Question 5.6.8: Photography from a spy plane In the late 1950s, the Soviets ...
Photography from a spy plane
In the late 1950s, the Soviets labored to develop a missile that could stop the U-2 spy plane. On May 1, 1960, Nikita S. Khrushchev announced to the world that the Soviets had shot down Francis Gary Powers while Powers was photographing the Soviet Union from a U-2 at an altitude of 14 miles. How wide a path on the earth’s surface could Powers see from that altitude? (Use 3950 miles as the earth’s radius.)
Figure 5.90 shows the line of sight to the horizon on the left-hand side and righthand side of the airplane while flying at the altitude of 14 miles.
Since a line tangent to a circle (the line of sight) is perpendicular to the radius at the point of tangency, the angle α at the center of the earth in Fig. 5.90 is an acute angle of a right triangle with hypotenuse 3950 + 14 or 3964. So we have
\begin{aligned} \cos \alpha &=\frac{3950}{3964} \\ \alpha &=\cos ^{-1}\left(\frac{3950}{3964}\right) \approx 4.8^{\circ} . \end{aligned}The width of the path seen by Powers is the length of the arc intercepted by the central angle 2α or 9.6°. Using the formula s = αr from Section 5.1, where α is in radians, we get
s=9.6 \mathrm{deg} \cdot \frac{\pi \mathrm{rad}}{180 \mathrm{deg}} \cdot 3950 \text { miles } \approx 661.8 \text { miles. }From an altitude of 14 miles, Powers could see a path that was 661.8 miles wide. Actually, he photographed a path that was somewhat narrower, because parts of the photographs near the horizon were not usable.
