Astronomy Problem Set 5 Solutions: Applying Kepler's Third Law, Assignments of Astronomy

Download Astronomy Problem Set 5 Solutions: Applying Kepler's Third Law and more Assignments Astronomy in PDF only on Docsity! Astr 1010 Problem Set #5, Solutions 1. For Mercury we have Pyr = 0.2409 and aAU = 0.387 (data is from Table 4 in Explorations). Comparing the two sides of Kepler’s Third Law, we find: P 2yr = 5.803 × 10−2 a3AU = 5.80 × 10−2 which agree to three figures. For the four given planets, we find: Planet Pyr aAU P 2yr a 3 AU Mercury 0.2409 0.387 5.803 × 10−2 5.80 × 10−2 Mars 1.8809 1.524 3.5377 3.540 Saturn 29.4577 9.539 8.67756 × 102 8.680 × 102 Neptune 164.793 30.06 2.71567 × 104 2.716 × 104 The results agree (out to the permitted number of significant digits). 2. Here, aAU = 3.3, so a3AU = (3.3) 3 = 35.9 By Kepler’s Third Law, we also have P 2yr = a 3 AU = 35.9 and solving for Pyr gives Pyr = √ 35.9 = 6.0 The period of the planet is 6.0 yr. 3. Here, Pyr = 11.18, so P 2yr = (11.18) 2 = 125.0 By Kepler’s Third Law, we also have a3AU = P 2 yr = 125.0 and solving for aAU gives aAU = (125.0)1/3 = 5.0 The semi–major axis of the planet is 5.0AU. 4. We are given the period of Halley’s Comet, so Kepler’s 3rd Law will give us the semi–major axis of the orbit. We get: P 2yr = (76.1) 2 = 5.79 × 103 = a3AU so that aAU = (5.79 × 103)1/3 = 18.0 1