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Next, find the length of the Martian day, in decimal Terran hours as: |
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Use this result to obtain the average number of degrees per hour that Mars rotates on its axis in one Martian Day, as: |
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Against the "fixed" backdrop of stars, Mars moves approximately another 0.524033 degrees in its orbit each day and must therefore rotate this additional 0.524033 degrees on its axis before the Sun is in the same relative position in the Martian sky as it was the day before. The time required to achieve this extra amount of rotation is given by: |
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This value is equivalent to about 129.0322278 seconds. |
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This amount of time must be added to the length of the sidereal day to obtain the length of the mean solar day, which is, therefore: |
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This value is about 24 hours 39 minutes 31.59 seconds, in terms of Terran time. The actual value used by astronomers includes a few more factors, such as precession, and is: |
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You now have the first of the necessary values. |
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9.13.2.1.3
Length of Martian Year |
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The other vital piece of information is the length of the Martian year, in terms of Martian days. For this, determine the relative lengths of Martian and Terran days. Calculate the length of the Martian day, in terms of Terran days, as: |
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The duration of the Martian year, in terms of Martian days, is then found as: 686.98/1.027491088 = 668.5994731 Martian days in one Martian year. This is the desired magic number defining the length of the Martian year, in terms of (local) Martian days, which is the length of the Martian mean solar year. These two calculated values are essential. From them, you may derive the rest of the calendar. |
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