| |||||||

SOLUTIONS TO EXERCISE 10
Between January 1 and July 1, Earth has moved through half of its orbit around the Sun. Thus, the two observations are separated by 2 AU, and half that distance (the "parallactic baseline," in some sense) is the number "l" that we use in the parallax equation. The distance measured, "r" in the parallax equation, is 10 light-years. Thus the angle "theta" can be given by tan(theta) = l/r = (1 AU) / (10 light-years) 1.5 * 10^8 km = ----------------------------------------- = 1.6 * 10^(-6) 10 * (3 * 10^5 km/s) * (3.16 * 10^7 s/yr) (theta) = 0.000091 degree = 0.33 arcseconds. Another way to do this question is by using radians; in this case, use the entire distance between Earth at Jan. 1 and Earth at Jul. 1 - but remember that this will be twice the "parallactic angle." Remember also that one radian is approximately 57.3 degrees. Thus, 2 * (theta) = (2 AU) / (10 light-years) = 0.0000032 radian (theta) = 0.0000016 radian = 1.6 * 10^(-6) * 57.3 = 0.000091 degree. A third way to do this question is to notice that 10 light-years is 10/3.3 = 3.0 parsecs. At a distance of one parsec, the parallactic angle is 1 arcsecond; at 3.0 parsecs, the angle is (1/3.0) = 0.33 arcsecond. If you use this technique, remember that the more distant the object, the smaller the angle. |