[" Find the height over the earth's surface at which the "],[" meight of a body becomes half of its value at the surface."]

(a) Find the acceleration due to gravity in a mine of depth 640 km if the value at the surface is 9.800 m//s^(2) . The radius of the earth is 6400 km (b) Find the height over the earth's surface at which the weight of a body becomes half of its value at the surface.

Find the height over the earth's surface at which the weight of a body becomes half of its value t the surface.

Find the height over the eart's surface at which the weight of a body becomes half of its alue t the surface.

Find the height over the eart's surface at which the weight of a body becomes half of its alue t the surface.

The height vertically above the earth's surface at which the acceleration due to gravity becomes 1% of its value at the surface is (R is the radius of the earth)

The height of the point vertically above the earth's surface, at which acceleration due to gravtiy becomes 1% of its value at the surface is (Radius of the earth =R)

The height of the point vertically above the earth's surface, at which acceleration due to gravtiy becomes 1% of its value at the surface is (Radius of the earth =R)

At what height above the earth's surface does the value of g becomes 36% of the value at the surface of earth ?