While the ideas of gravity and acceleration at and above Earth's surface are taught in first year physics, the acceleration due to gravity below the surface is often ignored or approximated with a homogeneous Earth. This essay describes a method of finding the gravitational acceleration within a spherical approximation of Earth using a variable density function. The gravitational acceleration at some distance away from a
point mass is given by
The acceleration due to gravity external to a spherically symmetric
uniform and continuous distribution of mass can be described by the same equation.Thus, if we make the assumption that the Earth is a sphere, we can use this equation to find the acceleration due to Earth's gravity for any position on or above the Earth's surface. But what of the acceleration below the surface? For a spherically symmetric uniform and continuous distribution of mass, the mass closer to the center of the distribution than the position in question provides a net gravitational acceleration at that position. The acceleration produced by the shell of mass further from the center, however, sums to zero. A good tutorial on that fact may be found here. If we make the assumption that the Earth's density is
constant with radius—i.e., that the Earth is homogeneous—then the acceleration
falls off in a linear fashion, reaching zero at the center. That is,
if a distribution of mass is spherically symmetric The solution is to characterize each layer's density as a
function of radius, then integrate the density of each layer over radius to
find mass. Since we're assuming the Earth to be spherically symmetric,
it'll then be easy to calculate the acceleration due to gravity for any
point within the Earth.
If we plot this function over the radius of the Earth, we obtain the following. Click on the plot thumbnail for a high resolution image
best viewed with a 1280x1024 screen resolution.
The Mathematica source code may be found here.We can find the mass within a sphere of any radius h. Note that h is a variable while h is the height of
the inner core to outer core boundary given in the table above._{1}where the values of the indexed constants come from the density function table shown above. The mass functions for the other layers may be similarly derived—but it's fairly obvious that the only difference will be in the indexed constants. The mass of the entire i^{th} layer is thereforewhere the values of the indexed constants come from the density function table shown above. To calculate the total mass "below" a given radius r, we sum the
masses of the layers that are completely "below" height r, then add the partial
mass of the layer that contains radius r. In the equation
below, the height of radius r lies within the n^{th} layer.where the values of the indexed constants come from the density function table shown above. Using the volumetric radius of the Earth, 6.3710×10 ^{6} m, this
function evaluates to a total Earth mass of 5.9727×10^{24} kg.
This is only 0.015% lower than the NASA figure of 5.9736×10^{24} kg. ^{3} That's very close given that the PREM densities
were inferred from the speed of sound within the Earth using seismographic
data.If we wish to approximate the Earth's internal structure using a constant, average density, we can simply use the volume of a sphere with radius r.which is what that triple integral reduces to when the integrand is a constant, and where and where r is the volumetric radius of the Earth._{v}The average density of the Earth evaluates to 5.5139×10 ^{3} kg·m^{-3}.Plotting the functions for mass over the radius of the Earth, we obtain the following. The blue line represents the true mass profile, while the dashed gray line represents 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. Now that we have a fairly accurate picture of the mass distribution within
the Earth, we can return to where we started; that is, to deriving a function for the
acceleration due to Earth's gravity that works as well below the surface as
it does above.
To account for a variable mass and variable density, we substitute our function for mass into the equation for gravitational
acceleration. We also provide a condition for radii outside the Earth. The value calculated for gravitational acceleration at Earth's volumetric radius is 9.8212 m·s ^{-2}.
According to the World Geodetic System 1984 (WGS84), the true surface
acceleration on Earth varies from 9.7803 m·s^{-2} on the equator to
9.8322 m·s^{-2} at 90° latitude;
the mean value is 9.7976 m·s^{-2}.^{ 4 } Thus, our
derivation is higher than the mean by 0.24%. A slight disagreement
with true surface acceleration is to be expected since our derivation does
not account for centripetal acceleration or the fact that the Earth is an
oblate spheroid, not a
sphere. Both phenomena are caused by the
Earth's rotation.For reference, the WGS84 gravity formula, which takes these phenomena into account, is: ^{4}where where
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