Question

If 'g' is acceleration due to gravity on the surface of the earth having radius'R' the height at which the acceleration due to gravity reduced to g/2 is

Answer 1

Simply apply the formula for g i.e

g=GM/R^2 first in the surface of Earth and then at a height 'h' above the Earth surface. Note that in the formula, the distance is calculated from the center. So, at a height 'h' above the ground, distance from center will be R+h which I took in the 2nd equation. Hope you got the point.

Answer 2

At surface;

g=GM/R^2_____(1)

At height h let gravity become g/2

->g/2=GM/(R+H)^2_____(2)

Dividing 1 and 2

g÷g/2=GM/R^2÷GM/(R+H)^2

2=[R+H/R]^2

√2=1+h/r

√2-1=h/r

h={√2-1}R//