Calculate the distance from the surface of the earth at which acceleration due to gravity is the same as below and above the earth

Answer

`((sqrt(5)-1)/(2))R`

Explanation

g below surface `g=(GM)/(R^(3))(R-x)`
g above surface `g=(GM)/((R+x)^(2))`
`implies(GM)/(R^(3))(R-x)=(GM)/((R+x)^(2))`
`impliesx=((sqrt5-1)/(2))R` or 0