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SOLUTIONS TO EXERCISE 10
(a) The distance between Earth and Voyager 2 was 35 AU, or 35 * 1.5 * 10^11 m = 5.25 * 10^12 m = 5.25 * 10^9 km. So the time taken by radio waves to arrive at Earth from Voyager 2 was distance 5.25 * 10^9 km t = -------- = -------------- = 17500 sec = 4 hrs and 52 min. velocity 3 * 10^5 km/s (b) At a speed of 40000 km/hr, Voyager 2 travelled a distance of 40000 km/hr d = speed * time = ----------- * 17500 sec = 1.94 * 10^5 km 3600 sec/hr during the time a radio signal travelled to get from Voyager 2 to Earth - more than half the distance between Earth and the Moon. (c) The flux of the transmitter on Earth's surface was Luminosity 550 Watts F = ----------------- = ----------------------------- 4*pi*(distance)^2 4 * 3.14 * (5.25 * 10^12 m)^2 = 1.59 * 10^(-24) Watts per square meter. |