Question

Assuming that the mean density of the earth is 5500kgm^-3, that the constant gravitation is 6.7 x 10^-11 Nm^2kg^-2, and that the radius of the earth is 6400km, find a value for the acceleration due to gravity at the earth's surface.

Answer 1

$$mg=G\frac{mM}{R^2}\\g=\rho G\frac{4\pi R^3}{3R^2}\\g=\rho G\frac{4\pi R}{3} \\g=(6.7\cdot10^{-11})(5500)\frac{4\pi (6.4\cdot10^6)}{3}\\ g=9.88\frac{m}{s^2}$$