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Old Oct 5th 2017, 07:24 AM   #1
Join Date: Jan 2015
Posts: 96
Calculate earths mass and moment of inertia

Hi again everyone,
I have a problem (in two parts) where I believe I am tackling the solution with the correct methodology but somewhere along the line I must be going wrong because cannot match my answers to the ones in the book, in fact I am way out (by a factor of 10 to the power 11 or 12 for each).
The problem reads as follows -

The moment of inertia of a sphere with uniform density about an axis through it's centre is .. 2/5 M R^2) = 0.400 M R^2.
Satellite observations show the Earth's moment of inertia is .. 0.3380 M R^2)
The geophysical data suggests that the earth consists of 5 main regions ..
1. Inner Core (R=0 to R=1220km) of average density 12,900 kg/m^3
2. Outer Core (R=1220 to R=3480km) average density 10,900 kg/m^3
3. Lower Mantle (R=3480 to R=5700km) average density 4,900 kg/m^3
4. Upper Mantle (R=5700 to R=6350km) average density 3,600 kg/m^3
5. Outer Crust (R=6350 to R=6370km) average density 2,400 kg/m^3

(a) Show that the moment of inertia about a diameter of a uniform spherical
shell of inner radius = R(1) and outer radius = R(2), and density 'Rho' is ..

I = (8/15 Pi) x Rho x ((R(2)^5) - (R(1)^5))
[Hint - form the shell by superposition of a sphere of density 'Rho' and
smaller sphere of density '-Rho' ]

(b) Check the given data by using them to calculate the mass of the earth.

(c) Use the given data to calculate the earths moment of inertia in terms of
'M R^2'.

I was able to do part (a) without much problem, but it is parts (b) and (c)
I am unable to solve to match the answers given in the book.
For part (b) I used the fact that Mass = density x volume and volume of
a sphere is .. 4/3 Pi R^3.
Now since we are using spherical shells one within the other then to calculate
the volumes we multiply (4/3) x Pi x ((R(2)^3 - R(1)^3) for each of the 5 'shells'
giving .. M = (4/3) Pi x Rho x((R(2)^3 - R(1)^3) for each 'shell'.
I then summed each incremental mass to get a total mass !
However my arithmetic gave individual masses of the order a number between
1 and 10 x 10^15 and when summed gave a larger number of the same order
(number x 10^15) .. the book gives the total mass as 5.97 x 10^24 !!
As you can see I am way out .. in the order of 10^9.
Similarly for part (c) .. M R^2 when taking the official figures for the earth is
M= 5.97 x 10^24 and R=6.38 x 10^6 .. so M R^2 = 4.07 x 10^37kgm^2
I used the expression ... I = (8/15) x Pi x Rho ((R(2)^5) - (R(1)^5)) for each 'shell' and then summed them. he correct answer should show.. (my answer/(4.07 x 10^37)) = 0.400 approximately, but my answer is of the order of .. number (between 1 and 9.99) x 10^23 .. which again is way way out.
Can anyone help show me where I am going wrong .. is it my method or is it my arithmetic ??

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Old Oct 5th 2017, 09:04 AM   #2
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Location: Morristown, NJ USA
Posts: 2,324
I'm guessing that you may have simply forgotten to convert the values of the radii from Km to meters. That would account for errors of 10^9 in part b and 10^15 in part c.
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Old Oct 5th 2017, 09:47 PM   #3
Join Date: Jan 2015
Posts: 96
Hi ChipB,
I can't believe I have done something as basic as that, and here I was begining to doubt my whole methodology on this one. I suppose I should look on the positive side of the situation in that it was a careless mistake on my part as opposed to a failure in the understanding the basic problem.
Thank you for your help. Again it is much appreciated.
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