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IPhO 1986, Теория — Задача 2. Q2

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IPhO 1986 — Theory

Условие

International Physics Olympiad 1956

Early this century a model of the earth was proposed in which it was assumed to be a sphere of radius $R$ consisting of a homogeneous isotropic solid mantle down to radius $R_c$ . The core region within radius $R_c$ contained a liquid. Figure 2.1

Earth as a large circle with a concentric core circle, centre O; radii to surface points E and X enclose angle 2θ; arrows mark the radius R and the core radius R_C. — Figure 2.1

The velocities of longitudinal and transverse seismic waves P and S waves respectively, are constant, $V_P$ , and $V_S$ within the mantle. In the core, longitudinal waves have a constant velocity $V_{CP}$ , $< V_P$ , and transverse waves are not propagated.

An earthquake at E on the surface of the Earth produces seismic waves that travel through the Earth and are observed by a surface observer who can set up his seismometer at any point X on the Earth’s surface. The angular separation between E and X, $2\theta$ given by

2\theta = \text{Angle } EOX

where O is the centre of the Earth.

(i) Show that the seismic waves that travel through the mantle in a straight line will arrive at X at a time t (the travel time after the earthquake), is given by

t = \frac{2R\sin\theta}{v}, \qquad \text{for } \theta > \arccos\left[\frac{R_c}{R}\right],

where $v = v_P$ for the P waves and $v = v_S$ for the S waves.

(ii) For some of the positions of X such that the seismic P waves arrive at the observer after two refractions at the mantle-core interface. Draw the path of such a seismic P wave. Obtain a relation between $\theta$ and $i$ , the angle of incidence of the seismic P wave at the mantle-core interface, for P waves.

(iii) Using the data

\begin{aligned} R &= 6370\ \mathrm{km} \\ R_C &= 3470\ \mathrm{km} \\ v_{CP} &= 10.85\ \mathrm{km\ s^{-1}} \\ v_S &= 6.31\ \mathrm{km\ s^{-1}} \\ v_{CP} &= 9.02\ \mathrm{km\ s^{-1}} \end{aligned}

and the result obtained in (ii),draw a graph of $\theta$ against $i$ . Comment on the physical consequences of the form of this graph for observers stationed at different points on the Earth's surface.
Sketch the variation of the travel time taken by the P and S waves as a function of $\theta$ for $0 \le \theta \le 90$ degrees.

(iv) After an earthquake an observer measures the time delay between the arrival of the S wave, following the P wave, as 2 minutes 11 seconds. Deduce the angular separation of the earthquake from the observer using the data given in Section (iii).

(v) The observer in the previous measurement notices that some time after the arrival of the P and S waves there are two further recordings on the seismometer separated by a time interval of 6 minutes 37 seconds. Explain this result and verify that it is indeed associated with the angular separation determined in the previous section.

Отговори

Решение

Answers Q2

Earth with core, centre O: surface points E and X near the top with angle θ θ between radii; a dashed straight ray from E tangent to the core reaches X', with angle φ at O and radius R to X'; label R on the radius to E. — Figure2.1

2.(i)

\mathrm{EX} = 2R\sin\theta \quad \therefore t = \frac{2R\sin\theta}{v}

where $v = v_P$ for P waves and $v = v_S$ for S waves.

This is valid providing X is at an angular separation less than or equal to X', the tangential ray to the liquid core. X' has an angular separation given by, from the diagram,

2\phi = 2\cos^{-1}\left(\frac{R_C}{R}\right),

Thus

t = \frac{2R\sin\theta}{v}\quad , \qquad \text{for } \theta \le \cos^{-1}\left(\frac{R_C}{R}\right),

where $v = v_{\mathrm{P}}$ for P waves and $v = v_{\mathrm{S}}$ for shear waves.

(ii) $\dfrac{R_C}{R} = 0.5447$ and $\dfrac{v_{CP}}{v_P} = 0.831.3$

Earth with core, centre O: ray from E enters the core at A with incidence angle i and refraction angle r, crosses to B and exits to X; C is the foot of the perpendicular from O to AB; angles α at E and θ at O, core radius R_c. — Figure 2.2

From Figure 2.2

\theta = A\hat{O}C + E\hat{O}A \Rightarrow \theta = (90 - r) + (1 - \alpha) \qquad (1)

(ii) Continued

Snell’s Law gives:

\frac{\sin i}{\sin r} = \frac{v_P}{v_{CP}}. \qquad (2)

From the triangle EAO, sine rule gives

\frac{R_C}{\sin x} = \frac{R}{\sin i}. \qquad (3)

