Assume that the earth is a uniform sphere of radius R. If the acceleration due to gravity at an altitude h is the same as its value at the same depth below the surface of the earth, then show that `h=1/2(sqrt(5)-1)R.`

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 .

Considering the earth as a uniform sphere of radius R show that if the acceleration due to gravity at a height h from the surface of earth is equal to the acceleration due to gravity at the same depth, then h=1/2(sqrt(5)-1)R.

Considering the earth as a uniform sphere of radius R show that if the acceleration due to gravity at a height h from the surface of earth is equal to the acceleration due to gravity at the same depth, then h= 1/2(sqrt5-1) R

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

Imagine earth to be a solid sphere of mass M and radius R. If the value of acceleration due to gravity at a depth d below earth's surface is same as its value at a height h above its surface and equal to (g)/(4) (where g is the value of acceleration due to gravity on the surface of earth), the ratio of (h)/(d) will be

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 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)