Съдържание

IAO 2006, Теория, Група α — Задача 4. Galaxy.

Оригинал на условието (изображение)Съобщи за грешка

Всички материали от IAO 2006

Theoretical round. Problems to solve · 10 ноември 2006 г.

Бележка

Maybe not all problems have correct questions. Some questions (maybe the main question of the problem, maybe one of the subquestions) may have no real sense. In this case you have to write in your answer (in English or Russian): «situation is impossible – ситуация невозможна». Of course, this answer has to be explained numerically or logically.

Source annotations

Printed source inconsistencies are retained: the English telescope key repeats Δx_m on the right of its displacement identity; the galaxy key uses both1.89×10⁻³ and1.86×10⁻³rad; the alpha eccentricity ratio has a sign inconsistency before the later positive0.07 result. The English polar key prints71°06′N/72°48′S; the Russian copy has handwritten70°06′N/71°48′S. No scientific correction has been silently substituted.

Условие

Figure 4.1. shows the distribution of hydrogen gas in the disk of the spiral galaxy NGC 1325 as a contour map. The contour image is superimposed on an optical image of the galaxy. Assume the disk of the galaxy to be nearly circular. Find the angular diameter of the galaxy. (As the angular size of the galaxy is small to an observer on the Earth, you can use plane geometry as an approximation.) Estimate the diameter and distance to the galaxy in kpc. Find the angle of inclination of the galaxy plane with respect to the plane of the sky.

Решение

On the horizontal axis scale,

20\ \mathrm s=300''\text{ in R.A.}=300''\times\cos(21^\circ33')=279''\sim43.5\ \mathrm{mm}.

On the vertical axis scale,

5'=300''\sim46.7\ \mathrm{mm}.

Drawing the right angle triangle with the galaxy being hypotenuse, two sides are $47\ \mathrm{mm}\sim301''$ and $37\ \mathrm{mm}\sim238''$ .

Thus, by Pythagoras theorem, the angular diameter of the galaxy is $384''\sim6'.5$ .

From the length scale drawn on the right hand top corner, $5\ \mathrm{kpc}\sim7\ \mathrm{mm}\sim45''$ .

Physical diameter of the galaxy $=5\times384/45=43$ kpc.

384''=1.89\cdot10^{-3}\ \mathrm{rad}.

Distance to the galaxy $=43\ \mathrm{kpc}/1.86\cdot10^{-3}=23$ Mpc.

We may find this distance also directly from the scale $5\ \mathrm{kpc}=45''$ , so the distance to the galaxy $=206265\times5\ \mathrm{kpc}/45=23$ Mpc.

Using a simple metre scale, in the first figure, the axes of the ellipse can be measured: $b=20$ mm, $a=60$ mm.

Thus, $i=\cos^{-1}(b/a)=\cos^{-1}(20/60)=\cos^{-1}(0.33)\approx70^\circ$ .

Note: From literature, $i=71^\circ$ .

Note for Jury members

Note for Jury members, problems4 and5. Due to difference in printers the scales of pictures were different in different copies of pictures. So the values of the lengths may be different from those written here. But the ratios, which are necessary for solutions, of the lengths should be the same.

Оригинал в Архива: theo1.jpg · официални решения: theosol1.jpg · допълнителни данни: theo2.jpg · допълнителни данни: theodata1.pdf · допълнителни данни: theodata2.jpg · допълнителни данни: constants.jpg · допълнителни данни: theosol2.jpg · допълнителни данни: theosol3.jpg · допълнителни данни: theosol4.jpg