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Equation of time
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=== Right ascension calculation === The right ascension, and hence the equation of time, can be calculated from Newton's two-body theory of celestial motion, in which the bodies (Earth and Sun) describe elliptical orbits about their common mass center. Using this theory, the equation of time becomes: : <math>\Delta t = M + \lambda_p - \alpha</math> where the new angles that appear are: * {{math|1=''M'' = {{sfrac|2π(''t'' − ''t''<sub>p</sub>)|''t''<sub>Y</sub>}}}}, is the [[mean anomaly]], the angle from the [[periapsis]] of the elliptical orbit to the mean Sun; its range is from 0 to 2{{pi}} as {{math|''t''}} increases from {{math|''t''<sub>p</sub>}} to {{math|''t''<sub>p</sub> + ''t''<sub>Y</sub>}}; * {{math|''t''<sub>Y</sub>}} = {{val|365.2596358}} days is the length of time in an [[anomalistic year]]: the time interval between two successive passages of the periapsis; * {{math|1=''λ''<sub>p</sub> = ''Λ'' − ''M''}}, is the ecliptic longitude of the periapsis; * {{math|''t''}} is [[dynamical time]], the independent variable in the theory. Here it is taken to be identical with the continuous time based on UT (see above), but in more precise calculations (of {{math|''E''}} or EOT) the small difference between them must be accounted for{{r|Hughes+|p=1530}}<ref name="computingGST"/> as well as the distinction between UT1 and UTC. * {{math|''t''<sub>p</sub>}} is the value of {{math|''t''}} at the periapsis. To complete the calculation three additional angles are required: * {{math|''E''}}, the Sun's [[eccentric anomaly]] (note that this is different from {{math|''M''}}); * {{math|''ν''}}, the Sun's [[true anomaly]]; * {{math|1=''λ'' = ''ν'' + ''λ''<sub>p</sub>}}, the Sun's true longitude on the ecliptic. [[File:EquationOfTimeGeom.svg|thumb|upright=2.5|right|The celestial sphere and the Sun's elliptical orbit as seen by a geocentric observer looking normal to the ecliptic showing the 6 angles ({{math|''M'', ''λ''<sub>p</sub>, ''α'', ''ν'', ''λ'', ''E''}}) needed for the calculation of the equation of time. For the sake of clarity the drawings are not to scale.]] All these angles are shown in the figure on the right, which shows the [[celestial sphere]] and the Sun's [[elliptical orbit]] seen from the Earth (the same as the Earth's orbit seen from the Sun). In this figure {{math|''ε''}} is the [[obliquity]], while {{math|1=''e'' = {{sqrt|1 − (''b''/''a'')<sup>2</sup>}}}} is the [[Eccentricity (mathematics)|eccentricity]] of the ellipse. Now given a value of {{math|0 ≤ ''M'' ≤ 2π}}, one can calculate {{math|''α''(''M'')}} by means of the following well-known procedure:{{r|Duffett-Smith|p=89}} First, given {{math|''M''}}, calculate {{math|''E''}} from [[Kepler's equation]]:{{r|Moulton|p=159}} : <math>M = E - e\sin{E}</math> Although this equation cannot be solved exactly in closed form, values of {{math|''E''(''M'')}} can be obtained from infinite (power or trigonometric) series, graphical, or numerical methods. Alternatively, note that for {{math|1=''e'' = 0}}, {{math|1=''E'' = ''M''}}, and by iteration:{{r|Hinch|p=2}} : <math>E \approx M + e\sin{M}</math> This approximation can be improved, for small {{math|''e''}}, by iterating again: : <math>E \approx M + e\sin{M} + \frac{1}{2}e^2\sin{2M}</math>, and continued iteration produces successively higher order terms of the power series expansion in {{math|''e''}}. For