Question

An object with mass m away from the center of the center of the earth with mass M by the relationship r = h + R where R the radius of the earth and h the object height from the surface of the earth.At any height h, the value of the acceleration earth gravity equal half its value at the surface?

Answer 1

here g' = g/2

here acceleration due to gravity at the earth surface g = GM/R^2

at any height g' = GM/(R+h)^2

if g' = g/2

GM/(R+h)^2 = 1/2*GM/R^2

R+h =R/sqrt(2)

h =R(1-sqrt(2)/sqrt(2))