PROBLEM 5. Earth's orbit around the Sun is elliptical, but in many cases it is sufficiently accurate to approximate the orbit with a circle of radius equal to the mean Earth-Sun distance of 1.49598 x 10^8 km. This distance is called the "Astronomical Unit" (AU). Listed in the chart that follows are actual Earth-Sun distances, given to five significant digits, on the first day of each month of a representative year. (The "American Ephemeris" lists daily distances and the actual times for these distances to seven significant digits.)
a. To how many significant digits is it reasonable to approximate the Earth-Sun distance as though the orbit were circular?
Solution: To two significant digits, each of the distances in the table can be given as 1.5 x 10^8 km.
b. What are the largest possible absolute and relative errors in using the Astronomical Unit as the Earth-Sun distance in a computation instead of one of the distances from the table?
Solution: (1.49598 - 1.4710) x 10^8 = 0.0240 x 10^8 km (smallest table value)
(1.49598 - 1.5208) x 10^8 = -0.0248 x 10^8 km (largest table value) absolute error less than/equal to 0.0248 x 10^8 km
relative error less than/equal to 0.0248/1.49598 = 0.0166, or 1.7 percent.
The procedure of "dimensional analysis", described earlier, is easily adapted and commonly used in science and technology for the task of unit conversion. Recall that in dimensional analysis the units are manipulated in accordance with the rules of algebra.
Suppose we wish to change a length of 623 cm to meters. The adaptation of dimensional analysis for unit coversion involves multiplication by a "factor unit" chosen according to the following simple principles: the factor unit is a fraction with a value of 1, whose numerator is expressed in terms of the unit we wish to have and whose denominator is expressed in terms of the unit we wish to change. Since 100 cm = 1 m, in order to change 623 cm to m, we perform the multiplication
(623 cm/1) X (1m/100 cm),
"canceling" the cm in numerator and denominator to get
623/100 m, or 6.23 m.
More complex conversions can be done using multiplication by several factor units and those readers wishing to convert between British and metric units can also use this method. For example, the speed of light, 3.00 x 10^5 km/sec, can be found in miles per hour:
(3.00 x 10^5/1) X (km/sec) X (1 mile/1.61 km) X (60 sec/1min) X (60 min/1 hour)
= 6.71 x 10^8 miles per hour.