The depth `d`, at which the value of acceleration due to gravity becomes 1/n times the value at the surface is (R = radius of the earth)

The depth at which the value of acceleration due to gravity is 1/n times the value at the surface, is (R=radius of the earth)

The depth 'd' at which the value of acceleration due to gravity becomes (1)/(n) times the value at the earth's surface is (R = radius of earth)

The depth 'd' at which the value of acceleration due to gravity becomes (1)/(n) times the value at the earth's surface is (R = radius of earth)

The depth 'd' at which the value of acceleration due to gravity becomes (1)/(n) times the value at the earth's surface is (R = radius of earth)

What is the depth at which the value of acceleration due to gravity become 1/n times the value that at the surface of earth ? (radius of earth = R)

What is depth at which value of acceleration due to gravity becomes 1/n times value that at surface of earth

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 at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth. (radius of the earth = 6400 km ) is