Added note, especially for those who have studied section (12a) "How Orbital Motion is Calculated"
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The simple pair of formulas derived in problem (8)
is very powerful and useful. For instance, it can reproduce rotations in 3-dimensional space by first rotating (x,y) by some angle α around the z axis,, giving the above result and producing (x',y'), while z remains the same, so that in the new system z' = z. Then we rotate (x',y',z') by an angle β around the x' axis, using the same formulas but with (y',z', β) in place of (x,y, α). Let the new variables be (x",y",z"). Then x' remains the same, i.e. x"=x', but (y',z') are replaced by
Suppose a satellite's position is given by its orbital elements. Knowing its semi-major axis a, eccentricity e and mean anomaly M, one can derive its true anomaly f (using appropriate formulas, not given in #12a) and from it the radial distance r
In cartesian coordinates (x',y',z'), with z' perpendicular to the orbital plane and x' along the major axis, the satellite's position is
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But what are its actual celestial coordinates (x,y,z)? From the illustration shown here (taken from section #12a), (x,y,z) can be reached by conducting two simple rotations by angles (ω, i, Ω). To connect the two systems, the simple transformation derived above is applied, consecutively--first a rotation around line N by an angle i giving an intermediate system (x",y",z"), after which z' = z" = z. And then another rotation by an angle ω+Ω around the shared z axis, bringing the x' axis, which originally passed the apogee P, into the celestial x axis.Special notations exist which simplify the job. Problems involving the orientation of satellites in 3-dimensional space (the "attitude" of those satellites) are handled in a similar way. |
Back to #M-11a Trigonometry Proficiency Drill
Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated (formatting) 9 January 2005
