A satellite revolves around the earth at a height of 1000 km. The radius of the earth is `6.38 xx 10^(3)`km. Mass of the earth is `6xx10^(24)kg and G=6.67xx10^(-14)"N-m"^(2)kg^(-2)`. Determine its orbital velocity and period of revolution.

An artificial satellite revolves round the earth at a height of 1000 km . The radius of the earth is 6.38xx10^(3)km . Mass of the earth 6xx10^(24)kg,G=6.67xx10^(-11)Nm^(2)kg^(-2) . Find the orbital speed and period of revolution of the satellite.

A satellite is revolving around the earth at a height of 600 km. find The speed of the satellite. Radius of the earth =6400 km and mass of the earth =6xx10^24kg .

An artificial satellite circles around the earth at a height of 2,200 km. Calculate its orbital velocity and period of revolution. Take, Radius of earth = 6.37 xx 10^(3) km Mass of earth = 6 xx 10^(24) kg G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)

A satellite orbits the earth at a height of 500km from its surface. Mass of the earth =6xx10^(24)kg Radius of the earth =6.4xx10^(6)m Mass of the satellite = 300kg G=6.67xx10^(-11)Nm^(2)kg^(-2) Calculate (i) kinetic energy (ii) potential energy and (iii) total energy of the satellite.

Calculate the height above the earth at which the geostationary satellite is orbiting the earth. Radius of earth = 6400km. Mass of earth = 6 xx 10^(24) kg. G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .

Calculate the force of gravitation due to earth on a ball of 1 kg mass lying on the ground.(Mass of earth =6xx10^(24) kg Radius of earth =6.4xx10^(3) km and G=6.7xx10^(-11)Nm^(2)/Kg^(2)