Question

the satellite aryabhatta is revolving around the earth at a height of 650 km from the surface of the earth and completes one revolution in 96 minutes. if the earth's radius is 6400 km, the acceleration

Answer 1

h = height of the satellite above the surface of earth = 650 km = 0.65 x 10⁶ m

R = radius of earth = 6400 km = 6.4 x 10⁶ m

r = orbital radius for the satellite = R + h = (6.4 x 10⁶) + (0.65 x 10⁶) = 7.05 x 10⁶ m

T = Time period of revolution = 96 minutes = 96 x 60 sec = 5760 sec

w = angular velocity of the satellite = 2π/T = 2(3.14)/5760 = 0.0011 rad/s

a = centripetal acceleration of the satellite

centripetal acceleration of the satellite is given as

a = r w²

inserting the values in the above equation

a = (7.05 x 10⁶) (0.0011)²

a = 8.53 m/s²