I am designing calendrics for an Earth-like planet located 1.1 AU from a Sun-like star. On Earth, on April 30th and August 13th, the Sun appears to pass directly overhead between the Tropics of Cancer and Capricorn. I am wondering how an additional 0.1 AU of distance would affect zenithal passages. Can you help? Mathematics welcome.
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1$\begingroup$ Please do not make edits which change the entire question. Ask a new question. (And anyway, the latitude of the tropics is determined by the tilt, or obliquity to use a fancier word, of the axis of rotation of the planet with respect to its orbital plane and has nothing to do with its orbit. Earth's axis is tilted 23°26′, so the tropics are at 23°26′ latitude north and south.) $\endgroup$– AlexPSep 20 at 19:58
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$\begingroup$ I see, thank you for your advice, and for this clarification. Very helpful! $\endgroup$– JM YadenSep 20 at 20:16
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$\begingroup$ Thank you for editing my question, L.Dutch. You have helped me before. I very much appreciate your time and effort. $\endgroup$– JM YadenSep 24 at 3:58
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2 Answers
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No Change
The tropical zone is defined by the tilt of the Earth's axis, not by our distance to the sun. Astronomically speaking, the tropics are the area of Earth where the sun can at some point of the year pass directly overhead. So, sure, your other Earth may be a bit colder, the years longer, and have an over all very alien climate... but the actual boundary line of the tropics would remain the same as long as other Earth has the same axial tilt.
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$\begingroup$ Thank you so much for your help, Nosajimiki. This is exactly what I needed to know. Thank you also to AlexP for helping me to clarify my question. $\endgroup$– JM YadenSep 21 at 2:17
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Kepler's third law of planetary motion, published in his book Harmonices Mundi, in 1619, four hundred years ago:
The ratio of the square of an object's orbital period with the cube of the semi-major axis of its orbit is the same for all objects orbiting the same primary.
Since ratio of the new orbit of the Earth is 1.1 times as large as the radius of our real-world orbit, the new orbital period period will be $\sqrt {1.1^3} = 1.1537$ times as long. Instead of 365.2422 solar days, the new year will be 421.3762 solar days long. (Solar days as they are now; the question does not say how long a solar day is in the new conditions.)
The point being that the year will be about 15% longer, so that our real-world Gregorian calendar makes no sense.
P.S.
I really do not understand what you mean when saying that "on April 30th and August 13th the Sun appears to pass directly overhead between the Tropics of Cancer and Capricorn". Each and every day of the year the sun appears to pass directly overhead at some latitude between the tropics. At equinoxes, that is around 20 March and around 22 September, that latitude is the Equator. At solstices, that is around 21 June and around 22 December, that latitude is one of the tropics -- the tropic of Cancer at the northern solstice in June and the tropic of Capricorn at the southern solstice in December. There is no day in the year when the sun passes directly overhead both on the northern and the southern tropic.
This a graph showing latitude of the subsolar point (the point on Earth where the sun is at the zenith at noon) as the date progresses from day 1 to day 365 of the year. (Note that the longitude is immaterial -- the map is there only as a visual reference. The Earth will go through a complete 360° rotation during each day.) Diagram made by user Cmglee, available at Wikimedia under the Creative Commons Attribution-Share Alike 4.0 International.
P.P.S.
The word tropic comes from Latin tropicus, which comes from Greek tropikos meaning "something to do with a turn", from tropē meaning "a turn" from the verb trepō meaning "I turn". The Greeks called tropics the latitudes where the sun appears to turn back from going north or south.
The word trope meaning "a motif" or "a turn of phrase" comes from the same Greek verb...
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$\begingroup$ I think OP meant "respectively", as in one date separate for each tropic. The dates do not correspond to zenithal passages over them, though. $\endgroup$ Sep 20 at 19:24
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$\begingroup$ Thank you so much for your response, AlexP. You have helped me before, and I very much appreciate your expertise. To be more specific, there are two zenithal passages on the 14.8 degree of latitude, which is where the ancient Maya city of Izapa is located. I appreciate and understand your comment that increasing Earth's semi-major axis renders the Gregorian calendar meaningless. My new calendrics are designed around a 421+ day year, yes. I am going to modify my main question, thank you. $\endgroup$– JM YadenSep 20 at 19:52