A body weighs 64 N on the surface of the Earth. What is the gravitational force on it (in N) due to the Earth at a height equal to one-third of the radius of the Earth?

A body weighs 64 N on the surface of the Earth. What is the gravitatio

1.	A body weighs 64 N on the surface of the Earth. What is the gravitational force on it (in N) due to the Earth at a height equal to one-third of the radius of the Earth?
Answer» Solution :LINEAR velocity v=72 km/h = `(72 xx 1000)/(60 xx 60) = 20` m/s Initial angular velocity of the wheel `=omega_(0) = v/r = 20/(0.25) = 80` rad/s Final angular velocity =`omega =0` Time =t= 16S Angular ACCELERATION = `alpha` = ? `alpha = (omega-omega_(0))/t = (0-80)/16 = -5 "rad"//s^(2)` Average torque `=tau = lalpha = 5 xx 5 = 25` Nm