Consider a simple model for the interior of the Earth: there is a spherical iron
ID: 2060608 • Letter: C
Question
Consider a simple model for the interior of the Earth: there is a spherical iron core with constant mass density ?0 and radius a; outside the core is "rock" with constant density ?1. Use these values for the densities: ?0= 8.90×103 kg/m3 and ?1= 4.30×103 kg/m3. The radius of the Earth is R = 6.40×106 m.(a) First, calculate the mass of the Earth. [Hint: Use G and g.]
(b) Calculate the radius a of the iron core.
(c) Derive the graviational field g(r) as a function of r. Hand in (in Room 1312) a numerically accurate graph of g(r). The value at r=R must be 9.81 m/s2.
(d) Calculate g(a).
Tries 0/25
Explanation / Answer
The problem suggests that you consider that the Earth is comprised of a central iron core of some radius r with a given density, surrounded by a concentric shell of another density out to a radius R. You should be able to write an expression for the total mass of the Earth that depends upon r and R. What you known is that the acceleration due to gravity at the Earth's surface (radius R) is given by Newton's law of gravity (can you write the appropriate expression given M and R?). You know the value of g. You know R. Plug in the mass expression from above into the Newton's law formula for g. Solve for the iron core radius. As for g(r), you will need to consider Newton's gravitation law again, and deal with the mass contained in the spherical volume below radius "r". 2. Relevant equations F=-G?(?(r')/r)dv' g=-?F 3. The attempt at a solution