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SOLUTIONS TO EXERCISE 12
(a) According to http://www.jpl.nasa.gov/jupiterflyby/ and its FAQ page, Cassini came to within 9,721,800 km of Jupiter's cloud tops. Jupiter's radius to its cloudtops is 71,400 km. Thus the distance between Cassini and Jupiter's center is 9721800 + 71400 = 9793200 km. The mass of Jupiter is 1.90 * 10^27 kg. Using Newton's Law of Gravity, the gravitational acceleration will be (get the units right!) GM 6.67 * 10^(-11) [mks] * 1.90 * 10^27 kg a = ----- = --------------------------------------- = 0.00132 m/s^2. r^2 (9.79 * 10^9 m)^2 (b) The distance between Jupiter and the Sun is 5.20 AU = 7.8 * 10^11 m. The projected area of Jupiter that faces the Sun's light at any given time is (pi * r^2), where r is the radius of Jupiter; from Part (a) above, this radius is 71400 km or 7.14 * 10^7 m. Also taking into account Jupiter's albedo of 50%, the luminosity of Jupiter in visible light is L = 0.50 * (flux of sun on Jupiter) * (projected area of Jupiter) 10^26 W = 0.50 * ---------------------------- * 3.14 * (7.14 * 10^7 m)^2 4 * 3.14 * (7.8 * 10^11 m)^2 = 1.05 * 10^17 W, about a billion times less luminous than the Sun. (c) Once again, we use the relationship between luminosity and flux. This time the distance is between Cassini and Jupiter's cloudtops. F = 1.05 * 10^17 W / (4 * 3.14 * (9.22 * 10^9 m)^2 ) = 9.83 * 10^(-5) W/m^2. |