Substituting (2) and (3) into (1)

\theta = \left[90 - \sin^{-1}\left(\frac{v_{CP}}{v_P}\sin i\right) + i - \sin^{-1}\left(\frac{R_C}{R}\sin i\right)\right] \qquad (4)

(iii)

For Information Only
For minimum $\theta, \dfrac{d\theta}{di} = 0$ . $\Rightarrow 1 - \dfrac{\left(\frac{v_{CP}}{v_P}\right)\cos i}{\sqrt{1 - \left(\frac{v_{CP}}{v_P}\sin i\right)^2}} - \dfrac{\left(\frac{R_C}{R}\right)\cos i}{\sqrt{1 - \left(\frac{R_C}{R}\sin i\right)^2}} = 0$

Substituting $i = 55.0^{\mathrm{o}}$ gives LHS=0, this verifying the minimum occurs at this value of $i$ . Substituting $i = 55.0^{\mathrm{o}}$ into (4) gives $\theta = 75.8^{0}$ .

Plot of $\theta$ against $i$ .

Plot of θ against i from 0 to 90°: θ starts near 90°, falls to a minimum of 75.8° at i = 55°, and rises to just above 90° at i = 90°; dashed guide lines mark these values.

Substituting into 4:

$i = 0$ gives $\theta = 90$

$i = 90^{\circ}$ gives $\theta = 90.8^{\circ}$

Substituting numerical values for $i = 0 \to 90^{\circ}$ one finds a minimum value at $i = 55^{\circ}$ ; the minimum values of 0, $\theta_{\mathrm{MIN}} = 75{\bullet}8^{\circ}$ .

Physical Consequence

As $\theta$ has a minimum value of $75{\bullet}8^{\circ}$ observers at position for which $2\theta < 151{\bullet}6^{\circ}$ will not observe the earthquake as seismic waves are not deviated by angles of less than $151{\bullet}6^{\circ}$ . However for $2\theta \le 114^{\circ}$ the direct, non-refracted, seismic waves will reach the observer.

Travel time against θ: curves S (upper) and P (lower) rising from 0 up to a dashed line at θ_C = cos⁻¹(R_C/R); after a gap, a flat P curve labelled 'Refracted waves' between θ_min and 90°; a second axis below runs from 90° to 180° with an arrow pointing left.

(iv) Using the result

t = \frac{2r\sin\theta}{v}

the time delay $\Delta t$ is given by

\Delta t = 2R\sin\theta\left[\frac{1}{v_S} - \frac{1}{v_P}\right]

Substituting the given data

131 = 2(6370)\left[\frac{1}{6.31} - \frac{1}{10.85}\right]\sin\theta

Therefore the angular separation of E and X is

2\theta = 17.84^{o}

This result is less than $2\cos^{-1}\left(\dfrac{R_C}{R}\right) = 2\cos^{-1}\left(\dfrac{3470}{6370}\right) = 114^{o}$

And consequently the seismic wave is not refracted through the core.

(v)

Earth with core radius R_C, centre O: ray from E to point D on the core surface reflected to X, radii R to E and X, angles θ θ at O, Y the foot of the perpendicular on chord EX.

The observations are most likely due to reflections from the mantle-core interface. Using the symbols given in the diagram, the time delay is given by

\Delta t' = (\mathrm{ED} + \mathrm{DX})\left[\frac{1}{v_S} - \frac{1}{v_P}\right]

\Delta t' = 2(\mathrm{ED})\left[\frac{1}{v_S} - \frac{1}{v_P}\right] \ \text{ as ED} = \text{EX by symmetry}

In the triangle EYD,

(\mathrm{ED})^2 = (R\sin\theta)^2 + (R\cos\theta - R_C)^2

(\mathrm{ED})^2 = R^2 + R_C^2 - 2RR_C\cos\theta \qquad \sin^2\theta + \cos^2\theta = 1

Therefore

\Delta t' = 2\sqrt{R^2 + R_C^2 - 2RR_C\cos\theta}\left[\frac{1}{v_S} - \frac{1}{v_P}\right]

Using (ii)

\Delta t' = \frac{\Delta t}{R\sin\theta}\sqrt{R^2 + {R_C}^2 - 2RR_C\cos\theta}

\Rightarrow 396.7s\ \text{ or 6m 37s}

Thus the subsequent time interval, produced by the reflection of seismic waves at the mantle core interface, is consistent with angular separation of $17.84^{0}$ .

Оригинал в Архива: th.pdf · официални решения: th_sol.pdf