Table D/2 Solar System Constants6.673(10 1") m³/(kg · s²)3.439(10 8) ft* /(lb - sec*)5.976(1024) kg4.095(1023) lb- sec² / ftUniversal gravitational constantGMass of EarthmePeriod of Earth's rotation (1 sidereal day)23 h 56 min 4 s23.9344 h0.7292(10 4) rad /s0.1991(10 °) rad /sAngular velocity of EarthMean angular velocity of Earth-Sun linew =Mean velocity of Earth's center about Sun107 200 km /h66,610 mi / hrBodySurface Gravitational Acceleration m/s²(ft/sec?)Mean Distance to Sun kmEccentricity of Orbit Period of Orbit solarMean Diameter kmMass Relative toEscape Velocity km/s(mi/sec)(mi)days(mi)EarthSun1392 000333 000274616(865 000)(898)(383)Мoon384 39840.05527.323 4760.01231.622.37(238 854)(2 160)(5.32)(1.47)Mercury57-3 x 10687.970.0540.2065 0003.474.17(35.6 x 106)(3 100)(11.4)(2.59)Venus108 x 1060.0068224.7012 4000.8158.4410.24(67.2 x 106)(7 700)(27.7)(6.36)Earth149.6 x 1060.0167365.2612 74229.821311.181.000(92.96 x 106)(7 918)2(32.22)3(6.95)Mars227.9 x 1060.093686.986 7880.1073-735.03(141.6 x 106)(4 218)(12.3)(3.13)Jupiter4778 x 1060.04894333139 822317.824.7959.5(483 x 106)(86 884)(81.3)(36.8)1 Mean distance to Earth (center-to-center)2 Diameter of sphere of equal volume, based on a spheroidal Earth with a polar diameter of 12 714 km (7900 mi) and an equatorial diameter of 12 756 km (7926 mi)3 For nonrotating spherical Earth, equivalent to absolute value at sea level and latitude 37.5°4 Note that Jupiter is not a solid body. A spacecraft Sis orbiting Jupiter in a circular path 1650 km above the surface with a constant speed. Using the gravitational law,calculate the magnitude v of its orbital velocity with respect to Jupiter. Use Table D/2 of Appendix D as needed.1650 kmAnswer: v =ikm/h

Given value--- mass of earth = 5.976 × 1024 kg. Gravitational constant = 6.673 × 10-11 m3/ kg s2 .…

A spacecraft S is orbiting Jupiter in a circular path 1650 km above the surface with a constant speed. Using the gravitational law, calculate the magnitude v of its orbital velocity with respect to Jupiter. Use Table D/2 of Appendix D as needed. 1650 km Answer: v = i km/h

A spacecraft S is orbiting Jupiter in a circular path 1650 km above the surface with a constant speed. Using the gravitational law, calculate the magnitude v of its orbital velocity with respect to Jupiter. Use Table D/2 of Appendix D as needed. 1650 km Answer: v = i km/h

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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