If the acceleration due to gravity at a altitude h `(h lt lt R)` above the earth's surface is `g_1` and at the same depth below the surface of the earth is `g_2`, then prove that `g_2=g_1(1+h/R)`. [R=radius of the earth]

Assume that the earth is a uniform sphere of radius R. If the acceleration due to gravity at a height h above the earth's surface is g_1 and at the same depth below the surface of the earth is g_2(h lt R) , then show that, g_2=g_1(1-h/R)(1+h/R)^2 .

The acceleration due to gravity at a height above the earth's surface is ……..than that at the same depth below the surface of the earth. [Fill in the blanks]

The acceleration due to gravity at a height R ( radius of the earth ) from the earth' s surface is

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 ]

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 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.

Where on the surface of the earth is the value of g least due to the diurnal motion of the earth ?

What will be the value of the acceleration due to gravity at a point 3 km above the earth's surface ? Diameter of the earth =12800 km and g=980cm*s^(-2) on the surface of the earth.