school of earth sciences examination semester 1 final examination… · examination semester 1...

Semester One Final Examination 2016 ERTH2020 Introduction to Geophysics

NOTE: This examination paper must be submitted inside your examina-tion booklet, otherwise your examination will not be marked.

School of Earth SciencesEXAMINATION

Semester 1 Final Examination, 2016ERTH2020 Introduction to Geophysics

Examination Duration: 150 minutes (2.5 hours)Reading Time: 15 minutes

Answer 4 questions only.All questions are of equal value (20 marks) with part marks as indicated.

Exam Conditions:Closed book examination.During reading time, writing is permitted only in this document, not in answer booklet.No electronic aids (laptops, phones etc). UQ approved calculators only.Materials supplied: 2 answer booklets.

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Question 1

(a) How does gravitational potential relate to gravitational potential energy. (1)

(b) Consider the derivation for gravitational potential at a distance r from a point mass m.

(i) The derivation begins with the statement

dU =Gm

r2dr

With the aid of a sketch explain the meaning of this statement.(ii) To determine the integration constant C we need to use a boundary condition. This

defines the sign of gravitational potential. What is that boundary condition? (2)

(c) Write down (do not prove) the resultant expression for gravitational potential at a dis-tance r from a point mass m. (1)

(d) Assume that to a first approximation the earth acts as if all its mass was at the centre.Using the formula from (c) we are able to calculate the potential at any distance from thecentre of the earth. Write down an expression which would then allow us to estimatethe earth’s vector gravitational acceleration (g) from the potential (U ). (1)

(e) Using a numerical approach, based on Part (d), estimate the gravitational acceleration(magnitude and direction) experienced by a satellite orbitting 2500km above the surfaceof the earth. Carefully explain the sign of your answer. (3)

(f) Gravitational acceleration can also be calculated directly using the relationship

g = −Gmr2

r̂

Use this direct form to verify your result in (e). (2)

In magnetics, the observed B field is made up of a component BH which would occur in avacuum, and an induced component BM resulting from the susceptibility of the material.That is

B = BH + BM

where BH = µ0H and BM = µ0M.

A magnetic survey is being carried out in a region of SE Qld where the BH vector has mag-nitude 52750 nT, and an Inclination of -56 ◦. A NS profile is run across a high susceptibilitydyke, which strikes EW. At a traverse point near the dyke, the BM vector has magnitude1250 nT, and is acting horizontally, partially cancelling BH.

(g) Sketch a section showing the dyke, and primary and induced fields. Illustrate thegeneral location along the profile where the induced field would be horizontal. (2)

(h) What would be the reading on a proton-precession magnetometer at that location? (3)

(Q1 continued over page)

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Question 1 (continued)

A magnetic survey is being carried out over a 2-day period. On Day 1, all drift readings aremade at Base Station A. For logistical reasons, a different base station (B) is used on Day 2.The following table gives base station readings made on Day 2. The two readings made atBase Station A were for the purpose of data integration.

Base Station Time Reading (nT)

A 0755 52461B 0800 53350B 1000 53373B 1400 53373B 1700 53410A 1705 52519

(i) Why is the morning reading at Base Station A different from the afternoon reading. (0.5)

(j) Why is the reading at Base Station A (at any given time) different from that at B. (0.5)

The following table gives two of the field readings taken on Day 2.

Station Time Reading (nT)

F1 0900 53393F2 1635 52224

(k) Calculate the drift-corrected values, relative to Base Station B, for stations F1,F2. (2)

(l) It is subsequently desired to adjust all Day-2 readings so that they can be merged withthe Day-1 data. What are the final corrected values for stations F1, F2 (relative to BaseStation A). (2)

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Question 2

(a) Explain the relationship between resistivity (ρ) and resistance (R) for a wire of length land cross sectional area A. (1)

(b) Figure 1 shows a point source of current (I) on a homogenous earth having resistivity ρ.Using the result in (a) deduce the resistance of a hemispherical shell of thickness dr at adistance r from the current source. (1)

Figure 1: Current flow (red lines) away from a point electrode

(c) Hence derive an expression for the change in electrical potential dV across the shell.(1)

(d) Integrate in the radial direction, applying a reasonable boundary condition to deducethe constant of integration. Hence derive the expression for the potential V at a distancer from a point source of current. (3)

(e) Sketch a general 4-electrode array (C1, C2, P1, P2), and use the result from (d) to derivean expression for the potential difference ∆V measured between the potential electrodes(P1, P2). (3)

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The gradient-array is a very efficient means of profiling, particularly suited for steeply-dipping resistive bodies. It is, however, quite different to most other arrays, in that thecurrent electrodes C1, C2 are fixed, while the potential electrodes P1, P2 move laterally.Figure 2 illustrates three stages of a particular gradient-array experiment, as the P1 andP2 electrodes move from left to right along the line.

