a satellite is revolving around the earth at a height of 600 km from the surface. find the speed of the satellite , its magnitude of acceleration and time period of its motion around the earth. (take Mass of earth = 6 x 10^24 kg) and ( radius of earth = 6400 km)
a satellite is revolving around the earth at a height of 600 km from the surface. find the speed of the satellite , its magnitude of acceleration and time period of its motion around the earth. (take Mass of earth = 6 x 10^24 kg) and ( radius of earth = 6400 km)
Let ‘m’ be the mass of the satellite circling the earth in an orbit of radius ‘r’. The gravitational force on the satellite is provides the centripetal force.
mv2/r = GMm/r2
=> v = [GM/r]1/2
=> v = (6.67 × 10-11)(6 × 1024)/[(6400 + 600) × 103] = 7561 m/s
=> v = 7.56 km/s
The centripetal acceleration is = v2/r = 75612/[(6400 + 600) × 103] = 0.015 m/s2
Let the time period be T.
Circumference of the orbit is = 2πr = 2 × 3.14 × [(6400 + 600) × 103] = 4.396 × 107 m
So, T = 2πr/v = (4.396 × 107)/7561 = 5814 s = 1.6 h
Let ‘m’ be the mass of the satellite circling the earth in an orbit of radius ‘r’. The gravitational force on the satellite is provides the centripetal force.
mv2/r = GMm/r2
=> v = [GM/r]1/2
=> v = (6.67 × 10-11)(6 × 1024)/[(6400 + 600) × 103] = 7561 m/s
=> v = 7.56 km/s
The centripetal acceleration is = v2/r = 75612/[(6400 + 600) × 103] = 0.015 m/s2
Let the time period be T.
Circumference of the orbit is = 2πr = 2 × 3.14 × [(6400 + 600) × 103] = 4.396 × 107 m
So, T = 2πr/v = (4.396 × 107)/7561 = 5814 s = 1.6 h
