(a) What linear speed must an Earth satellite have to be in a circular orbit at an altitude of 232 km above Earth's surface? (b) What is the period of revolution? (3) Number n I Units m/s V n n eTextbook and Media Assistance Used M Assistance Used What is the centripetal force that provides the uniform circular motion? How is that force related to the centripetal acceleration? How is the acceleration related the speed? Key Ideas . The motion of satellites, both natural and artificial, is governed by Kepler's laws: 1. The law of orbits. All planets move in elliptical orbits with the Sun at one focus. 2. The law of areas. A line joining any planet to the Sun sweeps out equal areas in equal time intervals. (This statement is equivalent to conservation of angular momentum.) 3. The law of periods. The square of the period T of any planet is proportional to the cube of the semimajor axis a of its orbit. For circular orbits with radius r, T2 4 TT2 GM 7-3 (law of periods), where M is the mass of the attracting body-the Sun in the case of the Solar System. For elliptical planetary orbits, the semimajor axis a is substituted for r.The distance of satelite from earth surface - > h = 232 Kon . Radius of earth > R = 6371 kon . distance from centre of earth ? 8 = h+R = (6371+ 232 ) " = 6603 x 103 m. orbital speed . V= GM G- gravitational count 6. 674 x 10-11 m3 kg152 M - mass of earth 6 ' 6 74 X 10 X 5 . 972 X 1024 650 3 X 103 m/s = 5.972 x10LY kg 7. 8 x 10 m/s . b ) Period of revolution : T = distance speed 2 V 2X TIX 6603X103 S 7.8 x 10 3 5.3 X10 s