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SOLUTIONS TO EXERCISE 30
(a) Kepler's Third Law can be expressed as P^2 = K r^3, where P is the orbital period and r is the distance from the center of the primary object to the orbiting object. As discussed in class, for objects orbiting the Sun, the proportionality constant K is equal to one when P is in units of years and r is in units of astronomical units (1 AU = 1.50 * 10^8 km). This newly discovered minor planet is orbiting the Sun at a distance of 300,000,000 km. This is 300,000,000 / 150,000,000 = 2 AU. So we can say that P^2 = 1 * (2^3) = 8 ; so P = sqrt(8) = 2.83 years. Convert this length of time to seconds: 2.83 years = 2.83 * 365.25 * 24 * 3600 = 8.93 * 10^7 sec And the distance the minor planet travels in one orbit is the circumference of its orbit, which is 2 * 3.14 * 3 * 10^8 = 1.88 * 10^9 km So the minor planet's orbital velocity is 1.88 * 10^9 km --------------- = 21.1 km/sec 8.93 * 10^7 sec (b) This solution is left as an exercise to the reader. The answer turns out to be about 30 km/s, or about 66,000 miles per hour. |