How fast would I need to climb to keep the setting sun at the horizon? (I originally asked this question over on Aviation Stack Exchange, only for it to be closed as off-topic. I was told this was a better place to ask.)
The title is basically the whole question. If I wanted to see the same sunset twice, how fast would I need to climb?
I'm currently at about 32.5 degrees North latitude, and would be starting my climb from about 2,000 feet above MSL, but what I'm really interested in is some equation or function that I could use no matter where I am.
(Note: Although I'm American, I tend to prefer the metric system. However, aviation in general is stuck using feet to measure altitude around the world (with very few exceptions), hence the altitude in feet above. So I guess what I'm saying is, use whichever system you prefer, and I can convert to the other when necessary.)
Edit: To simplify the question, let's just assume I start climbing the moment the exact center of the sun reaches the horizon. What vertical speed would I need to keep the sun in that position relative to the horizon? And then I know that I just need to exceed that speed to "see the sunset again". I would be flying West at about 73 KIAS* (which, assuming standard temperature, would be about 76 knots true, or 39 m/s), so I don't know that my horizontal speed is going to affect the answer all that much.
*That's the maximum rate-of-climb airspeed for the typical plane I rent (Cessna 172 Skyhawk, if anyone's curious).
 A: Here's a very quick-and-dirty estimate. Let's assume the earth is a sphere of radius $R$, that we're at latitude $\phi$, that the sun is a point source, and sets due west, in the plane containing the latitude circle of our location, i.e., it's an equinox.
Looking down from the nearest pole, we see our latitude as a circle of radius $R\cos\phi$. At sunset, we see the sun's ray as the line tangent to the latitude at our location. In polar coordinates $(r, \theta)$ with origin at the center of our latitude, this line has equation $r = R\cos\phi \sec\theta$. The shadow's distance from the latitude circle at (longitude) angle $\theta$ (in radians) is therefore
$$
r - R\cos\phi = R\cos\phi(\sec\theta - 1)
\approx \tfrac{1}{2}R\cos\phi \cdot \theta^{2}.
$$
(In the diagram, our location is the rightmost point on the latitude circle and the vertical line represents the sun's ray through our location, shining upward. The approximation is the second-order Taylor approximation of secant.)

Edit: As David K notes in the comments, the preceding direction makes angle $\frac{\pi}{2} - \phi$ with the surface of the earth. A more accurate estimate of the altitude of the shadow above the ground is therefore obtained if we multiply by $\cos\phi$:
$$
\text{Altitude} \approx \tfrac{1}{2}R\cos^{2}\phi \cdot \theta^{2}.
$$
(This formula agrees to second order with the estimate obtained by modeling the earth's shadow as a cylinder circumscribed on the earth with the sun lying on the axis at infinite distance; calculation omitted.)
The earth rotates one full turn in $86400$ seconds, so $t$ seconds after sunset the earth has rotated by an angle
$$
\theta = \frac{2\pi t}{86400},
$$
and the altitude of the shadow (adjusted as noted above) has increased to
$$
\tfrac{1}{2}R\cos^{2}\phi \cdot \theta^{2}
\approx 3175 \cos^{2}\phi \frac{(2\pi)^{2}}{86400^{2}} t^{2}
\approx 1.68 \cos^{2}\phi \times 10^{-5} t^{2}
$$
kilometers. At latitude $32.5^{\circ}$ this becomes about $1.2 \times 10^{-5} t^{2}$. Thus:
$$
\begin{array}{l|lllll}
  \text{Minutes after sunset at sea level} & 1 & 2 & 3 & 4 & 5 \\
  \text{Sun is setting at altitude (in m)} & 43 & 172 & 387 & 688 & 1075 \\
\end{array}
$$
The rotation speed of the earth at the given latitude is just about $390$ m/s, so to account for the plane's speed we can multiply $\theta$ by $0.9$, which multiplies the second row of the table by $0.81$.
Added: The graph shows the altitude of the earth's shadow in this model (neglecting the ground speed of the plane) as a function of latitude (in $2^{\circ}$ increments) and time after sea level sunset, with $\phi = 32.5^{\circ}$ in light blue.
The world being what it is, I should add that while every effort has been made to provide accurate qualitative estimates, this answer is provided for amusement only, and is not intended to be used as a recommendation for actual flight. :)

