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Wed, 04 Jan 2012
Mental astronomical calculations
The other day I was musing on this, and it is a nice mental calculation to compute the rate of increase.
The day length is given by a sinusoid with amplitude that depends on your latitude (and also on the axial tilt of the Earth, which is a constant that we can disregard for this problem.) That is, it is a function of the form a + k sin 2πt/p, where a is the average day length (12 hours), k is the amplitude, p is the period, which is exactly one year, and t is amount of time since the vernal equinox. For Philadelphia, where I live, k is pretty close to 3 hours because the shortest day is about 3 hours shorter than average, and the longest day is about 3 hours longer than average. So we have:
day length = 12 hours + 3 hours · sin(2πt / 1 year)Now let's compute the rate of change on the equinox. The derivative of the day length function is:
rate of change = 3h · (2π / 1y) · cos(2πt / 1y)At the vernal equinox, t=0, and cos(…) = 1, so we have simply:
rate of change = 6πh / 1 year = 18.9 h / 365.25 daysThe numerator and the denominator match pretty well. If you're in a hurry, you might say "Well, 360 = 18·20, so 365.25 / 18.9 is probably about 20," and you would be right. If you're in slightly less of a hurry, you might say "Well, 361 = 192, so 365.25 / 18.9 is pretty close to 19, maybe around 19.2." Then you'd be even righter.
So the change in day length around the equinox (in Philadelphia) is around 1/20 or 1/19 of an hour per day—three minutes, in other words.
The exact answer, which I just looked up, is 2m38s. Not too bad. Most of the error came from my estimation of k as 3h. I guessed that the sun had been going down around 4:30, as indeed it had—it had been going down around 4:40, so the correct value is not 3h but only 2h40m. Had I used the correct k, my final result would have been within a couple of seconds of the right answer.
Exercise: The full moon appears about the same size as a U.S. quarter (1 inch diameter circle) held nine feet away (!) and also the same size as the sun, as demonstrated by solar eclipses. The moon is a quarter million miles away and the sun is 93 million miles away. What is the actual diameter of the sun?
[ Addendum 20120104: An earlier version of this article falsely claimed that the full moon appears the same size as a quarter held at arm's length. This was a momentary brain fart, not a calculational error. Thanks to Eric Roode for pointing out this mistake. ]