USAAAO 2024, II кръг — Long 2. The Analemma

• (a) $l(L)\approx L+2e\sin(L+\phi)$

• (b) $\delta\approx\epsilon\sin(L)$

• (c) $\Delta\alpha\approx2e\sin(L+\phi)-\frac{\epsilon^2}{4}\sin(2L)$

  Native intermediate lines mix M and L; they remain literal in the worked key.

• (d) The sin(2L) term produces a six-month component in Δα while δ oscillates annually, giving the figure-8.

(a) A convenient variable to define (though you don’t have to) is the true anomaly $\nu=l+\phi$, such that we’d expect $\nu=M$ for a circular orbit. The quantity we want to know is the difference between $\nu$ and $M$, or $\Delta\nu$, which should be small in the limit where the orbit is nearly circular.

We can find it carefully by expanding Kepler’s equation in small $e$, but that’s an unnecessarily complex approach if we just want the lowest order term of the expansion. I’ll adopt the strategy of just making an educated guess based on a few points in the orbit. We can think of $\Delta\nu$ as a function of $M$, and guess the form of this function $\Delta\nu(M)$.

First we should come up with a qualitative expectation/“ansatz” of what the answer will be. $\Delta\nu$ must be periodic in $M$ with period $2\pi$, so the answer should be built out of trig functions. As the Earth is passing perihelion, its angular velocity about the sun is a little higher than its average angular velocity over the full orbit, so we’d expect $\Delta\nu$ to be increasing. As the Earth is passing aphelion, by the same logic, we’d expect $\Delta\nu$ to be decreasing. Thus, a reasonable guess is $\Delta\nu=C\sin M$, where $C$ is some coefficient to be determined.

At perihelion ($M=0$), by definition $\nu=M=0$, so $\Delta\nu(0)=0$.

At aphelion ($M=\frac\pi2$), by the symmetry of the orbit we must have $\nu=M=\pi$, so $\Delta\nu(\pi)=0$ again.

The other two points to consider are along the semi-minor axis. Specifically, let’s consider the point along the semi-minor axis where Earth is moving from perihelion to aphelion. By Kepler’s second law, the mean anomaly is

If we draw a triangle through the Earth, the center of the orbit, and the Sun (see the figure), it is a right triangle with base $ae$ and height $b=a\sqrt{1-e^2}$. The area swept out by Earth is $\frac14$ the area of the ellipse, minus this triangle, which is $\frac14\pi ab-\frac12abe$. Thus

The true anomaly $\nu$ is given by the Earth-Sun-perihelion angle, which using the small-angle approximation is $\frac\pi2+e\frac ab$. For a nearly-circular orbit we can say $\frac ab$ is close to 1, so

This gives $\Delta\nu(\pi/2)\approx2e$. At the opposite point in the orbit along the semi-minor axis, we can carry out a very similar argument to show $\Delta\nu(3\pi/2)\approx-2e$.

Considering all four of these points, as well as our guess for the form of $\Delta\nu$, we get

which turns out to be correct in the small $e$ limit. (For non-small $e$ there is no closed-form solution; we would have to write it as an infinite series with higher order terms like $e^2\sin2M$, etc.) We can rewrite this as

Finally plugging in the definitions of $l,L$ in terms of $\nu,M$:

(b) We can get $\delta$ as a function of $l$ from spherical trigonometry. Draw a spherical triangle with the ecliptic and celestial equator as two sides, and the projection of the sun’s position onto the celestial equator giving a right angle. Then the spherical law of sines gives

Since $\epsilon$ is small we can apply the small angle approximation:

But we want $\delta$ as a function of $L$. To get this, plug in the expression for $l(L)$ from part (a):

Since $e$ is much smaller than 1, we might want to use the small angle approximation here. However, this isn’t quite the right form for it. The trick is to first apply the trig identity for the sine of a sum of two angles:

and we can now apply a small-angle approximation in $e$ on both of the resulting terms:

(If you’re familiar with calculus, you may also recognize that you can get this from a first-order Taylor series expansion.)

The second term here is proportional to $e\epsilon$, while the first term is just proportional to $\epsilon$. Since we assumed both $e$ and $\epsilon$ are much smaller than 1, we know the $e\epsilon$ term is much smaller than the $\epsilon$ term, so we can ignore it.

(c) This is the hardest part of the problem. It really just comes down to expanding everything out with trig identities and carefully determining which terms we are allowed to ignore.

We can derive $\alpha$ in terms of $l$ by drawing the same spherical triangle as before. The spherical law of cosines gives:

Since $\epsilon$ is small, we can write

(Unlike in the usual small-angle approximation, we have to include the second-order term here, otherwise we’d lose all dependence on $\epsilon$!)

Write $\alpha=L+\Delta\alpha$:

We can assume $\Delta\alpha$ will be small. Therefore we can do a similar trick as in part (b). Applying the cosine angle addition identity and the small-angle approximation:

Rearranging, and using the other approximation given in the problem:

Plugging in the expression for $l(L)$:

Again using the small-angle approximation trick:

Plugging this back in:

As before, we can ignore the term of order $e\epsilon^2$, since we assumed both $e$ and $\epsilon$ were small. However, we shouldn’t ignore the term proportional to $\epsilon^2$ since we didn’t assume anything about how $e$ and $\epsilon^2$ compare! For Earth, it turns out that they are roughly the same size.

(d) The analemma looks like a figure-8 because of the presence of the $\sin(2L)$ term, which gives a contribution to $\Delta\alpha$ that oscillates with a period of 6 months rather than 1 year. Without this term, the analemma would just be an ellipse as both $\Delta\alpha$ and $\delta$ are oscillating at the same frequency.

If we plug in the parameters for Earth:

we can make a parametric plot of $\delta$ against $\Delta\alpha$:

Source note: The aphelion paragraph prints M=π/2 and ν=M=π in the same sentence. Two intermediate part(c) terms print sin M and sin²(M) where adjacent lines use L; native notation and the final boxed expression are retained. The final diagram has horizontal δ and vertical Δα, while the preceding prose says a plot of δ against Δα; the source plot is retained with its original axis labels.