GEOMETRY

PROBLEM 1.

Geometry is fundamental to space science. A multitude of activities, from the prediction of flight paths to the design of equipment, make use of geometric analysis. Geometry enters into many of the problems of the preceding and subsequent chapters. Most of the problems in this chapter fall into three categories: those involving areas and volumes of plane and solid figures, those that use similarity, and those that use properties of circles or spheres. The Sun-Earth-Moon system happens to exhibit a striking geometric coincidence, which we examine in the first problem.

PROBLEM 1. To an observer on Earth, the Sun and the Moon subtend almost the same angle in the sky. The average angle is 0.52 degrees for the Moon and 0.53 degrees for the Sun. Depending on the particular location in its elliptic orbit, the Moon's angle ranges between 0.49 degrees and 0.55 degrees, whereas that of the Sun ranges between 0.52 degrees and 0.54 degrees. This is why the Moon sometimes completely blocks the Sun, producing a total solar eclipse.

a.If the mean lunar and solar distances are respectively 3.8 x 10^5 km and 1.5 x 10^8 km, what is the ratio of the solar diameter to the lunar diameter, and what is the ratio of the solar volume to the lunar volume? Solution: The geometry of the eclipse is illustrated in Fig. 4.1. Since the angle at 0 is the same for both the large and the small triangles and the triangles are isosceles, they must be similar. Letting RM and RS denote the lunar and solar distances, respectively, and DM and DS the lunar and solar diameters, we have

DS/DM = RS/RM = 1.5 x 10^8/3.8 x 10^5 = 390

If VM and VS are the lunar and solar volumes, respectively,

VS/VM = [(4/3)pi(DS/2)^3]/[(4/3)pi(DM/2)^3]

= (DS/DM)^3 = (390)^3 = 5.9 x 10^7.

b. Since the angle at O is very small, we can approximate the lunar or solar diameter by the arc length of the circle with radius RM or RS where this arc length subtends the angle at 0. Use the relation s = r(theta) (angle theta is in radians) to determine the actual values of DM and DS.

Solution:

0.52 degrees = 0.52 degrees x pi rad/180 degrees = 0.0091 rad

DM = RM (O.OO91) = 3.8 x 10^5 x 0.0091 = 3.5 x 10^3 km

0.53 degrees = 0.0092 rad, by the same conversion just shown

DS = RS (0.0092) = 1.5 x 10^8 x 0.0092 = 1.4 x 10^6 km

(Note: The reader may prefer to avoid the approximation by using the tangent function - that is, tan (theta/2) = lunar radius/lunar distance = solar radius/solar distance; note, however, that to two significant digits the result is the same.)

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