GeCAA 2020, Теория — Задача 7. Radiant of a Meteor Shower

• a) $\mathrm{LST}=21^{\mathrm h}20^{\mathrm m}50^{\mathrm s}$.
• b) $(A,a)=(202^\circ,-14^\circ)$.
• c) $(\alpha,\delta)=(16^{\mathrm h}44^{\mathrm m},-67.2^\circ)$.
• d) Triangulum Australes (native spelling).

Source edition note: The English key’s reprinted first meteor description is: “One of the meteors appeared at horizon exactly due west and streaked to a point at 15° altitude directly above the northern horizon.” The primary bilingual question instead gives 15° above north to the eastern horizon. Both worked methods below use the latter endpoint coordinates. The key’s (d) additionally says to choose one constellation and write the same name in the answer box; that answer box/sheet was not obtained.

(a) Step 1: Determine the GST at the time of observation. Let us denote the GST on at 00:00 UT on 1st January 2020 as $\mathrm{GST}_0$. Total 268 days have elapsed since start of the year till $0^{\mathrm h}$ UT on 25 September. [1.0]

At 21:00 for GMT+8 timezone, UT will be $21^{\mathrm h}-8^{\mathrm h}=13^{\mathrm h}$. Hence,

As longitude of observer is $120.4^\circ$, the LST of observation is,

(b) For horizontal coordinates

[1.0] The spherical triangle has the radiant, initial position and final position of the $i$-th meteor to be $(A,a)$, $(A_i,a_i)$ and $(A'_i,a'_i)$, where $a$ and $A$ denotes the altitude and azimuth respectively. Using the four-parts (cotangent) equation on the left and right triangles, we have, for both meteors:

Similarly,

We can then eliminate $\cot\theta$ to yield:

Multiplying the whole equation by $\sin(A'_i-A_i)\sin(A_i-A)/\cos a_i$,

Plugging in $i=1,2$, and eliminating $\tan a$ by division, we have

where $k$ is the RHS defined above (which is a known value). Expanding the equation above and dividing by $\cos A$, gathering terms of $\tan A$, we get

And

Plugging in the respective values from the question,

We get

However, the student should realize that for $A=21.94^\circ$, the radiant is in the path of the first meteor, which is unrealistic. Hence, it is more likely that the azimuth takes the second value. [1.0] Plugging in for the altitude, we find the coordinate of the radiant to be

As the meteor track origin and end points are rounded to integer degrees, the horizontal coordinates of the radiant will be

Alternative solution

In the vector representation, let the starting and final locations of the meteroids be $\vec u_i$ and $\vec v_i$ respectively. For the subsequent calculations, the vectors are most conveniently represented in Cartesian coordinates. Letting the $x$-axis point east, $y$-axis north, and $z$-axis towards the zenith, we have

For a generic vector $\vec u$ with azimuth $A$ and altitude $a$. We can therefore compute

The meteors move along great arcs, and the two intersections of the great arcs correspond to the two possible locations of the radiant. A great arc is uniquely defined by its normal vector, denote it with $\vec n_i$. The normal vector is conveniently calculated through the cross product of the starting and final position, $\vec n_i=\vec u_i\times\vec v_i$. The cross product can be calculated via the determinant of the following matrix

This yields

Every point on the great circle $i$ is perpendicular to $\vec n_i$. This means that the intersections of the two great circles are perpendicular to both $\vec n_1$ and $\vec n_2$. The vector parallel to the intersections can therefore be calculated by the cross product of $\vec n_1$ and $\vec n_2$. Let the vector corresponding to the intersection be $\vec m$. Then

Now all that’s left is to calculate the azimuth and altitude corresponding to $\vec m$. Note that $-\vec m$ is also an intersection. The equations for the altitude and azimuth can be reverse-engineered from equation (1):

One has to take care that the arctans are signed. This gives the final values of the intersection points to be

As in the first solution, the $A=22^\circ$ solution can be eliminated due to it being on the path of the first meteor This gives the final answer of

(c) Now, it is easy to convert the horizontal coordinates of the radiant to equatorial coordinates.

[1.0] Using the cosine rule:

From the relationship between the LST and the Hour Angle ($H=\mathrm{LST}-\alpha$), we get the equatorial coordinates of the radiant to be

Thus, the equatorial coordinates of the radiant are,

(d) Triangulum Australes [2.0]

Source note: The native cotangent derivation’s intermediate multiplied line prints $\sin(A'_i-A)$ where the preceding multiplier is $\sin(A'_i-A_i)$, and its first line has the right-hand term on a separate row without a printed equality. That arrangement is retained in the aligned transcription. The numeric $k$ substitution prints $\sin15^\circ$ in the denominator, followed by $k=1.414$, although the listed second azimuth difference is $45^\circ$. The equatorial substitution prints $-13.4^\circ$ as well as $-13.96^\circ$. All printed forms, numerical values, constellation spelling and marking criteria are retained.