Periodic time of a satellite revolving above Earth's surface at a height equal to R, radius of Earth, is : [g is acceleration due to gravity at Earth's surface]

Let g be the acceleration due to gravity on the earth's surface.

At a height equal to earth's radius, above the earth's surface, the acceleration due to gravity is

A satellite is in a circular orbit round the earth at an altitude R above the earth's surface, where R is the radius of the earth. If g is the acceleration due to gravity on the surface of the earth, the speed of the satellite is

Calculate the work required to raise a body of mass m to a height h just above the earth's surface. Radius of the earth is R and acceleration due to gravity at the earth's surface is g.

A particle is taken to a height R above the earth's surface, where R is the radius of the earth. The acceleration due to gravity there is

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of escape velocity from the surface of earth. R is the radius of earth and g is acceleration due to gravity at the surface of earth. (R=6400 km). The time period of revolution of satellite in the given orbit is

A body of mass m rises to a height h=R//5 from the earth's surface where R is earth's radius. If g is acceleration due to gravity at the earth's surface, the increase in potential energy is

An artificial satellite is moving in circular orbit around the earth with speed equal to half the magnitude of escape velocity from the surface of earth. R is the radius of earth and g is acceleration due to gravity at the surface of earth (R = 6400km) Then the distance of satelite from the surface of earth is .

An earth satellite of mass m revolves in a circular orbit at a height h from the surface of the earth. R is the radius of the earth and g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by