Menu div.question,div.steps {font-size: 16px; padding: 10px;line-height: 20px; margin: 20px;} div.question{border: 1px solid #ccc;} div#solution .button{font-size: 20px; margin: 20px; width: 200px; padding: 10px; text-align: center;} div.hint,div.answer,div.steps{display: none;} div.pre_next{font-size: 30px; width:90%; margin:auto;} div.pre_next a.nav_pre{float:left;} div.pre_next a.nav_next{float:right;} div.title div.notes{ font-family:sans-serif, Arial; font-size: 18px; } 1. Knows: Mass of Earth: M_e = 5.97 times 10^(24)\ kg Radius of Earth: r_e = 6.37 times 10^6\ m (a) Use above information to calculate the average density of the Earth. (b) Where does the value fit among those listed in Table 14.1. Look up the density of a typical surface rock like granite in another source and compare it with the density of the Earth. Hint V=4/3πr^3 ρ=m/V Answer (a) 5.52 \times 10^3\ kg//m^3 (b) It is between the density of aluminum and that of iron and is greater than the densities of typical surface rocks. Show Steps (a) V_e = 4/3π(6.37 times 10^6)^3 = 1.0827 times 10^(21) m^3 Density of Eearth: ρ_e=(5.97 times 10^(24)kg)/(1.0827 times 10 ^21 m^3) = 5.52 times 10^3 kg//m^3