Question

you are a geologist assigned to estimate the density of the Earth. look at picture for the full question:

Answer 1

so, neverminding the units.


`\bf g=\cfrac{GM}{R^2}\implies R^2=\cfrac{GM}{g}\implies R=\sqrt{(GM)/(g)}\quad \begin{cases} G=6.67\cdot 10^(-11)\\ M=5.97\cdot 10^(24)\\ g=9.81 \end{cases} \\\\\\ R=\sqrt{\cfrac{{(6.67\cdot 10^(-11))(5.97\cdot 10^(24))}}{9.81}} \\\\\\ R=\sqrt{\cfrac{{6.67\cdot 5.97\cdot 10^(13)}}{9.81}} \\\\\\ R\approx 6371116.97~m\\\\ -------------------------------\\\\ V=\cfrac{4\pi R^3}{3}\implies V=\cfrac{4\pi 40591131498470.95^3}{3} \\\\\\ V\approx 1083266582306022184268.1~m^3`