Tilting the Oblate Earth

The earth is an oblate spheroid, with an equatorial radius R_E = 6,378,137 meters and a polar radius R_P = 6,356,752 meters ( GPS coordinates ). It projects an ellipse when viewed from the equatorial plane, and a circle when viewed from the poles. In between, it also presents an ellipse. What is the shape of that ellipse?

Our reference directions are x to the right, y to the back, and z to the top. When we tilt the oblate spheroid backwards, the front moves up and the back moves down. We can slice the spheroid into a series of elliptical slices along the x axis, each slice in a plane in y and z, perpendicular to the x axis. By analyzing the vertical projection of each slice, we can generalize to the vertical projection of the whole spheroid.

The equation for the ellipsoid is

1.0 = x^2/RE^2 + y^2/R_E^2 + z^2/R_P^2

The equation for the elliptical slice through the center of the spheroid is:

1.0 = y^2/R_E^2 + z^2/R_P^2 ~ ~ ~ x = 0 or alternately, x = R_E \cos( \theta ) and y = R_P \sin( \theta ) . We can rotate that ellipse by \theta_{eq} :

[ x_{rot}, y_{rot} ] = \left( \begin{array}{cc} \cos( \theta_{eq} ) & -\sin( \theta_{eq} ) \\ \sin( \theta_{eq} ) & \cos( \theta_{eq} ) \end{array} \right) \times [ R_E \cos( \theta ), R_P \sin( \theta ) ]

[ x_{rot}, y_{rot} ] = [ R_E \cos( \theta_{eq} ) \cos( \theta ) - R_P \sin( \theta_{eq} ) \sin( \theta ) ~ , ~ R_E \sin( \theta_{eq} ) \sin( \theta ) + R_P \cos( \theta_{eq} ) \cos( \theta ) ]

All we need to know is the maximum of y_{rot} = R_E \sin( \theta_{eq} ) \sin( \theta ) + R_P \cos( \theta_{eq} ) \cos( \theta ) as a function of \theta . Differentiating the above, we get

d y_{rot} / d \theta = R_E \sin( \theta_{eq} ) \cos( \theta ) - R_P \cos( \theta_{eq} ) \sin( \theta ) .

Setting that to zero, we get:

R_E \sin( \theta_{eq} ) \cos( \theta ) = R_P \cos( \theta_{eq} ) \sin( \theta )

and so:

\sin( \theta ) / \cos( \theta ) = tan( \theta ) = ( R_E / R_P ) ( \sin( \theta_{eq} ) / \cos( \theta_{eq} ) = ( R_E / R_P ) \tan( \theta_{eq} )

\sin( \theta ) = \tan( \theta ) / \sqrt{ \tan( \theta )^2 + 1 }

\cos( \theta ) = 1 / \sqrt{ \tan( \theta )^2 + 1 }

y_{rot, max} = \left( R_E \sin( \theta_{eq} ) \tan( \theta ) + R_P \cos( \theta_{eq} ) \right) / \sqrt{ \tan( \theta )^2 + 1 }

y_{rot, max} = \left( R_E \sin( \theta_{eq} ) R_E \sin( \theta_{eq} ) / R_P \cos( \theta_{eq} ) + R_P \cos( \theta_{eq} ) \right) / \sqrt{ \tan( \theta )^2 + 1 }

y_{rot, max} = ( R_E^2 \sin( \theta_{eq} )^2 + R_P^2 \cos( \theta_{eq} )^2 ) / ( R_P \cos( \theta_{eq} ) \sqrt{ R_E^2 \sin( \theta_{eq} )^2 / R_P^2 \cos( \theta_{eq} )^2 + 1 } )

y_{rot, max} = ( R_E^2 \sin( \theta_{eq} )^2 + R_P^2 \cos( \theta_{eq} )^2 ) / \sqrt{ R_E^2 \sin( \theta_{eq} )^2 + R_P^2 \cos( \theta_{eq} )^2 }

y_{rot, max} = \sqrt{ R_E^2 \sin( \theta_{eq} )^2 + R_P^2 \cos( \theta_{eq} )^2 }

y_{rot, max} = \sqrt{ ( R_E^2 - R_P^2 ) \sin( \theta_{eq} )^2 + R_P^2 }

and finally, as a function of x, the projected ellipse for the oblate earth is:

y = sqrt{ 1 - x^2 / R_E^2 } \sqrt{ ( R_E^2 - R_P^2 ) \sin( \theta_{eq} )^2 + R_P^2 }

\Large y = sqrt{ ( R_E^2 - x^2 ) ( ( 1 - ( R_P / R_E )^2 ) \sin( \theta_{eq} )^2 + ( R_P / R_E )^2 }

Anything behind this ellipse will be eclipsed.