From Kepler's third law, \\dfrac{T^2}{r^3}=\\dfrac{4\\pi^2}{GM} where r is radius, semi-major axis of the elliptical orbital of satellite. Near the planet surface r=R ; M=\\cfrac{4}{3} \\pi R^3 \\rho So, T=\\sqrt{\\dfrac{3\\pi}{G\\rho}} \\propto R^{0}