At what distance above and below the surface of the earth a body will have same weight? (Take radius of earth as $$R$$)

At height $$h$$ above surface: $$g_h = g\frac{R^2}{(R+h)^2}$$.

At depth $$d$$ below surface: $$g_d = g\left(1 - \frac{d}{R}\right)$$.

Setting equal and using $$h = d$$:

$$\frac{R^2}{(R+h)^2} = 1 - \frac{h}{R} = \frac{R-h}{R}$$

$$R^3 = (R-h)(R+h)^2$$

Let $$x = h/R$$: $$1 = (1-x)(1+x)^2 = (1-x^2)(1+x) = 1+x-x^2-x^3$$

$$x + x^2(-1) - x^3 = 0$$... Actually: $$x - x^2 - x^3 = 0$$, $$x(1-x-x^2) = 0$$.

$$x^2 + x - 1 = 0$$, $$x = \frac{-1+\sqrt{5}}{2}$$.

$$h = \frac{(\sqrt{5}-1)R}{2} = \frac{\sqrt{5}R - R}{2}$$.

The answer corresponds to Option (4).