Calculate angular momentum of earth rotating about its axis. Take `I = (2)/(5)MR^(2)`, where `M = 6 xx 10^(24) kg` and `R = 6400 Km`.

The value of angular momentum of the earth rotating about its own axis is

Calculate the angular momentum of the earth rotating about its own axis. Mass of the earth = 5.98 xx 10^(27) kg, mean radius of the earth =6.37 xx 10^(6)m,M.I of the earth = (2)/(5)MR^(2) .

Calculate the angular momentum of earth rotating about its own axis. Take Radius of earth =6.4xx10^(6)m Mass of earth =5.97xx10^(24)kg

Calculate rotational K.E. of earth about its own axis, taking it to be a sphere of mass 6 xx 10^(24) kg and radius 6400 km .

Calculate the angular momentum and rotational kinetic energy of the earth about its own axis. Given mass of the earth is 6 xx 10^(24) kg and its radius = 6.4 xx 10^3 km. [Moment of inertia of the earth about its axis of rotation is 2/5 mr^(2) ]

Calculate the rotation kinetic energy of the earth about its own axis. (M= 6 times 10^24 kg and R = 6400km)

Find the magnitude of the acceleration due to gravity at the surface of the earth. ( M = 6 xx 10^(24) Kg, R = 6400 Km)

Derive an expression for critical velocity of satellite. Calculate the acceleration due to gravity at a height of 300 km from the surface of the earth. M = 6xx10^24kg R = 6400 km.

Calculate the value of orbital velocity for an artifical satellite of earth orbiting at a height of 1000 km (Mass of the earth =6 xx 10^(24) kg, radius of the earth = 6400 km )