Assume that the earth is a ball of molten iron surrounded by rock. Estimate the
ID: 2270253 • Letter: A
Question
Assume that the earth is a ball of molten iron surrounded by rock. Estimate the radius at which the transition takes place to get a overall density that agrees with the average density of Earth. HINT: Calculate the mass of the core (Iron part) as a function of the radius r(t) (t for transition), and then add in the mass of the outer shell of rock by adding in the mass of the whole radius of the earth as rock minus the mass of rock that the core displaces ( or a sphere of radius r(t) of rock). Then find the value of r(t) so that this mass agrees with the mass of the earth.
Explanation / Answer
Iron density = 7000 kg/m^3
Rock density = 2500 kg/m^3
Radius of earth R = 6.4*10^6 m
Density of earth = 5500 kg/m^3
Volume of earth = (4/3)*pi*R^3
Mass of iron core = (4/3)*pi*R_t^3 *7000
Mass of earth assuming purely rock = (4/3)*pi*R^3 *2500
Mass of core assuming purely rock = (4/3)*pi*R_t^3 *2500
Net mass of shell = [(4/3)*pi*R^3 *2500] - [(4/3)*pi*R_t^3 *2500]
= (4/3)*pi*(R^3 - R_t^3) *2500
Total mass of earth = Total mass of iron core + rock shell
= (4/3)*pi*R_t^3 *7000 + (4/3)*pi*(R^3 - R_t^3) *2500
= (4/3)*pi*[2500*R^3 + 4500*R_t^3]
Density of earth = mass / volume
= (4/3)*pi*[2500*R^3 + 4500*R_t^3] / [ (4/3)*pi*R^3]
= 2500 + 4500*(R_t / R)^3
Equating, 2500 + 4500*(R_t / R)^3 = 5500
4500*(R_t / R)^3 = 3000
(R_t / R)^3 = 2/3
(R_t / 6.4*10^6)^3 = 2/3
R_t = 5.59*10^6 m
or 5590 km
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