So, how much energy would it take to disintegrate the Earth? Marvin the Martian wants to know! So how do we know how much energy it would take to move all
those bits apart? It's simply the change in potential energy as the bits are
separated. So how do we find the change in potential energy? Like
this.
To apply this relationship to the problem of finding the
gravitational binding energy of the Earth, we imagine removing one
infinitesimally thick outer shell after the other from a spherical
distribution of mass that gets smaller and smaller as shells are removed.
Each mass shell is moved an infinite distance away through the diminishing
force of gravity between the shell and sphere. It will take some energy
to move all these shells; that energy is the negative of the binding energy of the system.
To find energy needed to remove all these shells, we will find the change in
the system's potential energy. To find the change in potential energy, we need to know the mass of the shell, the mass of the
remainder of the sphere, and the force through which we're moving the shell. If, on the other hand, we do not wish to assume a constant
density throughout the Earth, we'll need a density that varies with radius.
We can approximate the Earth as a sphere having several layers, each layer
having a quadratic density function as described in the Earth Gravity essay. In short, we have a
piecewise density function.
Going back to our expression for the mass of the shell and the mass of the sphere inside the shell, we now have the following infinitesimal potential energy, this time using the real density instead of an average density. Same as before, to sum all the infinitesimal bits of binding energy we integrate over the radius of all the shells from the outside in. This time, however, we can only integrate one layer at a time since the piecewise density function is not differentiable across the layer boundaries. We'll first derive a relationship for all the infinitesimal shells within one layer, then we'll sum the layers. For a point having a radius within a given layer, we will need to find the potential energy change that occurs as that layer is stripped, shell by shell, from the radius of the point in question to the inside of the layer. For a point h in a
layer having lower and upper bounds of h_{1} and h_{2} respectively, we haveWhich can be seen to reduce to the equation for average density by assuming a sphere of one layer where a = b = h_{1} = 0
and c = <ρ>We can also see that for the entire i^{th} layer we haveand thus for any point r along the radius of the sphere, we havewhere ΔU _{0} = 0 and the indexed constants for each piece of the function are described in
the table above.And since the gravitational binding energy is the same magnitude but opposite in sign to the change in potential energy as the system is disassembled, we have Using the average density of the Earth = 5.51394×10 One place in which gravitational binding energy becomes
important is in the Nordtvedt effect, which is a test of the equivalence
principle envisioned by Kenneth Nordtvedt.
Thus, the Earth's gravitational mass might be less that its inertial mass by up to 320 parts per trillion. So, how might we go about detecting this difference? When Nordtvedt discussed the effect, he came up with the idea of using the Moon as a reference. We can perform the same calculations for the Moon and obtain
^{3,7}
In this case, the average density method figure is only 1.80% above the dual density figure. This is expected; the core is dense but very small; and, we're assuming a constant density in the core and the mantle. Having a better density profile for the Moon would make the variable density function more accurate. If we plot the average and variable density binding energy
functions over the radius of the Moon, we obtain the following; the blue
line represents the dual density function and the dashed line is the
average density approximation. The
Mathematica source code may be found here.
Click on the plot thumbnail for a high resolution image best viewed with a 1280x1024 screen resolution. Even though the lunar density profile isn't very accurate,
it's accurate enough to show that the gravitational binding energy mass
fraction of the Moon—20 parts per trillion—is
over an order of magnitude smaller than that of the Earth. That being
the case, Notwithstanding possible violations of the
equivalence principle, Marvin has his work cut out for him. References |