Figure 2: Gradient-array profiling exercise. The three sketches (top to bottom) show the situation at the start of the profile(x=100m), mid-way through the profile (x=500m), and at the end of the profile (x=900m). Current electrodes are fixed(C1 at 0m, C2 at 1000m). Potential electrodes have a fixed spacing (P1P2 = 50m) and are moved horizontally between thecurrent electrodes, moving by 10m between each reading. The plotting point (x) is mid-way between the P1 P2 electrodes.

(f) For the experiment shown in Figure 2, imagine first that the earth is homogenous, andthat the current is constant. The voltage ∆V measured along the profile will still change,because the electrode geometry changes from reading to reading. Using your formulafrom (e) show that the voltage reading (∆V ) will be about an order of magnitude largerat the start and end of the profile, compared to the centre. (3)

(g) Because of this changing electrode geometry, a simple formula for apparent resistivity(ρα) is not possible. It must be calculated using an expression

ρα = K∆V

I

where the geometric factor K varies continuously (and smoothly) along the line.

For the experiment in Figure 2, calculate the geometric factor K for x=100m, 500m, 900m.Then, assuming K changes smoothly along the line, sketch the general form of K as afunction of x (4)

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Now consider a real experiment based on Figure 2, where the earth is actually changingalong the profile. Readings taken along the central part of the profile are shown below. (Thelisted x coordinate is at the plotting point, mid-way between P1 and P2.)

x (m) ∆V (mV) I (mA)

400 13 420410 12 420420 13 419430 12 419440 12 419450 12 419460 23 418470 42 418480 27 418490 12 418500 11 417510 12 417220 11 417530 12 417540 12 417550 12 417560 12 417570 12 417580 13 416590 13 416600 13 416

(h) The current values seem relatively consistent from reading to reading. Why is thisreasonable? (0.5)

(i) Give a possible cause for the slight decrease in current over time . (0.5)

(j) What is the apparent resistivity of the country rocks in this area. (1)

(k) A resistive dyke is present on this part of the profile. Where is it, and what is itsapparent resistivity? (2)

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Question 3

Resistivity sounding is being performed using the Schlumberger array. An expression con-venient for field use is

ρα = πG∆V

I

where G is a geometric factor given by

G =(L2 − l2)

2l

Here L and l are the distances from the centre of the array to the current and potentialelectrodes respectively.

(a) A reading is being taken with L = 220 m and l = 10 m. With the current off, a steadyreading of 9 mV is observed on the voltmeter. What phenomenon is being observed.

(1)

(b) When the current is switched on the voltmeter reads 47 mV, and the current is 353mA.Calculate the apparent resistivity. (2)

(c) For the purposes of error estimation we will assume that L2 � l2. Write down the for-mula for a simplified geometric factor G′ to be used in error estimation. The estimated er-rors in the current and potential electrodes are ± 0.1m and ± 0.01m respectively. Hence,estimate the approximate percentage error in the geometric factor. (2)

(d) The accuracy of the meters is such that the estimated error in each voltage reading is± 1mV. The error in each current reading is ± 1mA. Calculate the absolute error in theapparent resistivity estimated in Part (b). (3)

(e) The Schlumberger sounding is being recorded using the following equipment:

• 1 x 12V battery• 1 x 12-500V converter• 2 x current cables• 2 x brass spikes• 1 x ammeter• 2 x potential cables• 1 x voltmeter• 2 x porous pots

Draw a sketch which indicates how these items are connected. (3)

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Figure 3 shows the Schlumberger sounding curve obtained.

Figure 3: Schlumberger sounding curve

(f) Explain how many layers are indicated in this location. Estimate the layer resistivities.(You may need to use ≤ or ≥ symbols) (3)

(g) Assume that for the Schlumberger array a nominal depth of investigation is 0.125 *(total array length). Estimate approximate depths to the interfaces, and sketch yourresultant model. (3)

(h) With reference to the concept of equivalence, give a specific numerical example of howthe resistive layer in (g) could be subject to non-uniqueness. (3)

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Question 4

A short seismic refraction line has been recorded perpendicular to the direction of a pro-posed road, to define the best location for a road cutting. Figure 6 shows the measuredarrival times at each geophone. The times are also tabulated below.

