Discussion
Consider a spherical earth of radius 6000 km and surface velocity of 6 km/sec, with a velocity distribution of the type v (r) = \(a / \sqrt{r}\). At the distance reached by a wave emerging at a take-off angle of 45° from a focus on the surface. Calculate the interval between the arrival time of the directed P- and reflected PP-waves (the PP-wave is one that is reflected at the surface at the midpoint between the focus and the point of observation) :
Answer: C
The interval between P and PP is about 78 seconds.
With \(v = a/\sqrt{r}\), the ray quantity \(\eta = r/v\) varies as \(r^{3/2}\). For \(\eta \propto r^{n}\), rays satisfy \(r^{n}\cos n\theta = \) constant, so a ray leaving the surface at angle i from the vertical covers a distance \(\Delta = 2(90^\circ - i)/n\) and takes \(T = (2/n)\,\eta_0 \cos i\), where \(\eta_0 = R/v_0 = 6000/6 = 1000\) s.
\[ \begin{aligned} \Delta_P &= 2(90^\circ - 45^\circ)/1.5 = 60^\circ \\ T_P &= 1333.3 \cos 45^\circ = 942.8\ \text{s} \end{aligned} \]
Each leg of PP covers 30 degrees, so \(90^\circ - i = 1.5 \times 15^\circ\), giving \(i = 67.5^\circ\):
\[ \begin{aligned} T_{PP} &= 2 \times 1333.3 \cos 67.5^\circ \\ &= 2 \times 510.2 = 1020.5\ \text{s} \\ T_{PP} - T_P &= 77.7 \approx 78\ \text{s} \end{aligned} \]
The other values do not follow from this velocity law and geometry.
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