solution to problem set 2

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California Insititute of Technology

Solar System Science

Problem Set 2 Solutions

Monday 14 April 2013

Problem 1: Cratering Ages

During Apollo 16’s mission to the Moon, astronauts John Young and Charlie Duke traversed

about 20 km, picking up about 96 kg of Moon rocks along the way. One of those rocks has

now been classified as sample 61015. It was picked up near Nectaris basin on the Moon, and

has been characterized as an impact melt created during the Nectaris basin impact.

Figure 1: From lpi.usra.edu

In this problem, you will apply the concepts of radioactive dating and crater densities to

determine the average flux of impacts on the Moon since the formation of Nectaris basin.

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A) Suppose you measure the amount of samarium and neodymium in sample 61015. You

determine that one portion of 61015 has14762

Sm

14460

Nd= 0.176 and

14360

Nd

14460

Nd= 0.104752 and another

has14762

Sm

14460

Nd= 0.2516 and

14360

Nd

14460

Nd= 0.107603. How old is sample 61015?

Note : The decay chain of the samarium-neodymium system is 14762

Sm → 143

60Nd + 4

2α with

a half life of about 1011 years. 144

60Nd is a stable isotope of neodymium. Scientists actually

used argon-argon dating to determine the age of sample 61015.

B) The majority of other collected rocks which are identified as impact ejecta on the Moon

have dates which are very similar to that of sample 61015. Discuss possible implications of

this finding.

C) The LOLA (Lunar Orbiter Laser Altimeter) instrument aboard the Lunar Recon-

naissance Orbiter has determined that the crater density at Nectaris Basin is about 135±14

craters with diameters greater than 20 km per 106km2. Another large basin, the Imbrium

basin, has an estimated age of about 3.8 Gyr and ∼28 20-km-or-larger craters per 106 km2.Neither basin appears to have lost craters due to erosion or volcanic fill since the time of

formation. How can you reconcile these two measurements?

Figure 2: Locations of Nectaris and Imbrium impact basins on the Moon

D) Sketch a cratering history of the moon.

E) How might you adjust your calculated cratering rate for the Moon so that it could be

applied to Mars?

Solution to A To determine the age of this sample we use,

daughter

stable

t

=

daughter

stable

0

+

parent

stable

t

(eλt − 1) (1)

where t is time since formation of the sample, and λ is the decay constant and is given by

λ = ln2

τ where τ is the half-life of the decay chain. Here, the parent isotope is 147

62Sm, the

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daughter isotope is 143

60Nd, and the stable isotope is 144

60Nd. If we were to plot the data given

for the isotopic ratios of the sample we would find that they fall on a line having a y-intercept

of

14360

Nd

14460

Nd

0

and a slope of (eλt − 1). We will calculate the slope and solve for t:

eλt − 1 =

∆

14360

Nd

14460

Nd

∆

14762

Sm

14460

Nd

(2)

Therefore, the age of sample 61015 is 3.84 Gyr.

Solution to B The fact that the majority of other collected rocks which are identified

as impact ejecta on the Moon have dates which are very similar to that of sample 61015

suggests that there was a pronouned period with a high impact rate. In fact, this is calledthe Late Heavy Bombardment.

Solution to C Based on the data, Nectaris seems to be slightly older than Imbrium.

This is further evidence that there was a period of high impact flux around the time of the

formation of Nectaris which tapered off shortly thereafter when Imbrium was formed.

Solution to D See figure 3

Figure 3: Sketch of the cratering history of the Moon.

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Solution to E To be applied to Mars, the Moon’s cratering rate will have to be adjusted

to account for Mars’ different location in the Solar System, its gravity, its atmosphere, and

its varying surface conditions over time.

Problem 2: The Great Escape

In this problem, we’ll see why the Moon has a very thin atmosphere of sodium atoms but

virtually no atmospheric hydrogen.

A) Derive an expression for the most common velocity of (or the velocity at the peak of)

the Maxwell-Boltzman distribution.

