precalc question

Question

An aircraft is flying at the speed of 500 mph at an elevation of 10 miles above the earth, beginning at the North pole and heading South along the Greenwich meridian. A spy satellite is orbiting the earth at an elevation of 4800 miles above the earth in a circular orbit in the same plane as the Greenwich meridian. Miraculously, the plane and satellite always lie on the same radial line from the center of the earth. Assume the radius of the earth is 3960 miles.

(a) When is the plane directly over a location with latitude 74◦30′18′′ N for the first time?

(b) How fast is the satellite moving?

(c) When is the plane directly over the equator and how far has it traveled?

(d) How far has the satellite traveled when the plane is directly over the equator?

Answers (please explain why): a) 2.147 hours, b) 1103 mph, c) 12.47 hrs, 6236 miles, d) 13760 miles

Answer 1

First note that the aircraft is traveling in a circle of radius 3960 + 10 miles. Likewise, the satellite is moving in a circle of radius 8760 miles. We can calculate the circumference of each circular path through the relation C = πD = 2πR (D is diameter, R is radius). We will multiply the circumferences by the proportion of the circle that the objects have traveled in parts a, c, and d.

We first need to know how to convert 74°30'18" (which is read as 74 degrees, 30 minutes, and 18 seconds) into just a degree measure. To do this, note that there are 60 minutes in 1 degree, and 60 seconds in 1 minute. So 74°30'18" is equal to 74+(30/60) + (18/60)/60 = 74.505°. This converts the minutes and seconds into their degree equivalents and we add up each contribution. This is exactly like converting

The North pole is at 90°N, so for part a the aircraft has traveled 90°-74.505° = 15.495°. Since there are 360° in a complete circle it has traveled 15.495°/360° of a complete circle. A complete circle would have a circumference of 2π*(3970 mi) so the partial arc length from 90° to 74.505° would be given by 2π*(3970mi)*(15.495/360) = 1073.6419 miles. This expression can be interpreted as the circumference of the circle multiplied by the fraction of the circle that the aircraft transits. But part a asks us for the time it takes, not the distance, so we divide that distance by the aircraft's speed of 500 mph to get 2.147 hours.

The same reasoning will answer parts b-d. Just recognize that since the satellite remains directly above the aircraft it will reach any particular latitude (and therefore the equator) at the same time as the aircraft. (it may be easier to answer part c before part a). Don't forget that the satellite travels at a larger radius than the aircraft. Also, note that the equator is at 0°, so the aircraft and satellite will have traveled 90° upon reaching the equator which is also 1/4th of a full circle (90°/360° = 1/4).