10. A satellites is in a circular orbit around the earth at a height of 360 km above the earth's surface. What is its time period? What is its orbital speed?

Orbital speed=8102.39m/s

Time period=2935.98seconds

Explanation:

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g (R2 (R+h) 2)

V2R+h=g (R2 (R+h) 2)

V=√g (R2R+h)

V = sqrt (9.8 * (6371000) ^2 / (6371000+360000)

V = sqrt (9.8 * (4.059*10^13/6731000)

V=sqrt (65648789.18)

V = 8102.39m/s

Time period, T = sqrt (4 * pi*R^3) / (G * Mcentral)

T = sqrt (4*3.142 * (6.47*10^6) ^3 / (6.673*10^-11) * (5.98*10^24)

T=sqrt (3.40*10^21) / (3.99*10^14)

T = sqrt (0.862*10^7)

T = 2935.98seconds