Calculus Applications in Physics 2 Homework 5 Due Friday, February 9, 2018 1. Ea

Question

Calculus Applications in Physics 2 Homework 5 Due Friday, February 9, 2018 1. Earth Problem (worth two regular problems) The mass of the earth is M 5.98 x1024 kg, and its radius is R-6.37x106 m. Furthermore, its moment of inertia about the rotational axis has been found to be 0.33MR2. With this information and a bit more, you're going to model the density of the interior of the earth as follows. From the center out to about 3470 km from the center, we'll assume the density is constant. Call it p,. (This is the earth's core.) From there out to the surface, we'll assume the density varies linearly with distance from the center. That is, = ar + b . We know that near the surface the density averages 3.3 g/em (3300kg/m) so this can be rewritten: = 3300 kgm' +a(r-6, 370, 000m) (Notice, this varies linearly with r, and when r is 6,370,000 m, the density is 3300, just as we said.) Based on these assumptions find the values of p and a that best fit the mass and moment of inertia quoted in the first paragraph. (Geologists who try to figure out what the earth's interior is like have to make sure that whatever model they believe describes it is consistent with measured properties such as the mass and the moment of inertia.) To get you started: We can write the mass of the earth as Here you know what M is, but you have some unknowns lurking in p. When you do the integral, you'll have to do it in two separate pieces, one for the core and one for the rest (which is mostly the earth's mantle). This is because the function that you're integrating (p) has a different mathematical form in different parts of the earth. Similarly, the integral for the moment of inertia, also has to be separated into two parts. When you've done everything you can do with these integrals, you should end up with two equations (one from each integral) and two unknowns, pe and a.

Explanation / Answer

given

rho = 3300 + a(r - 6370,000)

M = integral(rho*dV)
I = ingtegral(r^2*rho*dV)

now dV = 4*pi*r^2*dr

hence

also, rhoc = constant till r = Ri

so
M = rhoc*4*pi*Ri^3/3 + 4*pi*integral(3300r^2 + a(r - 6370,000)r^2)dr (consider inner radius of this integral, Ri = 3470,000 m)
5.98*10^24 = rhoc*4*pi*3470,000^3/3 + 4*pi*(1100(6370,000^3 - 3470,000^3) + a([6370,000^4 - 3470,000^4]/4 - 6370,000(6370,000^3 - 3470,000^3)/3)
5.98*10^24 = 1.750157*10^20rhoc + 4*pi*(238362223000000000000000 - 84735936166666666666666666.66667a)
2.9846*10^4 = 1.750157rhoc - 10648231.78225a

from moment of inertia
I = integral(r^2*rho*4*pi*r^2*dr)
0.33*5.98*10^24*(6370000^2) = 4*pi*rhoc*(3470,000^5)/5 + 4*pi*(3300(6370000^5 - 3470000^5)/5 + a[(6370,000^6 - 3470,000^6)/6 - 6370,000(6370000^5 - 3470000^5)/5 )
-1.35467*10^7 = 1.2644rhoc -2.3587077*10^6 a

solving
rhoc = -1.54591*10^7
a = -2.543

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