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It is usually asserted that the shape of the Earth is a prolate ellipsoid because of the effect of its rotation. The analysis below investigates this matter and concludes that while the shape is approximately ellipsoidal it is not precisely ellipsoidal.
The problem of determining the shape is a matter of determining the equipotential surfaces. This can be reduced to finding the shape of the cross section cruve of an equipotential surface. The relevant cross section is with a plane that passes through the polar axis of the Earth.
The form of the cross section curve may be found by determining the curve whose tangent is perpendicular to the force vector at each point. Two lines are perpendicular if the product of their slopes is equal to negative one. Let y(x) be the equation of the cross section curve. Its slope is dy/dx. The force vector through a point has components Fx and Fy. The slope of the force vector is Fy/Fx.
For perpendicularity
The force at a point (x,y) is the combination of the gravitational force and the centrifugal force associated with the rotation of the Earth. and moon The gravitational attraction of the Earth on a unit mass inversely proportional to the square of the distance r2=x2+y2 and is directed toward the center of the Earth at (0,0). The centrifugal force is directed outward in the x-direction only.
The forces per unit mass are:
where M is the gravitational constant times the mass of the Earth and ω is the rotation rate of the Earth.
The condition of perpendicularity is then
This is the general differential equation for the shape of an equipotential curve for a rotating planet. It has a solution but not necessarily one that is simple enought to be expressed analytically. Consider first the special case of ω=0. For this case the equation reduces to:
This is the equation of a circle. In this case the equipotential curves would be concentric circles.
Consider now the equation of an ellipse
where a and b are the axes in the the x and y directions, respectively.
In differential form the equation of the ellipse is
The general equation is not exactly of this form and thus the equipotential curve is not exactly an ellipse. However, if the curve is nearly a circle so r is nearly constant then
This equation thus gives the eccentricity of the ellipse of an equipotential curve.
Since M = Gρ(4/3)πR3 where ρ is the mass density and R is the average radius of the Earth it follows that
Since r/R = 1 at the surface of the Earth
The values of the parameters of this equation in SI units are:
The RHS of the above equation evaluates to 1 - 0.00343. The value of b/a is 7900/7926 = 1 - 0.00328 so (b/a)2 = 1 - 0.00655. The error in terms of the difference in the RHS versus the LHS is small, about 0.3 of 1 percent.
At the surface of the hydrosphere (r/R) is slightly larger than 1 and the eccentricity of the ellipse for the hydrosphere will slightly larger than for the surface of the Earth. For the tops of the troposphere and stratosphere the values of (r/R) will be greater still and consequently the eccentricities of the ellipses larger.
The general equation as derived above is:
A change of variables to u=y2 and v=x2 reduces the general equation to:
A further change of variable to w=u+v reduces the general equation to
This equation has the formal solution
but the integral does not have a closed form value.
The role of ω can be removed from the equation by making the change of variables s = ω4/3w and q=ω4/3v so the equation is
(To be continued.)
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