Question

An artificial satellite is in a circular orbit d = 680.0 km above the surface of a planet of radius r = 5.55 x 10 km. The period of revolution of the satellite around the planet is T = 1.15 hours. What is the average density of the planet? density kg/m

Answer 1

бщФ Powe eta 7a d= 680 km = 680 x 103 m 5.55 x 103 km =5,55 x 106m Time period T = 1.15 hours = 1.15 x 3600 S. Gravitational force serves the ordy 2 centripetal force m2 Coud) F: Gmm 2 (Ptujua T 2 (pte) 1 x ve X Т2 F-Fc 417² (ortd) GM (ntd] 2 72 CS Scanned with CamScanner

3 417² totd) → W T2x XG 2 3 4 / 37 m3x S T d x 41² (nto) T2XG i (btdj3 T2, G. 3 3 3 ) an 3 n E = (5.55X10693 *{(5550+680) xro} (1.15*360012 6.67x10 = 1.17X104 {eelbox CS Scanned with CamScanner