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SOLUTIONS TO EXERCISE 2
(a) Using Kepler's third law, p^2 = Kr^3. For our solar system, we know already that K = 1 if p is measured in years and r is measured in AU. So, the orbital period of Jupiter around the Sun is p = sqrt( 1 * 5.2 * 5.2 * 5.2) = 12 years. Going from the near side of the Sun-Jupiter system to the far side would take half an orbit, so the time required would be about 6 years. (b) v = r/t, and from Exercise 1, Part (c) above, we know that r = 1.5 * 10^6 km. The time, in seconds, is about (6 yr) * (525,600 min/yr) * (60 sec/min) = 1.9 * 10^8 s. So the average velocity during this motion of Jupiter would be (1.5 * 10^6 km) / (1.9 * 10^8 s) = 0.0079 km/s. (c) How fast does a human being run? In the Olympics, we often hear of sprinters running the 100-meter dash in about 10 seconds. These sprinters are thus going about (100 m)/(10 s) = 10 m/s, or 0.01 km/s. This is 0.0021 km/s faster than the velocity of the motion calculated in Part (b) above; so for short distances, a person could actually outrun the motion of our solar system's center of mass. |