Съдържание

IAO 2019, Теория, α+β — β-2. NGC of the Year.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2019

XXIV International Astronomy Olympiad — Theoretical round, α and β · 19 октомври 2019 г. · 8 т.

For translation only.

Note for translators: in texts for Group β by the red color marked the text identical to that used for Group α.

Source availability

The three photographed sheets contain the combined English translator edition and the bilingual diagram for β-5. The archived data sheets contain Solar-System parameters, geographical coordinates, constants and the Hertzsprung–Russell diagram. The star-data table referred to in the key (including Ruchbah, Vega, Altair and UY Scuti) and the additional mid-round formula promised to Group α are not included in these archived theory/data sources. The surviving key supplies its own working values; no missing handout has been reconstructed.

Условие

NGC 2019 is a globular cluster in the constellation Mensa, observed against the background of the Large Magellanic Cloud. The cluster has a visible magnitude $m=10.9^{\mathrm m}$ , colour index $B-V=0.04^{\mathrm m}$ , and is located at a distance of approximately $L=50\ \mathrm{kpc}$ from us. To the right you can see a $4\times4$ arcmin negative picture of NGC 2019. Assume that the cluster consists for the most part of approximately the same stars located on the main sequence, as well as a small percentage of white dwarfs, and that all these stars are more or less evenly distributed in the space of the central “sphere” region, the image of which is clearly visible in the center of the image against the background of the LMC. Estimate, at what distance from the center of the cluster an observer need to be located in order to distinguish separately the white dwarfs visible in the projection on the center of the cluster with the naked eye.

Original reference sheet: Solar-System parameters and geographical coordinates

Решение

Absolute magnitude of NGC 2019 is

M=m-5^{\mathrm m}\times\lg(50\ \text{кПк}/10\ \text{Пк})=10.9^{\mathrm m}-18.5^{\mathrm m}=-7.6^{\mathrm m}.

Compared to main sequence stars, white dwarfs make a very insignificant contribution to luminosity. We know color indexes of Vega and Altair (see the Table) and the position of these stars in the Hertzsprung-Russell diagram, by interpolation it can be obtained that the stars of the NGC 2019 cluster have a spectral class A2 and absolute magnitude of about $M_{\mathrm{A2}}=+1^{\mathrm m}$ . Thus, the number of stars of the main sequence $N$ in the cluster is related to the difference in absolute stellar magnitudes as

M_{\mathrm{A2}}-M=2.5^{\mathrm m}\times\lg N,\qquad N=10^{8.6/2.5}\approx2800.

Since the percentage of white dwarfs is insignificant, we can assume that the total number of stars is also equal to 2800.

By comparing the cluster sizes in the image with the total image size of $4'$ , we can find that the angular diameter of the central region of the “sphere” is approximately $0.4'$ – $0.5'$ . We may take the value $\delta=0.5'$ for the solution.

The physical diameter of the central part of the cluster is:

D=L\times\delta=50\ \mathrm{kpc}\times0.5'/3438'\approx7.3\ \mathrm{pc}

(where 3438 is the number of arc minutes in radians).

Thus, the density of stars in the cluster is equal to:

\rho=2800/(\tfrac16\pi D^3)\approx14/\mathrm{pc}^3.

Along the line of sight passing through the center of the cluster, the number of stars in the picture plane per 1 square parsec is:

n=7.3\ \mathrm{pc}\times14/\mathrm{pc}^3\approx100/\mathrm{pc}^2.

Therefore, the average distance between the images of stars in the picture plane is:

d=(1/n)^{1/2}\approx0.1\ \mathrm{pc}.

In order to distinguish the white dwarfs with the naked eye, it is necessary that:

Their apparent magnitude of the white dwarfs would be no more than $6^{\mathrm m}$ .
The angular distance between the stars (any stars) would be at least $50''$ .

The first condition gives a distance of no more than $S_1=10\ \mathrm{pc}\times10^{(6-11)/5}\approx1\ \mathrm{pc}$ .

