Calculate the height from earth’s surface at which value of g is 1% than that of the earth surface.

Calculate the value of ‘g' at height ‘h' from the earth’s surface.

The value of the acceleration due to gravity at a height h from the surface of the earth is g_1 and that at a depth h below the earth’s surface is g_2 . Show that g_2/g_1 = (1+h/R) (Radius of earth R >> h) I.

The value of the acceleration due to gravity at a height h from the surface of the earth is g_1 and that at a depth h below the earth's surface is g_2 .Show that (g_2)/(g_1)=(1+h/R) [Radius of earth Rgt gt h ]

What is the condition for setting up an artificial satellite at a height h from the earth’s surface? (Radius of the earth = R, mass of the earth= M).

If a seconds pendulum is taken to the surface of the moon from the earth, its time period would be (acceleration due to gravity on the surface of the moon is 1/6 th that on the earth's surface)

If the change in the acceleration due to gravity (g) at an altitude h above the earth's surface is equal to the change in g at a depth x below the earth's surface [assume that both x and h are significantly smaller than the radius of the earth], then

The ratio of the mass of the moon to that of the earth is 1:81. If the radius of the moon is assumed to be one-fourth of that of the earth , then what is the acceleration due to gravity on the surface of the moon in terms of that on the earth's surface ? Hence, find the escape velocity from the surface of the moon in terms of that from the earth's 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)

An object of mass m is raised up to a small height h above the earth's surface. If the acceleration due to gravity on the earth's surface is g, prove that the weight of the object will decreases by (2mgh)/R with respect to that on the earth's surface . R =radius of the earth.