small values of {{math|''e''}} (much less than 1) two or three terms of the series give a good approximation for {{math|''E''}}; the smaller {{math|''e''}}, the better the approximation. Next, knowing {{math|''E''}}, calculate the [[true anomaly]] {{math|''ν''}} from an elliptical orbit relation{{r|Moulton|p=165}} :<math>\nu=2\arctan\left(\sqrt{\frac{1+e}{1-e}}\tan\tfrac12 E \right)</math> The correct branch of the multiple valued function {{math|arctan ''x''}} to use is the one that makes {{math|''ν''}} a continuous function of {{math|''E''(''M'')}} starting from {{math|1=''ν''<sub>''E''=0</sub> = 0}}. Thus for {{math|0 ≤ ''E'' < π}} use {{math|1=arctan ''x'' = arctan ''x''}}, and for {{math|π < ''E'' ≤ 2π}} use {{math|1=arctan ''x'' = arctan ''x'' + π}}. At the specific value {{math|1=''E'' = π}} for which the argument of {{math|tan}} is infinite, use {{math|1=''ν'' = ''E''}}. Here {{math|arctan ''x''}} is the principal branch, {{math|1={{abs|arctan ''x''}} < {{sfrac|π|2}}}}; the function that is returned by calculators and computer applications. Alternatively, this function can be expressed in terms of its [[Taylor series]] in {{math|''e''}}, the first three terms of which are: : <math>\nu \approx E + e\sin{E} + \frac{1}{4} e^2\sin{2E}</math>. For small {{math|''e''}} this approximation (or even just the first two terms) is a good one. Combining the approximation for {{math|''E''(''M'')}} with this one for {{math|''ν''(''E'')}} produces: : <math>\nu \approx M + 2e\sin{M} + \frac{5}{4} e^2\sin{2M}</math>. The relation {{math|''ν''(''M'')}} is called the [[equation of the center]]; the expression written here is a second-order approximation in {{math|''e''}}. For the small value of {{math|''e''}} that characterises the Earth's orbit this gives a very good approximation for {{math|''ν''(''M'')}}. Next, knowing {{math|''ν''}}, calculate {{math|''λ''}} from its definition: : <math>\lambda = \nu + \lambda_p</math> The value of {{math|''λ''}} varies non-linearly with {{math|''M''}} because the orbit is elliptical and not circular. From the approximation for {{math|''ν''}}: : <math>\lambda \approx M + \lambda_p + 2e\sin{M} + \frac{5}{4}e^2\sin{2M}</math>. Finally, knowing {{math|''λ''}} calculate {{math|''α''}} from a relation for the right triangle on the celestial sphere shown above{{r|Burington|p=22}} : <math>\alpha = \arctan \left(\cos{\varepsilon}\tan{\lambda}\right)</math> Note that the quadrant of {{math|''α''}} is the same as that of {{math|''λ''}}, therefore reduce {{math|''λ''}} to the range 0 to 2{{pi}} and write : <math>\alpha = \arctan \left( \cos{\varepsilon}\tan{\lambda} + k\pi \right)</math>, where {{math|''k''}} is 0 if {{math|''λ''}} is in quadrant 1, it is 1 if {{math|''λ''}} is in quadrants 2 or 3 and it is 2 if {{math|''λ''}} is in quadrant 4. For the values at which tan is infinite, {{math|1=''α'' = ''λ''}}. Although approximate values for {{math|''α''}} can be obtained from truncated Taylor series like those for {{math|''ν''}},{{r|Whitman|p=32}} it is more efficacious to use the equation{{r|Milne|p=374}} : <math>\alpha = \lambda - \arcsin \left( y\sin\left( \alpha + \lambda \right) \right)</math> where {{math|1=''y'' = tan<sup>2</sup><big>(</big>{{sfrac|''ε''|2}}<big>)</big>}}. Note that for {{math|1=''ε'' = ''y'' = 0}}, {{math|1=''α'' = ''λ''}} and iterating twice: : <math>\alpha \approx \lambda - y\sin{2\lambda} + \frac{1}{2}y^2\sin{4\lambda}</math>.
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