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Someone I know said that Christmas is on the horizon. If that is the case we should be able to calculate its height given that we are on a curved sphere of known circumference and know the pace at which Christmas is approaching.

Given the earth's circumference at the equator at 24901 miles and Christmas arriving every 365 days we calculate Christmas approaching at a rate of 2.84 miles per hour $(24901/(365/24))$.

Now that we have the circumference and the speed, and given that we spotted it today (Friday August 21), how tall is Christmas? We can take our height at 5'9" and need to factor in Christmas at a perpendicular angle to the center of the earth as well.

To keep things simple we can ignore gravity's curvature on light and assume a straight line for the travel of light, and we don't have to factor in our own angle to center of the earth.

Please show your work :)

cmdematos
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2 Answers2

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This is really a kooky question.

In an ordinary, non-leap-year like $2015$, Christmas is the $359$th day of the year. Today, August $21$, is the $233$rd day of the year. Christmas is therefore $359-233 = 126$ days away. That equates to about $126/365\cdot360 = 124$ degrees. Unfortunately, that is more than $90$ degrees, so unless Christmas is crooked (gasp!) and doesn't extend straight out from the center of the Earth, Christmas cannot yet be on the horizon.

If I misunderstand your question, please point it out, and I'll try it again.

By the way, when it does become relevant, Andrew Young at San Diego State told me years ago, back in the Usenet group sci.astro.amateur (which I used to frequent, and still go back and visit from time to time), that the atmospheric refraction is a non-trivial consideration, but can be accounted for in many equations, approximately, by simply assuming the Earth to have a diameter about $15$ percent larger than it really is. In other words, just assume the usual figure of $7920$ miles (or so) is really nautical miles, and not statute miles.

Brian Tung
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The ray from Christmas to us (I'm going to assume that Xmas travels in a circle above the equator, and we are on the equator) is tangent to the equator at some point $P$ that's on our horizon. The angle from us to $P$ is $$ \arccos((24901/(2\pi)) / ((24901/(2\pi)) + 5.75/5280) ) \approx 0.000741333199 $$ (where that's in radians). The radius of the earth is $$ 24901/(2\pi) $$ so the circumferential distance from us to $P$ is about $$ (24901/(2\pi)) * 0.000741333199 \approx 2.94 $$ miles.

August 21 (the day I'm answering) is the 233 day of the year, while christmas is the 359th; the difference is 126 days, which is 126/365.25 of the year. (Since "today" is vague within a 1-day timescale, I figure that calling a year 365.25 days long is within the range of vagueness). That means that the angle between us and the terrestrial position of Xmas (i.e., the point on earth directly below the thing we're calling Xmas) is $$ \theta = \frac{126}{365.25} 2 \pi = approx 2.168 $$ radians. I'm going to call this 2.17 radians, again figuring that puts me well within the unspecified-ness of the problem. Subtracting off the 0.000741333199 radians from us to $P$ ...introduces an error that can again be ignored.

So all we need to know is "how high must an object that's 2.168 radians away be for us to see it?", or, put differently, at what distance from the center of a circle of radius $r = 24901/(2\pi)$, along a ray in direction 2.17 radians, must a point be for the line from $Q$ to $P = (r, 0)$ to be tangent to the circle at $P$.

Unfortunately, the angle $2.17$ radians is larger than $pi/2$, and hence it's in the second quadrant. There's no distance sufficient to make it visible.

John Hughes
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