Viewing Server Sky from Northern and Southern Latitudes

The m288 server sky orbit, at about 1 Earth radius altitude, is below the horizon from the north and south poles, as is any equatorial orbit. From any point on earth, even on the equator, only part of the orbit to the east and west is visible. The portion of the orbit directly south of any given point on the Earth is highest in the sky. But at midnight, for the m288 orbit, that portion of the orbit is in eclipse. What portions are visible at what latitudes, and at what times?

First, how far north can the due south portion of an equatorial orbit be seen? Wider orbits are visible from higher latitudes. The oblate earth makes orbits visible slightly more to the north.

The oblate earth is 6378.1 km radius at the equator, and 6356.8 km radius at the pole. Assume that \vec x and \vec y are vectors in the equatorial plane, and \vec z is a vector in the direction of the north pole. x, y, and z are the associated coordinates. Define the radial vector h such that h^2 ~ = ~ x^2 + y^2 . A cross section of the shape of the oblate earth can be defined as :

\large 1 = \Large ( { h \over q } )^2 ~ + ~ ( { z \over p } )^2

. . . where p = 6356.8 km ( Pole radius ) and q = 6378.1 km ( eQuator radius ).

The orbit is at radius r in the \vec h direction. and is just barely visible at the horizon at point (h,z) on the ellipse where the tangent of that point points at (r,0). The slope at point (h,z) can be found from:

\large 0 = \Large { {\partial} \over {\partial h } } ( ( { h \over q } )^2 ~ + ~ ( { z \over p } )^2 ) { \partial h } ~ + ~ { {\partial} \over {\partial z} } ( ( { h \over q } )^2 ~ + ~ ( { z \over p } )^2 ) { \partial z }

\large 0 = \Large ( { { 2 h } \over { q^2 } } ) \partial h ~ + ~ ( { { 2 z } \over { p^2 } } ) { \partial z }

\Large { { \partial z } \over { \partial h } } ~ = ~ - ( { {p^2} \over {q^2} } )^2 ~ ( { h \over z } )

The same slope describes the triangle formed by z and r - x :

\Large { { \partial z } \over { \partial h } } ~ = ~ - ( { {p^2} \over {q^2} } )^2 ~ ( { h \over z } ) ~ = ~ - { z \over { r - h } }

\Large { { h ( r - h ) } \over { q^2 } } ~ = ~ { { z^2 } \over { p^2 } }

\Large { { h r } \over { q^2 } } ~ = ~ { { h^2 } \over { q^2 } } ~ + ~ { { z^2 } \over { p^2 } } ~ = ~ 1 . So:

\large h = \Large { { q^2 } \over r }

Find z:

\large z^2 ~ = ~ \Large ( { { p^2 } \over { q^2 } } ) ~ ( { { q^2 } \over r } ) ~ ( r - { { q^2 } \over r } )

\large z ~ = ~ \Large { p \over r } \sqrt { r^2 - q^2 }

Find latitude:

\large \tan( latitude ) ~ = ~ \Large { { -1 } \over {slope} } ~ = ~ ( { { q^2 } \over { p^2 } } ) ~ { z \over h }

\large \tan( latitude ) ~ = ~ \Large ( { { q^2 } \over { p^2 } } ) ~ ( { r \over { q^2 } } ) ( { { p \sqrt { r^2 - q^2 } } \over r } )

\large \tan( latitude ) ~ = ~ \Large { { \sqrt { r^2 - q^2 } } \over p }

For p = 6356.8 km and q = 6378.1 km, here is the maximum latitude for various M orbits, without J_2 correction:

But that is not the whole story. Our latitude defines a plane that touches the edge of the orbit at one point. In other words, satellites in that orbit are visible for only one instant. Rather than using radius r , we are more interested in the y value, where y = h ~ cos( \theta ) . \theta it the true anomaly, the position along the orbit, and a function of time. If the eccentricity e is zero, the function is linear.

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