A particle is dropped from a height h=3R above earth surface where R is the radius of earth. If g is the acceleration due to gravity on earth surface, then speed with which the particle strikes the earth surface is

If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is

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

If g is the acceleration due to gravity on the surface of the earth , its value at a height equal to double the radius of the earth is

The acceleration due to gravity at a depth R//2 below the surface of the 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

A particle of mass 'm' is raised to a height h = R from the surface of earth. Find increase in potential energy. R = radius of earth. g = acceleration due to gravity on the surface of earth.

A particle is projected vertically upwards with a velocity sqrt(gR) , where R denotes the radius of the earth and g the acceleration due to gravity on the surface of the earth. Then the maximum height ascended by the particle is

A particle is projected vertically upwards with a velocity sqrt(gR) , where R denotes the radius of the earth and g the acceleration due to gravity on the surface of the earth. Then the maximum height ascended by the particle is

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]

The height above the surface of the earth where acceleration due to gravity is 1/64 of its value at surface of the earth is approximately.