Figure 4: Refraction arrival times. Geophone spacing 10m. Shots at 0m and 120m.

Coordinate (m) tF (ms) tR (ms)

0.0 82.010.0 23.0 78.020.0 35.0 74.030.0 37.5 70.540.0 41.0 68.050.0 46.5 67.560.0 54.0 68.070.0 59.5 65.580.0 64.5 61.590.0 69.5 58.5100.0 73.0 53.0110.0 79.0 23.0120.0 82.0

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(a) Assuming a simple overburden - bedrock situation, estimate the velocity in the surfacelayer (v1). (2)

(b) Which interpretation method is appropriate for the bedrock layer, and why? (1)

(c) The reciprocal method cannot necessarily be applied at all geophones on the spread.Why? (1)

(d) With the aid of a sketch (or sketches), explain how to calculate:

(i) the velocity function at a geophone (tV )

(ii) the refractor velocity at a particular point along the spread

(iii) the time depth function (tG) at a geophone

(iv) the refractor depth at that geophone (4)

(e) Compute and plot the velocity function (tV ) at relevant geophones, and hence com-ment on any variations in the bedrock velocity (v2). (You may use the grid below.)

(4)

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(f) Compute and plot the time-depth function (tG). (You may use the grid below.) Calculatethe overburden thickness at its shallowest and deepest points. (4)

(g) The road cutting needs to be 20m deep, and 30m wide. Suggest the optimum location,giving two reasons. (4)

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Question 5

Figure 5 shows part of a Vibroseis shot record from central Australia. More details are givenin the figure caption.

(a) Examine the approximately linear events highlighted in blue and yellow. Identify thetype of event (e.g. direct wave, refraction, reflection etc..). Extend the marking along theevents (Blue: approximately from Geophones 40-87. Yellow: 94-135). (2)

(b) Estimate the apparent velocities of the blue and yellow events. Hence, comment on theorientation of the refracting interface (approximately horizontal, dipping towards high-numbered geophones etc.). Estimate the true velocity below this interface (v2). (3)

(c) Draw in an axis at the shotpoint (between Geophones 90 and 91). Estimate the interceptsfor the yellow and blue events (which should be similar). Direct arrivals around Geo-phones 92-94 indicate an approximate surface velocity (v1) of about 1200 m/s. Estimatethe depth to the interface. What is the likely geological significance of this interface? (3)

(d) What type of events are highlighted in pink and green? Extend the markings as far aspossible in each direction (Pink: Geophones 35 -70, Green: 52-72). (1)

Because of strong surface waves, it is hard to track these events back towards the centre ofthe record. Hence, we cannot use our normal method of determining velocity. We need todevise an alternative approach.

(e) Figure 6 shows reflection ray paths to two geophones at distances x1 and x2 from a shot.It also shows the zero-offset ray path. Using the NMO concept, write down an equationwhich relates the reflection time tx1, for the geophone at x1, to the zero-offset time t0 andthe layer velocity v. Write a second equation for the reflection time tx2, for the geophoneat x2. (1)

(f) Eliminate t0 from the equations in (e), and hence derive an expression for the layer veloc-ity v in terms of the parameters x1, x2, tx1, tx2. (4)

(g) Use the result in (f) to estimate the average velocity down to the reflecting interface forthe pink event. Now, use the standard NMO equation to estimate the zero-offset time (t0),and hence the depth to the interface. (4)

(h) Repeat this process for the green event, to determine the velocity above the reflectinginterface and its depth. (2)

Appendix: Formulas and Constants

G = 6.67 ∗ 10−11 N m2 kg−2ME = 5.97 ∗ 1024 kgRE = 6371 km

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Figure 5: Part of Vibroseis shot record from central Australia. Here the distance axis is geophone number (30-150shown). The geophone spacing is 20m. The shot point is midway between Geophones 90 and 91. The time axis goes from0s to 1s in increments of 0.1s. Selected events are highlighted. Note that, for these Vibroseis data, the true arrival time ofan event corresponds to a black peak, as marked.

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Figure 6: Reflection ray paths for two geophones at distances x1 and x2 from a shot. The zero-offset ray path is alsoshown. The layer velocity is v.

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