The second condition. The angular distance $50''$ for the length of 0.1 pc turns out at the distance $S_2=206265''/50''\times0.1\ \mathrm{pc}\approx400\ \mathrm{pc}$ .

Thus, to fulfill both conditions, it is necessary to approach the cluster stars by a distance $S\approx1\ \mathrm{pc}$ . Given that this distance is comparable to the size of the cluster (and even less than it), and we need to distinguish any stars, for example, the nearest stars of the cluster, this distance is the distance to the near edge of the cluster. Accordingly, the distance to the center is $X=S+D/2\approx5\ \mathrm{pc}$ .

Basic criteria

(The same criteria for α and β)

Number of stars in cluster – 1.5 pt.
Size of the cluster – 0.5 pt.
Distance to the cluster – 0.5 pt.
The density of stars (3D) – 1 pt.
The number of stars in the picture plane per square parsec (or similar) – 1 pt.
Distance, enough to see the (sunlike for α and white dwarfs for β) stars as $6^{\mathrm m}$ (criterion 1) – 1 pt.
Distance to resolve (separate) two stars (criterion 2) – 1 pt.
The answer using the both criteria – pt.
Using (adding) half of size of the cluster – 0.5 pt.

Additionally:

Wrong number of significant digits in the answer: α – minus 0.5 pt, β – minus 1 pt.

Theoretical round. Sketches for solutions. For jury job ONLY

Note. The given sketches are not full; the team leaders have to give more detailed explanations to students. But the correct solutions in the students’ papers (enough for 8 pts) may be shorter.

Note. Jury members should evaluate the student’s solutions in essence, and not by looking on formal existence the mentioned sentences or formulae. The formal presence of the mentioned positions in the text is not necessary to give the respective points. Points should be done if the following steps de facto using these positions.

Theoretical round. Basic criteria. For work of Jury

Note. The given sketches of solutions are not full; the team leaders have to give more detailed explanations to students. The correct solutions in the students’ papers (enough for 8 pts) may be shorter.

Note. Jury members should elaborate more detailed criteria, and also create criteria for other correct ways of the student’s solutions.

Crossed-out source passage

A red handwritten cross on solution page 2 deletes the following duplicated block between the first estimate $\delta=0.5'$ and “The physical diameter of the central part of the cluster is”. It is recorded here rather than treated as an active step of the α solution:

“The physical diameter of the cluster is:

D=L\times\delta=50\ \text{кПк}\times0.5'/3438'\approx7.3\ \text{Пк}.

M_{\mathrm{A2}}-M=2.5^{\mathrm m}\times\lg N,\qquad N=10^{8.6/2.5}\approx2800.

By comparing the cluster size in the image with the total image size of $4'$ , we can find that the angular diameter of the central region of the ‘sphere’ is approximately $0.4'$ – $0.5'$ . We may take the value $\delta=0.5'$ for the solution.”

Printed-source discrepancies

The original key is retained literally, including its inconsistent numerical steps: the comet-radiation derivation prints $F=2F_p=2(E/c)\pi r^2$ ; the interstellar-comet key repeats $3.2\cdot10^{-5}$ pc/year for different speeds and ends with “only stars $7.2^{\mathrm m}$ and brighter” followed by “No”. The sunset key calls the Chukchi correction “sunrise” and the Eskimo correction “sunset”. The satellite question uses $\zeta$ , while the key uses $\xi$ , and its Fig. 3 is also called Figure 5.3. These are source discrepancies, not silently corrected calculations. The NGC rubric prints “The answer using the both criteria – pt.” without a numerical score; no value is invented.

Оригинал в Архива: theo1.JPG · официални решения: theosol.pdf · допълнителни данни: theo3.JPG · допълнителни данни: hrd.JPG · допълнителни данни: data_2.JPG · допълнителни данни: data_1.JPG · допълнителни данни: theo2.JPG