4 2 b starting with newtons version of keplers 3rd

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Unformatted text preview: the corresponding physical characteristic. 4π 2 b. Starting with Newton’s version of Kepler’s 3rd Law ( P 2 = a3 G ( m1 + m2 ) [equation 2.37]), derive the relationship between orbital period of a “low orbit” and this physical characteristic from part (a) assuming a “low orbit” essentially has an semimajor axis equal to the planet’s radius and the satellite has much lower mass than the planet. c. Given your derivation in part (b), should a low Jovian orbit around Jupiter have a period higher or lower than 90 minutes. Appendix C Table 1: Planetary Physical Data from Carroll and Ostlie Equatorial Sidereal Mass Average Density Radius Rotation Period Planet (M⊕) (kg m-3) (R ⊕ ) (days) Mercury 0.055 0.383 5427 58.6462 Venus 0.815 0.949 5243 243.018 Earth 1.000 1.000 5515 0.997271 Mars 0.107 0.533 3933 1.02596 Jupiter 317.83 11.209 1326 0.4135 Saturn 95.159 9.449 687 0.4438 Uranus 14.536 4.007 1270 0.7183 Neptune 17.147 3.883 1638 0.6713 Pluto (dwarf planet) 0.002 0.178 2110 6.3872 Eris (dwarf planet) 0.002 0.188 – Page 2 of 2 – 2100?...
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This document was uploaded on 03/04/2014 for the course ASTROPHYS 362 at Minnesota State University Moorhead .

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