Homework #1 Humphrey Geology 4880 Fall 2013

Some notes on the questions and on the answers: This is a great example of why you have to be careful when finding “answers” on the Web. I generally assume anything on the web is incorrect until checked by an independent means.

1 Ok, so this is one example of a problem that (done correctly) you should be able to reproduce on an exam… hint hint

a

shear or driving stress

2500 kg m^-3 x 10 m s^-2 x 2 m x sin(35) = 29,000Pa

normal stress

2500 kg m^-3 x 10 m s^-2 x 2 m x cos(35) = 41,000pa

resisting stress is

2500 kg m^-3 x 10 m s^-2 x 2 m x cos(35) x tan(30) = 24,000

F.S. = 0.84

b for stability F.S. = 1, therefore S must make up the difference between the frictional resisting stresses and the driving stresses:

therefore S = 29,000Pa – 24,000Pa or S ~ 5,000Pa

c From class notes, a = g ( sin(35) – cos(35) x tan(30) ) ~ 1 m/s2

d and v2 = 2aL, where L is 100m/sin(35) ~ 175m

At the road it is going 19 m/s, and the horizontal deceleration is

a = g x (-tan(30)) ~ 5.7 m/s2

So L = 350/2a, or about 30m

The width of a five lane road is about 50m, so I figure it would stop in the middle of one of the lanes.

2 a) area of continents ~ 1.5 x 10¹⁴m², from taking 30% of earths surface. Volume is that area times 1 m thickness. Answer is volume divided by (volume) output per time ~ 15,000yrs.

b) a little less than 0.1mm/year

c) from the USGS web page the drainage area is about 4000 square miles. The volume per year is just the depth times the erosion rate (all in the same units). Change volume to mass, using density. Answer on the order of 5 x 10⁹ kg/year

d) Discuss in class

3 a) b) discuss in class

c) no

4 for plots see:

fig1 fig2 fig3 (also fig 4 gives the relative shapes of 2 common curves.)

This exercise was for several reasons. I wanted to check your graphing abilities, but I also wanted you to think about the shape of some of the important curves we will be using. Finally I wanted you to see that the shapes of curves vary hugely with the coordinate axes.

5

a) Chance of an impact in one year is 1/75,000,000, assume you will live for 75 years (makes the calculation easy), then your chances are about 1/1,000,000, or your chance of dying by asteroid is about 1 in a million per lifetime.

b) There are many ways to calculate your chance of dying by a landslide. One approach would be to note that the death rate in the US is about 25 per year, this leads to a probability of 25/300,000,000 per year or less than 1/100,000 per lifetime. Of course statistics are only as good as the data and interpretation. For example, I would suspect that the number of deaths in Kansas from landslides is 0, so in Kansas you are much more likely to die from a meteorite.

However, you should note that probabilities of infrequent natural events are actually quite difficult to estimate, and even common events such as river floods are much more difficult to treat statistically than is commonly believed (and please don’t believe most of the numbers quoted by “media” sources)

c) From same web site as 2c the yearly average discharge is about 1000 cubic ft per sec. This is 8.6 x 10⁸ m³ per year. Since a cubic meter of water is about 1000kg, the average calculated sediment load by weight is about 0.6%. Note this is more typically expressed in units of weight per volume, for kg per liter the answer is the same (0.006kg/liter). The actual question can be answered several different ways. The easiest is just to recognize that 15% of the answer in 2c is just a load of 900mg/liter.

The data in this question, both the dissolved load and the erosion rate, is very rough, nevertheless it is often the case that much of the erosion in passive regions is by chemical dissolution, and the percentage usually increases as the erosion rate decreases.