QQuestionEarth Science
QuestionEarth Science
Earth and Planetary Surface Processes
Winter 2017
Problem set 6
Due in class Wed 8 March, 3pm.
Office hours 10am- 11am, Hinds 467 Monday 6 March (note change in time from 9am).
Q^1. Plot the expected steady-state pattern (as a function of distance from the crest) of discharge of regolith on a convex hillslope in a climate in which the production rate of regolith is 20 mm/yr. Report your answer in m^2 /year. If the regolith was 0.8 m thick, what would be the spatial pattern of the mean downslope velocity of the regolith? How would these numbers change if 25% of the mass loss occurred in solution (dissolved in rainwater)? (From Anderson & Anderson)
Q^2. From Hartley et al., Nature Geoscience, 2011:
Seismic reflection profiling between Iceland and the UK shows a drowned landscape that was above sea level 55 Ma. The landscape was cut into mudstones and is now buried beneath >1 km of sediment, plus 700m of water. The landscape (black square in the below figure) was uplifted and then reburied due to the arrival of the head of mantle plume whose “stalk” now fuels volcanism at Iceland:
River profiles (gray is data, black is model):
What is the slope of the knickpoint marked β?
What is the (range of) approximate drainage areas at β?
Assuming n= 1 in the streampower law (so that knickpoint shape is preserved on retreat), what is the knickpoint retreat rate vAm ? You can assume v = 2.75 Myr- 1 and m = 0.5 for this question.
Approximately how many years prior to the landscape becoming “drowned” (frozen in shape) was the knickpoint initiated?
Refer to Figure 1. Assume the initiation of knickpoint β corresponds to the arrival of the “mushroom head” of the mantle plume, and the drowning of the landscape corresponds to the switch to the modern, narrow “plume tail” configuration.
What is the area over which uplift occurred?
Assuming the red-circled area was horizontal and below sea level before the plume arrived, what is the volume of uplifted rocks?
Assuming the drowned landscape under investigation is representative of the fractional erosion of the uplifted landscape across the whole red-circled area, what is the volume of rocks fluvially eroded in the plume area during the plume event?
What is the erosion flux in km^3 yr- 1?
How does this compare to the Holocene pre-dam global sediment flux of 8 km^3 yr- 1?
Q^3. In lecture we discussed accumulation of a stable cosmogenic isotope in a rock undergoing exhumation, and we also discussed accumulation of a radioactive cosmogenic isotope in a rock that has no erosion. In this question we will combine exhumation and decay.
Assume a production rate of 10Be of 5 atoms/(gram quartz)/yr at the surface, with an e-folding depth of 1 m (so the production rate is 1.84 atoms/(gram quartz)/yr at 1m depth). 10Be decays exponentially with a half-life of 1.4 Myr. Plot the expected concentration of 10Be at the surface, as a function of erosion/exhumation rate, for a quartzite rock. Make sure to consider the endmembers of very slow erosion/exhumation (for which the 10Be concentration will reach equilibrium between production and decay), and the endmember of very fast erosion/exhumation (for which decay is unimportant in setting the 10Be concentration)
3 months agoReport content
Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
Step 1:Calculate the regolith production rate in m²/yr
20 \text{ mm/yr} = 0.020 \text{ m/yr}
Convert the regolith production rate from millimeters per year to meters per year to standardize units.
Step 2:Steady-state discharge pattern of regolith
q_s(x) = R \cdot x
At steady state, the discharge of regolith per unit hillslope width at a distance x from the crest is the production rate times the distance from the crest.
Step 3:Report steady-state regolith discharge
q_s(x) = 0.020 x
The discharge pattern as a function of distance from the crest, in units of m²/yr.
Step 4:Calculate mean downslope velocity of the regolith
v(x) = \frac{q_s(x)}{h} = \frac{0.020 x}{0.8}
Divide the local discharge rate by the regolith thickness to get mean downslope velocity, in m/yr.
Step 5:Downslope velocity pattern
v(x) = 0.025 x
The mean regolith velocity at a location x is found by dividing discharge by thickness.
Step 6:Adjust for 25% mass loss in solution
q_{s,solids}(x) = 0.75 \times 0.020 x = 0.015 x \\ v_{solids}(x) = \frac{0.015 x}{0.8} = 0.01875 x
If 25% of mass is lost in solution, only 75% of regolith discharge is transported as solids.
Final Answer
q_s(x) = 0.020 x \\ v(x) = 0.025 x \\ q_{s,solids}(x) = 0.015 x \\ v_{solids}(x) = 0.01875 x
Discharge (m²/yr) and downslope velocity (m/yr), both as functions of distance from crest, for all scenarios.
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students