Calculate earths mass and moment of inertia

Jan 2015
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 ??



PHF Helper
Jun 2010
Morristown, NJ USA
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.
Jan 2015
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.