B) Calculate the average thermal velocity of a hydrogen atom on the Moon. The Moon’s

daytime temperature is ∼ 400 K.

C) Calculate the average thermal velocity of a sodium atom on the Moon.

D) Derive an expression for the escape velocity of a planetary body. Calculate the escape

velocity of the Moon.

E) Sketch the Maxwell-Boltzmann velocity distribution curves for hydrogen and sodium

on the Moon. Label the Moon’s escape velocity. Caption your graph, explaining why the

Moon has a tenuous atmosphere of Na but no H.

Solution to A The Maxwell-Boltzmann distribution is given by

f (v) =

m2πkT

3

2

4πv2e−mv

2

2kt (3)

where f (v) is the probability of finding a particle with a speed near v per unit speed, m is

the particle mass, k is the Boltzmann constant, and T is the temperature of the system. For

simplicity, let’s define A = 4π( m2πkT

)3

2 . To find the most likely (or most common) velocity,

we will calculate df

dv |v=vth= 0.

df

dv |v=vth= 0 = A

2ve−

mv2

th

2kT − v2th(2vth)

m

2kT e−

mv2

th

2kT

(4)

2Avthe−

mv2th

2kT

1 − v2th

m

2kT

= 0 (5)

v2th = 2kT

m (6)

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Therefore, the most common velocity vth =

2kT

m .

Figure 4: Sketch of the Maxwell-Boltzmann distribution.

Solution to B For hydrogen on the Moon, T = 400K, k = 1.38 · 10−23J/K, and

m = 1.67 · 10−27kg.

Therefore, vth = 2.6 km/s.

Solution to C For sodium on the Moon, T = 400K, k = 1.38 · 10−23J/K, and

m = 1.67 · 23 · 10−27kg.

Therefore, vth = 0.5 km/s.

Solution to D To find the escape velocity on the Moon, we equate gravitational potential

energy with kinetic energy.GMm

R =

1

2mv2esc (7)

where G is the gravitational constant, M is the mass of the Moon, m is the mass of the

particle escaping, and R is the radius of the Moon. This yields

vesc =

2GM R (8)

Therefore, for the Moon, vesc = 2.4 km/s.

Solution to E See Figure 5

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Figure 5: Most hydrogen atoms move faster than vesc while the vast majority of sodium

atoms move slower than vesc. Hence, there is no hydrogen on the Moon, but there is a

tenuous atmosphere of sodium.

Problem 3: Order of Magnitude Planetary Science

Order of magnitude estimation is a powerful tool to assess whether a hypothesis is worth

pursuing; if rough calculations can show that the predictions are far from the observations,

then detailed calculations aren’t really necessary and one can move on to more plausible

hypotheses. This method is extremely useful in planetary science, where observations are

usually no more precise than a factor of a few. We will now use it to tackle a simple

hypothesis: The oceans of Mars are all locked up in subsurface ice deposits.

A) Consider Figure 6, which shows the possible ancient shorelines of the primordial

Martian ocean. Given what is known of the elevations on Mars, e.g. from MOLA, estimate

the ocean’s total volume.B) Now consider Figure 7, which shows the global distribution of subsurface ice. Estimate

the thickness of the subsurface ice layer necessary to account for all the water in the ancient

Martian ocean, assuming the water mass fraction is constant with depth. Remember: this is

order of magnitude estimation; an integral over the entire surface is probably not necessary.

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http://mola.gsfc.nasa.gov/images/topo_labeled.jpg

http://mola.gsfc.nasa.gov/images/topo_labeled.jpg



C) How plausible are your results from part b? Is our hypothesis viable?

Figure 6: The possible shorelines of ancient Mars, as shown in lecture 1.17 (Di Achille andHynek, Nature Geosciences 3, 2010).

