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Can you write down an expression for the surface gravity of a homogeneous sphere in terms of its density \(\rho\)? What material would give you an Earth-sized planet, what would give you the planet of the Petit Prince, both with the terrestial value of \(g\)?

As somebody working on cosmology I feel compelled to give a weird solution to this question. Working with Newtonian gravity gives \(g = GM/r^2\) for the surface acceleration \(g\), which should assume Earth’s value of \(10\mathrm{m}/\mathrm{s}^2\), and the mass \(M=4\pi/3\:\rho r^3\) of a homogeneous sphere with size \(r\), such that \(4\pi/c\:G\rho r\). Thinking of cosmology with the acceleration being proportional to distance and the suggestive prefactor, one could write this as \(r = \frac{\rho_\mathrm{crit}}{\rho}\frac{2g}{c^2}\chi_H^2\) with the Hubble distance \(\chi_H = c/H_0 = 10^{25}\mathrm{m}\) and the critical density \(\rho_\mathrm{crit} = 3H_0^2/(8\pi G) = 10^{-26}\mathrm{kg}/\mathrm{m}^3\). A planet made from metal with \(\rho=10^4\mathrm{kg}/\mathrm{m}^3\) would be of size \(r = 10^4\mathrm{m} = 10\mathrm{km}\), as a rough estimate. The radius for other materials follow directly, \(r \propto 1/\rho\).