For a satellite orbiting just above the surface of the Earth, we can use the formula for the period T of the satellite in terms of the Earth's density d and the gravitational constant G:

$$T = 2\pi\sqrt{\frac{a^3}{GM}}$$

where a is the semi-major axis of the orbit (which is approximately equal to the Earth's radius R for a satellite orbiting just above the surface) and M is the mass of the Earth.

We can express the mass of the Earth M in terms of its density d and volume:

$$M = dV = d\times\frac{4}{3}\pi R^3$$

Now, substitute this expression for M into the equation for T:

$$T = 2\pi\sqrt{\frac{a^3}{Gd\times\frac{4}{3}\pi R^3}}$$

Since the satellite is orbiting just above the Earth's surface, we can approximate a ≈ R:

$$T = 2\pi\sqrt{\frac{R^3}{Gd\times\frac{4}{3}\pi R^3}}$$

Simplify the equation:

$$T = 2\pi\sqrt{\frac{1}{Gd\times\frac{4}{3}\pi}}$$

Now, square both sides of the equation:

$$T^2 = 4\pi^2\frac{1}{Gd\times\frac{4}{3}\pi}$$

Simplify further:

$$T^2 = \frac{3\pi}{Gd}$$

Thus, the quantity $\frac{3\pi}{Gd}$ represents $T^2$.