Tiêu đề GRAVITATION - 4.pdf ✅

CLASS VIII PHYSICS orbital velocity of a satellite: Let a satellite of mass m revolving around the earth of mass M and radius R. Let h be the height of the satellite from the surface of the earth. Let v0 be the velocity of the revolving satellite. The necessary centripetal force for the satellite is provided by gravitational force of attraction. Now, mv0 2 r = G mM r 2 Where, r = R + h mv0 2 (R + h) = G mM (R + h) 2 v0 = √ GM (R + h) Note: Orbital velocity is independent of mass of the satellite. If the satellite is very close to the surface of the earth, then h << R orbital velocity is vo = √gR Orbital velocity near the surface of earth is vo = 7.98 km/s. Orbital angular velocity ω = √ GM (r+h) 3 Time period of the satellite T = 2π√ (R+h) 3 GM Kinetic Energy of an orbiting satellite: The kinetic energy of an orbiting satellite is given by$ K ⋅ E = 1 2 mv0 2 K. E = 1 2 GMm (R + h) PotentialEnergyofanorbitingsatellite: The potential energy of an orbiting satellite is given by$ U = − GMm (R + h) Total Energy of an orbiting satellite: Gravitation SYNOPSIS - 4