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IOAA 2022, Теория — Задача 7. Georgia to Georgia

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International Olympiad on Astronomy and Astrophysics 2022 Georgia — Theoretical Competition · 16 август 2022 г. · 20 т.

Условие

Astronomer Keto was flying west overnight along the shortest possible route from Tbilisi (the capital of Georgia) to Atlanta (the capital of the US state of Georgia). She noticed that she is able to observe the star Furud ( $\zeta$ CMa) throughout the entire flight from one of the jet’s windows (although it did touch the horizon at one point, when it also happened to be exactly due South).

Calculate the latitude $\phi_B$ and longitude $\lambda_B$ , of Atlanta where she landed. (20pt)

It is given that:

The journey had a duration of 11 hours 25 minutes, and the jet travelled at an average speed of $875\,\mathrm{km/h}$ .
Furud has a declination of $\delta_F = -30^{\circ}4'$ .
The coordinates of Tbilisi are $\phi_A = 41^{\circ}43'$ N and $\lambda_A = 41^{\circ}48'$ E.
You should ignore the effect of the rotation of the Earth during the flight, the altitude of the jet, atmospheric refraction, and any wind.

Отговори

Решение

Spherical triangle with the pole P at the top and points B (lower left) and A (right) joined by a great-circle arc; sides labelled 90 − φ2 (P to B) and 90 − φ1 (P to A); the angle at P labelled L2 − L1; a meridian from P meets the arc AB at right angles at the point N.

In the diagram, $N$ is the northernmost point of the path $\widehat{\mathrm{AB}}$ . Triangle APB, triangle APN and triangle BPN are all spherical triangles. We shall use the convention that North and West are positive, and South and East are negative. We notice,

\begin{aligned} \widehat{\mathrm{PN}} &= -\delta_F = 30^{\circ}4' \\ \widehat{\mathrm{PA}} &= 90^{\circ} - \phi_A = 48^{\circ}17' \\ \widehat{\mathrm{PB}} &= 90^{\circ} - \phi_B \\ \sphericalangle PNA &= \sphericalangle PNB = 90^{\circ} \\ \sphericalangle BPA &= \lambda_B - \lambda_A && \mathbf{5.0} \\ x &= \widehat{\mathrm{AB}} = vt = 4375 \times 2\frac{17}{60} = 9990\,\mathrm{km} \\ \therefore x &= \frac{9900}{R_{\oplus}} = 1.5663\,\mathrm{rad} \\ &= 89.74^{\circ} = 89^{\circ}44' && \mathbf{2.0} \end{aligned}

In $\triangle APN$ , using sine rule,

\begin{aligned} \frac{\sin\sphericalangle A}{\sin\widehat{\mathrm{PN}}} &= \frac{\sin\sphericalangle N}{\sin\widehat{\mathrm{PA}}} \\ \therefore \sin\sphericalangle A &= \frac{\sin 90^{\circ} \sin(30^{\circ}4')}{\cos\phi_A} = \frac{\sin(30^{\circ}4')}{\cos(41^{\circ}43')} = 0.67119 \\ \therefore \sphericalangle A &= 0.7358\,\mathrm{rad} = 42.16^{\circ} = 42^{\circ}10' && \mathbf{3.0} \end{aligned}

(Recognise that to have a northern point on the path, it must be the acute solution)

In $\triangle PBA$ , using cosine rule,

\begin{aligned} \cos\widehat{\mathrm{PB}} &= \cos\widehat{\mathrm{PA}}\cos\widehat{\mathrm{AB}} + \sin\widehat{\mathrm{PA}}\sin\widehat{\mathrm{AB}}\cos\sphericalangle A \\ \sin\phi_B &= \sin\phi_A \cos x + \cos\phi_A \sin x \cos A \\ &= \sin(41^{\circ}43')\cos(89^{\circ}44') + \cos(41^{\circ}43')\sin(89^{\circ}44')\cos(42^{\circ}10') \\ \sin\phi_B &= 0.5563 \\ \therefore \phi_B &= 0.5900\,\mathrm{rad} = 33.80^{\circ} = 33^{\circ}48' && \mathbf{4.0} \end{aligned}

(Only one solution in the valid values of latitude)

Again using cosine rule,

\begin{aligned} \cos\widehat{\mathrm{AB}} &= \cos\widehat{\mathrm{PA}}\cos\widehat{\mathrm{PB}} + \sin\widehat{\mathrm{PA}}\sin\widehat{\mathrm{PB}}\cos\sphericalangle P \\ \therefore \cos\sphericalangle P &= \frac{\cos x - \sin\phi_A \sin\phi_B}{\cos\phi_A \cos\phi_B} \\ &= \frac{\cos(89^{\circ}44') - \sin(41^{\circ}43')\sin(33^{\circ}48')}{\cos(41^{\circ}43')\cos(33^{\circ}48')} \\ \cos\sphericalangle P &= -0.5891 \\ \therefore \sphericalangle P &= 2.201\,\mathrm{rad} = 126.13^{\circ} = 126^{\circ}8' \\ \lambda_B &= \sphericalangle P + \lambda_A = 126^{\circ}8' - 41^{\circ}48' \\ \lambda_B &= 1.472\,\mathrm{rad} = 84.33^{\circ} = 84^{\circ}20' && \mathbf{5.0} \end{aligned}

(The other solution of $\lambda_B - \lambda_A = -126.13^{\circ}$ gives $\lambda_B = -167.93^{\circ}$ which is clearly not valid)

Therefore, the coordinates of Atlanta are

(33^{\circ}48'\mathrm{N}, 84^{\circ}20'\mathrm{W}) \qquad \mathbf{1.0}

Alternative method to find $L_2$

Using the spherical sine rule in triangle PBA,

\frac{\sin A}{\sin(90^{\circ} - \phi_2)} = \frac{\sin(L_2 - L_1)}{\sin x}

\therefore \sin(L_2 - L_1) = \frac{\sin x \sin A}{\cos\phi_2}

\therefore \sin(L_2 - L_1) = \frac{\sin(89^{\circ}51')\sin(42^{\circ}10')}{\cos(33^{\circ}43')} = 0.80689

\begin{aligned} \therefore L_2 - L_1 &= 126.21^{\circ} \quad (= 126^{\circ}12' = 2.20\mathrm{rad}) \\ \therefore L_2 &= 126^{\circ}12' + L_1 = 126^{\circ}12' + (-41^{\circ}48') = 84.41^{\circ} \quad (= 84^{\circ}24' = 1.47\mathrm{rad}) \end{aligned}

(The other solution of $L_2 - L_1 = 53.79^{\circ}$ gives $L_2 = 11.99^{\circ}$ which is clearly not valid)

Accept all other valid solutions (such as the four parts rule or using Napier’s rules) that give $\phi_2 = 33.71^{\circ}$ and $L_2 = 84.41^{\circ}$ . Students may do both methods for calculating $L_2$ to confirm which solution is common to both as a quicker check than trying to do the reverse calculation with a given pair of co-ordinates.

Оригинал в Архива: Theory Problems.pdf · официални решения: Theory Solutions.pdf