w hr ρ c g = w hr ρ m g
ρ c h + hr ρ c = hr ρ m
⎛ ρ − ρ c
h = hr ⎜⎜ m
⎝ ρ c
hr ⎛ ρ c
= ⎜
h ⎜⎝ ρ m − ρ c
Combining these two relations:
⎞
⎟⎟
⎠
⎞
⎟⎟
⎠
⎛
ρc
h = TL ⎜⎜1 +
⎝ ρ m − ρ c
⎛ ρ m
h = TL ⎜⎜
⎝ ρ m − ρ c
⎛ ρ − ρ c
h = TL ⎜⎜ m
⎝ ρ m
⎞
⎟⎟
⎠
−1
(C
f
− 1)
−1
⎞
⎟⎟ (C f − 1)
⎠
⎞
⎟⎟(C f − 1)
⎠
Assume a representative Venusian crustal thickness of 70km
When Cf =2, ρm ~3300 kg m-3 ρc ~2750 kg m-3 then h=0.17 TL. If we take a
representative crustal thickness of 70km then h=11.6km.
The very large lunar south pole Aitken basin is currently about 8km deep and
has no major gravity anomaly associated with it. If this impact originally
excavated all the way through the crustal material (density 2800 kg m-3) to the
mantle (3300 kg m-3) then how thick is the lunar crust?
The mantle rebounded somewhat to fill the crater. As the mantle is denser it doesn’t
need to fill up the impact basin to produce the same gravity signature. Think of two
columns, one in the crater and one outside. There’s no gravity anomaly, so they contain
the same amount of mass.
ρ c gTL = ρ m g(TL − 8km)
1−
$ ρm '
ρ c 8km
=
or TL = 8km&
)
ρm
TL
% ρm − ρc (
Plugging in the numbers gives a crustal thickness of ~53km. If there were some crustal
material left in the basin then we need to add that to the final answer e.g. if there were
still 7km of crustal material in the crater above the mantle then the crustal thickness
€60km not 53km.
would be
Why does doing a Bouguer correction over a compensated feature result in a
large anomaly?
In a compensated situation the mass sticking out above the reference surface is
balanced by a low-density crustal root that sticks into the denser mantle. There is no
gravity anomaly associated with a structure like this. When a Bouguer correction is
done only half of this structure is subtracted, the Bouguer correction removed the
attraction of the mass that sticks out above the reference surface. It does not take into
account the low-density root and that density deficit produces a large negative gravity
anomaly (and vice-versa for a correction done to a compensated depression).
Northern Europe is still rebounding from the last ice age. What free-air and
Bouguer anomalies would you expect to see over this region. Which would be
larger?
The ice sheet both depressed the surface and pushed crustal material into the denser
mantle so that it could be isostatically compensated. After the ice-sheet disappears
there is a large negative free-air anomaly from both the deflected surface and the
displaced mantle material. Doing a Bouguer correction removes the part of the anomaly
associated with the surface deflection, but not the part associated with the displaced
mantle material. So the Bouguer anomaly is also negative, but not as strong as the
free-air anomaly.
If the signs of these anomalies were both reversed then what plausible
geological scenario would you be looking at?
If you see a situation were both anomalies are positive then you could have a
depression that was compensated, but has since been filled i.e. the mantle has bowed
up in response to removal of the crustal material and then the crustal material was
redeposited in the depression. This is similar to the lunar mascons and Utopia impact
basin on Mars both of which have been filled with later sediments. Such a feature
should sink slowly for the same reason that post-ice-age Europe is rebounding. The fact
that these features haven’t sunk billions of years after they were filled is a testament to
the strength of the lithospheres on these bodies.
2) Planetary thrust faults. As Mercury’s core solidifies and cools it contracts. If
the core volume decreases by a factor F, then show that the surface area of
the planet decreases by a factor:
3
2(1 − F ) ⎛ Rc ⎞
⎜
⎟
1−
3 ⎜⎝ RP ⎟⎠
Assume F is 0.995, how many square kilometers did Mercury loose?
