Question

A satellite is in a circular orbit around the Earth at an altitude of 2.24 x 106 m. (a) Find the period of the orbit. (Hint: Modify Kepler's third law so it is suitable for objects orbiting the Earth rather than the Sun. The radius of the Earth is 6.38 x 106 m, and the mass of the Earth is 5.98 x 1024 kg.) h (b) Find the speed of the satellite. km/s (c) Find the acceleration of the satellite. m/s2 toward the center of the earth

Answer 1

soin Given! Altitude th] =7.24 x 100 m Here, radius of Satellite orbit will be (8) => Radius of earth + Altitude is 82 (6.378106)+ (2.2.4 x108) =8-61 X 106 m a> The period of satellite is T = 2T VG Me Where, me = mass of earth 5.970 x 1024 T = 271 (8.61 X 100) 17667x10")[5.972 X1024) kg 3 T = 2.209 hours by speed of satellite: v=7 G Me ४ =7 6.67X1041) (5.972 X1024) 8:61 X 106 V-7 6-801 x 103 mis V7 6.801 km is c) Acceleration of Satellite as v2 ४ (6.861 X 1032 8.61 xlo 6 a=75.37 mis