The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R form the surface of the earth is (g=acceleration due to gravity at the surface of the earth)

What is the value of acceleration due to gravity on the surface of earth ?

The satellite is moving round the earth (radius of earth = R) at a distance r from the centre of the earth. If g is the acceleration due to gravity on the surface of the earth. The acceleration of the satellite will be

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

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

The value of acceleration due to gravity 'g' on the surface of a planet with radius double that of the earth and same mean density as that of the earth is ( g_(e) =acceleration due to gravity on the surface of the earth )

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

The ratio of acceleration due to gravity at a height 3 R above earth's surface to the acceleration due to gravity on the surface of earth is (R = radius of earth)

Calculate the acceleration due to gravity at a height of 1600 km from the surface of the Earth. (Given acceleration due to gravity on the surface of the Earth g_(0) = 9.8 ms^(-2) and radius of earth, R = 6400 km).