Solution to A Knowing the elevations of the Martian surface, we can estimate an average

depth to the Northern Ocean. For example, the shoreline appears to be at ∼ −2500 m, thedeepest parts are at ∼ −4500 m, and the shallower parts are at ∼ −3500 m; from these

numbers we can say that the average depth of the ocean is around 1500 m. Note that we

are not including any other bodies of water on Mars, such as Hellas and Argyre Planitia,

as they are smaller than the Northern Plains; including them would affect our result by less

than a factor of 2.

The extent of the ocean can be found from looking at Figure 6. We see that the shorelines

go as south as the equator, but on average they seem to oscillate around 30◦N. We can then

model the ocean as a 1500 m thick layer of water covering the planet north of 30 ◦N. As this

thickness is far less than the planetary radius, we can model the volume of this ocean as

V = Adr (9)

where A is the surface area covered by the ocean, and dr is the average depth of the ocean

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Figure 7: The percentage by mass of ice in Martian soil, as shown in lecture 1.18 (From

http://www.jpl.nasa.gov/news/news.php?release=2004-209)

(1500 m calculated above).

To find A, let us consider the area of a spherical shell in spherical coordinates:

A =

2π

0

π3

0

R2sinθdθdφ (10)

where R is the planetary radius of ∼3400 km, θ is the colatitude, and φ is the longitude.

Solving this integral gives area A = πR2. Note that, since this is order of magnitude

estimation, you can also arrive at a similar answer by just looking at the area and guessing.

Putting all these numbers together, we get an ocean volume V of:

V = πR2dr (11)

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Therefore, the volume of the primordial Martian ocean is about 5·1016 m3.

Solution to B To determine the thickness of a susbsurface ice layer, we need to know

density of ice, which is ∼ 900 kg/m3, which is close enough to water’s density of 1000 kg/m3

that we won’t care the result will not be drastically impacted by the difference. Therefore,we must find a way to fit ∼ 5·1016 m3 of ice into the Martian subsurface. First, we must find

the area where most of the ice lies.

From Figure 7, we see that there is subsurface ice across the whole planet, but most of

it is concentrated to areas poleward of 60◦. Using the same integral as in part a, but with π6

taking the place of π3

, we find the area north of 60◦N to be:

A60◦N =

2π

0

π6

0

R2sinθdθdφ (12)

A60◦N ≈ 0.25πR2 (13)

Similarly, the ice trappend southward of 60◦ covers an area of ∼ 0.25πR2. The total area

of significant subsurface ice is ∼ 0.5πR2, or half of the area of the Northern Ocean.

On average, these areas have soil that is ∼ 33 % water ice by mass. If we assume that

the rest of the soil is basalt with density ∼ 3000 kg/m3, then we can write:

M w

M b=

1

2 (14)

Where M w is the mass of water in a volume of Martian surface material and M b is the

mass of basalt in the same volume. For a volume of 1 m3, we have:

V total = V w + V b (15)

V total = M w

ρw+

M b

ρb(16)

V total = M w

1

ρw+

2

ρb

(17)

M w

V total=

1

ρw+

2

ρb

−1

(18)

and we get a pseudo-density of water ice in the Martian soil poleward of 60◦ of 600 kg/m3.

As before, V = Adr = 0.5πR2dr in our case. To take into account the new pseudo-

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density, we must increase the total volume that will be taken up by the ice frozen from the

Northern Ocean by a factor f of 1000600

. We now calculate

dr = 2fV

πR2 , (19)

and find that the thickness of the required subsurface ice layer dr is about 6 km.

As this all comes from estimation, the exact answer is not important, but it is clear that

we will need a several kilometer thick layer of entrained ice in order to contain all the water

that may have once existed in the Northern Ocean.

Solution to C Does a layer of ice 5-7 km in thickness beneath the surface seem plausible?

The answer is likely no. One reason is that the temperature probably increases with depth

due to internal heating, raising the temperature above the stability point of water ice. Also

important is the permeability of the material − can water penetrate down to 5-7 km whensubduction is not present? It is unlikely that our hypothesis is viable. In fact, it is more

likely that most of the ocean was lost to space.

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