#format jsmath = Satellite Distance = || Thinsat arrays will not always be directly above the ground antenna, but will be at an elevation angle $ e $ above the horizon. If the array orbits at a radius $ A \bullet R_E $, what is the distance $ D \bullet R_E $ between the array and the ground antenna? Note that $ D = 1 $ when elevation angle $ e $ is $ \pi / 2 $ or 90°.<
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>Signals between satellite and ground are attenuated by distance, though at 70 GHz attenuation due a longer slant path through the atmosphere is more important. For the M288 orbit, where A ≈ 2, $ D = \sqrt{ 3 } $ if $ e = 0 $, attenuating signals by -10 log 3 or -4.8 dB. The atmospheric attenuation is far greater. Unless there are multiple ground antennas pointed at different parts of the sky, a flat antenna pointed at zenith talking to a satellite at 20° elevation will get -10 log( sin( 20°)) or -4.7 dB less signal due to smaller projected signal capture area, while the distance attenuation is -3.1 dB . See PathAttenuation for the sum of all effects. || {{ attachment:SatDistance.png | | width=200 }} || The sin law states that: $ { { sin( d ) } \over D } = { { sin( a ) } \over A } = { { sin( c ) } \over 1 } $ $ sin( a ) = sin( e + { \pi \over 2 } ) = \cos( e ) $ $ sin( c ) = { { sin( a ) } \over A } = { { cos( e ) } \over A } $ $ cos( c ) = \sqrt{ 1 - \left( { cos( e )^2 } \over A \right)^2 } $ The sum of the corners of a triangle is $ \pi $, so $ \pi = a + c + d = e + { \pi \over 2 } + c + d \;\;\;\;\;\;\;\;\; { \pi \over 2 } = e + c + d \;\;\;\;\;\;\;\;\; d = { \pi \over 2 } - ( e + c ) $ $ sin( d ) = sin( { \pi \over 2 } - ( e + c ) ) \;\;\; = cos( e + c ) \;\;\; = cos( e ) cos( c ) - sin( e ) sin( c ) $ $ D = sin( d ) / sin( c ) \;\;\; = cos( e ) cos( c ) / ( cos( e ) / A ) - sin( e ) sin( c ) / sin( c ) \;\;\; = A cos( c ) - sin( e ) $ $ D = A \sqrt{ 1 - \left( { cos( e )^2 } \over A \right)^2 } - sin( e ) \;\;\; = \sqrt{ A^2 - cos( e )^2 } - sin( e ) $ || $ D = \sqrt{ ( A^2 - 1 ) + sin( e )^2 } - sin( e ) $ ||