The volume of the planet’s core changes from Vcore to FVcore, causing the surface area
of the whole planet to change from Sp to XSp, we would like to find X. As all the volume
change is caused by the change in core size. The change in the planets volumes ΔVp =
(1-F)Vcore. We can relate volume and surface area of a sphere:
S p = 4 πR p2
so :
3
1 32 1
S p = ( 4 π ) 2 R p3
3
3
rearrange :
Sp
3
2
3 4π
= 4 3 πR p3 = V p
We’ll make the 1st order approximation that: ΔV p =
3
€
Sp 2
since : V p =
3 4π
δV p
ΔS p
δS p
1
4 πR p2 R p
δV p
S 2
then :
= p
=
=
δS p 2 4 π
2
2 4π
Rp
2ΔV p
ΔS p or ΔS p =
2
Rp
The change in surface area, ΔSp, is (1-X)Sp so:
ΔS
2 ΔV p
2 (1− F)Vcore
X = 1− p = 1−
= 1−
Sp
S p Rp
(4πRp2 ) Rp
€
3
(1− F) 4 3 πRcore
X = 1−
2πR p3
⇒ ΔV p =
3
2(1− F) % Rcore (
X = 1−
'
*
3
& RP )
If F=0.995, (and Rcore/RP = 0.75 for Mercury) then X=0.9986. So Mercury lost 0.14% of
its surface area, which is (Rmercury = 2440km) roughly 105,200 km2.
€
Mercury’s lithosphere
broke along many thrust faults during this episode. If
each fault is about 500km long and has a displacement of 2km, how many
faults does Mercury need to accommodate this shrinkage? Assume a typical
thrust fault dip of 30°.
If we assume a typical fault dip of about 30 degrees then a displacement of 2km causes
~1.7km of surface to be overridden by the thrust sheet. If the faults are 500km long
then that corresponds to ~866 km2 to be lost. As Mercury lost 105,200 km2 in total that
corresponds to about 121 faults.
3) Bingham flows. The driving stress in a Bingham fluid must exceed a finite
yield value in order for flow to start. Equate the stresses at the base of the
flow to this yield value to get an expression for the thickness of the flow.
Let’s say that the flow has a thickness H, density ρ, and is on a slope α. If the yield
stress is σy then:
σ y = ρgH sin α
σy
ρgsin α
Note that σy is not a frictional force that depends on normal stress. The material
moves by internal deformation (flow), it doesn’t slide down the slope. There’s a
minor inconsistency here that often gets glossed over because slopes are usually
€
low.
The sine in the above expression should be a tangent if H is measured normal to the
slope.
H=
The flow spreads out laterally until the internal frictional forces balance
pressure. Since pressure increases with depth show that the flow will take on
a parabolic cross-section.
Different depths within the flow can spread laterally because they are under different
pressures. In all cases, the yield stress counters the forces that cause this spreading
and sets the width of the flow at each depth.
Flow coming out of
the page.
Consider a column (shaded above in the flow)
The mean pressure on face A is ½ρg(h+Δh)
The outward force on this face per unit length on the flow is ½ρg(h+Δh)2
The mean pressure on face B is ½ρgh
The outward force on this face per unit length on the flow is ½ρgh2
The difference in these two forces is balanced by the resistance to shear, which is
the yield stress times shearing area = σy Δx (again per unit length of the flow)
Equating these gives:
1 ρg ( h + ∂h ) 2 − h 2 = −σ y ∂x
2
[
ρgh ∂h = −σ y ∂x
€
]
as ∂h << h
(minus sign on dx is because the force is directed inward while dx increases
outward). Integrating this and setting h=H when x=0 gives:
ρg
x=
(H 2 − h 2 )
2σ y
i.e. a parabolic cross-sectional profile.
σ
(where σ is the yield
ρg sin 2 (α )
stress and α is the slope). How are flow width and flow thickness and the
slope related?
