The period of a satellite orbiting just above the surface of the earth is given by $T = 2\pi \sqrt{\frac{R}{g}}$. Since $g = \frac{GM}{R^2}$ and $M = \frac{4}{3}\pi R^3 d$, we have $g = \frac{G}{R^2} \cdot \frac{4}{3}\pi R^3 d = \frac{4}{3}\pi G R d$. Substituting this into the formula for $T$, we get $T = 2\pi \sqrt{\frac{R}{\frac{4}{3}\pi G R d}} = 2\pi \sqrt{\frac{3}{4\pi G d}} = \sqrt{\frac{4\pi^2 \cdot 3}{4\pi G d}} = \sqrt{\frac{3\pi}{Gd}}$. Therefore, $T^2 = \frac{3\pi}{Gd}$.