An airplane is flying above the Earth's surface at a height of 10.6km?
An airplane is flying above the Earth's surface at a height of 10.6km. The centripetal acceleration of the plane is 13.9m/s^2. What is the angular velocity of the plane moving in uniform circular motion? Take the radius of the Earth to be 6400km.
- oldprofLv 75 years agoFavorite Answer
We have Ap = g' + Af = 13.9 = g(r/R)^2 + w^2 R; where r = 6400 km and R = 6410.6 km, g = 9.8 m/s^2 at Earth's surface (r = 6400 km). Find w = ? rad/sec the angular motion. g' is the gravity field strength at 10.6 km.
w^2 R = Ap - g(r/R)^2; so w = sqrt((Ap - g(r/R)^2)/R ) = sqrt((13.9- 9.8*(6400/6410.6)^2)/6410600) = 8.028805E-04 rad/s. ANS.
The physics is this. The centripetal force includes the force of gravity and whatever force is necessary from lift to counter the centrifugal force. As these forces act on the same mass, the airplane, all the masses cancel out leaving just the accelerations Ap, g', and Af. Note there is a slight reduction in g to g' < g due to extended distance from the gravity source.