Show that the maximum
€ width of the flow is
€
€
ρg
H 2 − h 2 ) and that
From the previous parts of the question we know that x =
(
2σ y
σy
H=
. The width w, is equal to 2x and when h=0, then w=wf (the full width of
ρgsin α
the flow). Combining these relations gives:
€
ρg 2
wf =
H
σy
2
ρg % σ y (
wf =
'
*
σ y & ρgsin α )
σy
ρgsin 2 α
We can also use the above relationships to eliminate the yield stress to figure out
another relation between the flow’s slope, thickness, and width.
σy
wf =
€
ρgsin 2 α
H
wf =
sin α
i.e. flow thickness and width scale together with the sine of the slope.
wf =
If you assume Newtonian rheology, show that the mean velocity of a flow is
€ squared.
proportional to its thickness
For a Newtonian fluid stress is equal to the viscosity times the strain rate. Strain
occurs at some rate because the downhill velocity (U) varies with depth (i.e. this
material is not sliding down the slope as a coherent block, but is instead deforming
internally). Writing this down in equation form:
•
σ = ηε
ρg( H − y ) sinα = η
€
∂U
∂y
Integrating this and finding the constant of integration by setting U=0 when y=0 (i.e.
no basal sliding) yields:
ρg
U=
sin α Hy − 1 2 y 2
η
Integrating the velocity times flow width (assumed constant) over the thickness of
the flow gives the discharge (Q).
(
)
H
€
Q=
∫ Uw∂y
0
ρg
sin α 1 3 H 3
η
We could also set the discharge equal to the cross-sectional area of the flow (WH)
times some mean velocity.
ρg
HwV = Q = w sin α 1 3 H 3
€
η
ρgsin α 2
V=
H
3η
So mean velocity in a Newtonian flow is proportional to thickness squared.
Q=w
(
)
(
)
(Adapted from Melosh 2011)
€
Steep flow fronts are observed at the edges of broad, extensive lava flows on
the lunar Mare. These lava flows are considerably thicker than terrestrial lava
flows, often reaching 100 m in height. The average slope of one such flow is
about 0.5°.
Compute the Bingham yield stress of this lava flow.
Typical
terrestrial basaltic lava flows have yield stresses of several thousand Pa. Is
lunar magma substantially stronger than terrestrial magma?
σy
Putting in numbers for lunar gravity (1.62 ms-2),
ρgsin α .
H(100m), slope (0.5°) and density (3000 kgm-3) we find that the yield stress is: 4240
Pa. i.e. not all that dissimilar from terrestrial basalt.
We’ve seen that H =
€
4) Io’s mountains
We discussed in class the maximum
shear stress generated by a surface
load is just a fraction (usually about a
third to a half) of the peak load itself.
For example, the rectangular block
mountain and stress contours shown
schematically here generates a peak
shear stress in the subsurface of 0.352
ρgh at a depth of 0.865w (w=mountain
width which we’ll assume to be equal
to h for now) (Melosh 2011). For the
mountain to be supported then this
stress must be less than the typical
strength of rocks (~100 MPa).
Show that the maximum topography than can be supported like this is:
𝟎. 𝟕 𝝈𝒚
𝑮𝝆 𝝆𝒄
𝒉≈
𝑹𝒑𝒍𝒂𝒏𝒆𝒕
Where 𝝆 𝝆𝒄 are the planet’s mean density and crustal density respectively and
𝝈𝒚 is the strength of rock.
The shear strength of the rock must be more than the peak shear stress (0.352 ρcgh) for
the mountain to be supported. For the highest possible mountain then these quantities
are just equal.
𝝈𝒚 = 0.352𝜌! 𝑔ℎ
Gravitational acceleration (g) depends on the mass and size of the planet i.e. 𝒈 =
Combining and rearranging:
𝝈𝒚 𝑹𝟐 𝒑𝒍𝒂𝒏𝒆𝒕
𝒉=
𝟎. 𝟑𝟓𝟐 𝑮𝑴 𝝆𝒄
Replace the mass of the planet (M) with the volume times the mean density (𝝆)
𝒉=
𝝈𝒚 𝑹𝟐 𝒑𝒍𝒂𝒏𝒆𝒕
𝟎. 𝟑𝟓𝟐 𝑮 𝟒 𝟑 𝝅 𝑹𝟑 𝒑𝒍𝒂𝒏𝒆𝒕 𝝆 𝝆𝒄
𝒉=
𝟑 𝟒𝝅 𝝈𝒚
𝟎. 𝟑𝟓𝟐 𝑮𝝆 𝝆𝒄
𝒉≈
𝑹𝒑𝒍𝒂𝒏𝒆𝒕
𝟎. 𝟕 𝝈𝒚
𝑮𝝆 𝝆𝒄
𝑹𝒑𝒍𝒂𝒏𝒆𝒕
!"
