What is the height from the surface of earth, where acceleration due to gravity will be 1 4 of that of the earth? (R_E=6400 ~km )

Step-by-step solution

Acceleration due to gravity at height h is aligned & g_h= g (R+h)^2 g_h 1 (R+h)^2 & g_ h₂ g_ h₁ = ( R + h ₁ R+h₂ )^2 g 4 g = ( 6400+0 6400+h₂ )^2 & 1 2 = 6400 6400+h₂ h₂=6400 ~km aligned