The value of acceleration due to gravity `(g)` at height `h` above the surface of earth is given by
`g^'=(gR^2)/((R+h)^2)`. If `hlt ltR`, then prove that `g^'=g(1-(2h)/(R))`.

The value of acceleration due to gravity (g) at height h above the surface of earth is given as g^()=(gR)^(2)/((R+h))^(2) If h lt lt R , then prove that g^()=g(1-(2h)/(R))

Show that acceleration due to gravity at height h above the Earth’s surface is g_h = g ((R )/(R+h))^2

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 ]

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.

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 value of the acceleration due to gravity is g1 at a height h=(R)/(2)(R= radius of the earth) from the surface of the earth.It is again equal to (g1/2) at a depth d below the surface of the earth.The ratio d/R equals:

The value of the acceleration due to gravity is g_(1) at a height h=(R)/(2) (R=radius of the earth) from the surface of the earth.It is equal to 2g_(1) at a depth d below the surface of the earth.The ratio ((d)/(R)) is equal to:

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.

Given, acceleration due to gravity on surface of earth is g. if g' is acceleration due to gravity at a height h above the surface of earth, then