!!
In class we discussed how well (or not) this works for the terrestrial planets.
Using the above relationship, how high are the highest mountains on Jupiter’s
moon Io predicted to be? (crustal density is ~3000 Kg m-3)
Plugging in properties for Io i.e. 𝜌 3530 Kg m-3, Rplanet 1821 Km and 𝜎! of 100 Mpa.
We find that the predicted height of mountains on Io is 54.4 Km.
Io has prodigious amounts of volcanic activity, but also possesses nonvolcanic mountains that appear to be tilted crustal blocks. In reality, these
mountains top out at only ~17km. So something else is limiting their height.
Io’s average heat flux is a whopping 2.5 W/m2 (Earth’s is a comparatively
measly 0.08 W/m2), but most of that comes though local areas of volcanic
activity. In general, only a few percent (let’s say about 0.1 W/m2) is conducted
through the lithosphere.
When rocks get to about half their melting
temperature then they stop being able to support elastic stresses for long
periods.
With this info, and the above diagram, in mind, how high can mountains on Io
get? (Thermal conductivity is about 3 W/m/K, rock melts at ~1200K and Io’s
surface temperature is ~100K.)
The above figure tells us that most of the elastic stress is supported at a depth of 0.865
times the width of the mountain (assumed to be close to the height of the mountain
here). Bigger mountains have stress supported at greater depths, but if these deep
rocks are too warm (> Tm/2, where Tm is the melting temperature) then they stop
behaving elastically and can’t support the mountain.
With the conductivity (k) and heat flux (Q) supplied above we can estimate when the
temperature reaches 600K (Tm/2). Heat flux is given by: 𝑄 = 𝑘 𝑇! − 𝑇!"#$%&' 𝑧
Rearranging for z: 𝑧 = 𝑘 𝑇! − 𝑇!"#$%&' 𝑄 and substituting in values from the question
then z = 15km.
If this maximum depth of elastic behavior is roughly equal to the depth at which the
highest mountains (height h) are supported then: z =0.865*h or h = 17.3km. Not that far
off!
If Io’s mantle has a density of 3300 Kg m-3 then how deep of a crustal root
would be required to support a 17km high mountain through Airy Isostasy?
Knowing what you now know about Io’s internal temperatures is this a
reasonable way to support these mountains?
If mountains on Io are floating because they have crustal roots that displace mantle
material then we can balance the weight of the mountain with the buoyancy force on the
root per unit area.
ρc gh = ( ρ m − ρc ) ghr
" ρc %
hr = $
'h
# ρ m − ρc &
If h is 17km and the crustal and mantle densities are 3000 and 3300 kg m-3 respectively
then the depth of the root must be 170km (in addition to whatever the average crustal
thickness is).
Airy Isostasy is therefore very inefficient on Io because the crust/mantle density contrast
is small. We’ve seen in the last question that rocks reach half their melting temperature
on Io at depths of only ~15km. Extrapolating this further downwards would imply
reaching the melting temperature at depths of only about 33km. So maintaining a
coherent crust root to depths > 170km would be impossible.
In summary, we looked at a few options for supporting Io’s 17km high mountains.
Support by strength alone is limited by the very thin elastic lithosphere on Io. So even
though gravity there is low, mountains cannot reach their full potential height. Similarly
support by Isostasy seems unlikely, as distinct crustal roots are unlikely to survive to the
great depths needed to support these mountains.
Support by material strength where the depth of support is limited to be within the
elastic lithosphere provides the best explanation.