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IOAA 2021, Теория — Задача 8. IOAA Logo (15 points).

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Всички материали от IOAA 2021

Theory — Q8 — IOAA Logo (15 points). — English (Official) · 15 т.

Solar System

Object	Mean radius (km)	Mass (kg)	Semi-major axis (au)	Eccentricity
Earth	$6\ 378$	$5.972\times10^{24}$	$1.000000$	$0.016\ 710$

Условие

The IOAA2021 logo is formed by the acronym IOAA, where the first letter is represented by the silhouette of the building of the National Astronomical Observatory (OAN) of Colombia, the oldest observatory in America. This observatory is located in Bogota, where it was founded in 1803. The capital city of Colombia is bordered by two famous hills, Monserrate and its neighbor Guadalupe, which are icons of Bogota's cityscape that decorate the logo's background.

Night aerial/satellite view of Bogota: the city grid in the foreground, dark green hills behind with markers 3 (Monserrate, left) and 2 (Guadalupe, right), the Moon between them in the sky, and marker 1 (Observatorio Astronomico) in the city. — Aerial view of Bogota City. Numbers show locations for the quoted places: 1 is for OAN; 2 is for Guadalupe; and 3 is for Monserrate.

Point	Latitude	Longitude	Elevation (m.a.s.l)
1	$4^\circ\,35'\,53''\ N$	$74^\circ\,04'\,37''\ W$	2607
2	$4^\circ\,35'\,30''\ N$	$74^\circ\,03'\,15''\ W$	3296
3	$4^\circ\,36'\,18''\ N$	$74^\circ\,03'\,19''\ W$	3100

8.1 Estimate the distance (in km), between points 2 (Guadalupe) and 3 (Monserrate). [3 т.]

8.2 Estimate the angular separation (in degrees) between Guadalupe (2) and Monserrate (3) as observed from the National Astronomical Observatory of Colombia (1). [6 т.]

8.3 From the OAN, on September 21 at 8:00 p.m. the Moon was observed towards the eastern hills (between Monserrate and Guadalupe). The measured ecliptic coordinates (longitude and latitude) of the Moon are shown in the table. Determine the equatorial coordinates of the Moon at the time of observation.

Photograph of the Moon low above the eastern hills at dusk, with the white sanctuary buildings of Monserrate on the hilltop.

Panoramic night-sky photograph over the horizon labeled N, E, S from left to right, with the Moon near the eastern horizon and bright stars scattered above.

Local Time: 8:00 p.m.
$Az:\,+90^\circ42'59''\ /\ Alt:\,+19^\circ01'42''$
$\lambda:\,+12^\circ20'16''\ /\ \beta:\,-04^\circ24'14''$

Note: Azimuth measured from North to East. [6 т.]

Отговори

Решение

8.1

Assuming $R=6378\ \mathrm{km}$ (as per the table of constants)

\Delta x=R\cdot[(4^\circ36'18'')-(4^\circ35'30'')]\cdot\frac{\pi}{180}=1.4842\ \mathrm{km}

\Delta y=R\cdot[(74^\circ3'19'')-(74^\circ3'15'')]\cdot\frac{\pi\cdot\cos(4{,}35{,}54)}{180}=0.1237\ \mathrm{km}

\Delta z=(3.100-3.296)=-0.196\ \mathrm{km}

[2 points]

d_{2-3}=\sqrt{(x^2+y^2+z^2)}=1.502\ \mathrm{km}

[1 point]

8.2

Estimate the angular separation (in degrees) between Guadalupe (2) and Monserrate (3) as observed from the National Astronomical Observatory of Colombia (1). Take point 1 as the vertex.

Using the same method as part 8.1:

d_{1-2}=2.722\ \mathrm{km}

[2 points]

d_{1-3}=2.580\ \mathrm{km}

[2 points]

Using Cosine rule for spherical triangle,

\cos(A)=\frac{(d_{1-2}^2+d_{1-3}^2-d_{2-3}^2)}{2d_{1-2}d_{1-2}}=0.841

A=32.78^\circ

[2 points]

8.3

\alpha=\tan^{-1}\left(\frac{-\sin(\beta)\cdot\sin(\varepsilon)+\cos(\beta)\cdot\cos(\varepsilon)\cdot\sin(\lambda)}{\cos(\beta)\cdot\cos(\lambda)}\right)

\alpha=12^\circ43'3''=0h\ 50m\ 52s

[3 points]

\delta=\sin^{-1}(\sin(\beta)\cdot\cos(\varepsilon)+\cos(\beta)\cdot\sin(\varepsilon)\cdot\sin(\lambda))

\delta=1^\circ33'43''

[3 points]

Оригинал в Архива: TQ-8-Q.pdf · официални решения: TQ8.pdf · допълнителни данни: TQ-8-A.pdf · допълнителни данни: Standard-Table-of-constants.docx.pdf · допълнителни данни: TQ-G.pdf · допълнителни данни: data.pdf