Съдържание

IOAA 2014, Теория, дълги задачи — Задача 14. From Romania …. to Antipod! …with a ballistic messenger

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Всички материали от IOAA 2014

THEORETICAL TEST Long problems

Внимание: Бележка към темата

The final key has blank pages 6 and 14 and broken maps on page 15. Eagles A2 (delay derivation and Fig. 4), C1–C2 final formulas, and Antipod a), marking scheme and maps are recovered from the archived preliminary key (original pages 7–8, 14 and 24–25). Antipod a) in that key uses latitude 43° although both questions give 44°; its printed answer is retained as an alternate-edition answer. Other printed numerical discrepancies are preserved.

Indications

1. The problems were elaborated concerning two aspects:

a. To cover merely all the subjects from the syllabus;

b. The average time for solving the items is about 15 minutes per a short problem;

2. In your folder you will find out the following:

a. Answer sheets

b. Draft sheets

c. The envelope with the subjects in English and the translated version of them in your mother tongue;

3. The solutions of the problems will be written down only on the answer sheets you receive on your desk. PLEASE WRITE ONLY ON THE PRINTED SIDE OF THE PAPER SHEET. DON’T USE THE REVERSE SIDE. The evaluator will not take into account what is written on the reverse of the answer sheet.

4. The draft sheets is for your own use to try calculation, write some numbers etc. BEWARE: These papers are not taken into account in evaluation, at the end of the test they will be collected separately . Everything you consider as part of the solutions have to be written on the answer sheets.

5. Each problem have to be started on a new distinct answer sheet.

6. On each answer sheet please fill in the designated boxes as follows:

a. In PROBLEM NO. box write down only the number of the problem: i.e. 1 to 12 for short problems, 13 to 14 for long problems. Each sheet containing the solutions of a certain problem, should have in the box the number of the problem;

b. In Student ID – fill in your ID you will find on your envelope, consisted of 3 letters and 2 digits.

c. In page no. box you will fill in the number of page, starting from 1. We advise you to fill this boxes after you finish the test

7. We don’t understand your language, but the mathematic language is universal, so use as more relationships as you think that your solution will be better understand by the evaluator. If you want to explain in words we kindly ask you to use short English propositions.

8. Use the pen you find out on the desk. It is advisable to use a pencil for the sketches.

9. At the end of the test:

a. Don’t forget to put in order your papers;

b. Put the answer sheets in the folder 1. Please verify that all the pages contain your ID, correct numbering of the problems and all pages are in the right order and numbered. This is an advantage of ease of understanding your solutions.

c. Verify with the assistant the correct number of answer sheets used fill in this number on the cover of the folder and sign.

d. Put the draft papers in the designated folder, Put the test papers back in the envelope.

e. Go to swim

GOOD LUCK !

Note on marking

ALL THE CORRECT SOLUTIONS WHICH ARE DIFFERENT FROM THE AUTHOR’S SOLUTION WILL BE EVALUATED AND MARKED ACORDINGLY TO THE MARKING SCHEME.

Условие

The 8th IOAA organizers plan to send to the antipode (the point on the Earth’s surface diametrically opposite to the launch position) the official flag using a ballistic projectile. The projectile will be launched from Romania, and the rotation of the Earth will be neglected.

a) Calculate the coordinates of the target-point if the launch-point coordinates are: $\varphi_{\mathrm{Romania}} = 44^{\circ}\ \mathit{North}$ ; $\lambda_{\mathrm{Romania}} = 30^{\circ}\ \mathrm{Est}$ .

b) Determine the magnitude of the velocity and the launch angle, with respect to the horizon at launch site, in order that the projectile should hit the target.

c) Calculate the velocity of the projectile when it hits the target.

d) Calculate the minimum velocity of the projectile on its trajectory.

e) Calculate the flying –time of the projectile, from the launch to the impact. You may use the value of the gravitational acceleration at Earth surface as $g_0 = 9.81\ \mathrm{ms^{-2}}$ ; the Earth radius $R = 6370\ \mathrm{km}$ .

f) Will it be possible that the projectile will be seen by the naked eye when it is at the maximum distance from the Earth. You will use the following values: The Moon albedo $\alpha_M = 0.12$ ; The Moon radius $R_M = 1738\,Km$ ; the Earth –Moon distance $r_{EM} = 385000$ km; the apparent magnitude of the full moon $m_M = -12.7^{\mathrm{m}}$ . You assume that the projectile is perfectly metallic sphere with radius $r_{projectile} = 400\,x\,10^{-3}\,m$ and with perfectly reflective surface.

Отговори

Решение

Внимание: Непълно решение

The final key omits part a). The recovered preliminary key uses latitude 43° although both question editions give 44°; no matching 44° worked answer is supplied.

Archived preliminary marking scheme

Part	Points
a)	10
b)	10
c)	10
d)	10
e)	10
f)	10

a) — archived preliminary key

The two places are represented in the figure.

Earth globe with diametrically opposite launch point X and target Y, latitude and longitude angles, poles and equatorial plane. — Fig.

\varphi_Y=\varphi_{Y,\mathrm{Sud}}=\varphi_{X,\mathrm{Nord}}=\varphi_X;

\lambda_{Y,\mathrm{Vest}}+\lambda_{X,\mathrm{Eest}}=180^\circ;\quad\lambda_Y+\lambda_X=180^\circ.

\varphi_{\mathrm{Romania}}=43^\circ\ \mathrm{Nord};\quad\lambda_{\mathrm{Romania}}=30^\circ\ \mathrm{Est},

The landing point is

\varphi_{\mathrm{Antipod}}=43^\circ\ \mathrm{Sud};\quad\lambda_{\mathrm{Antipod}}=150^\circ\ \mathrm{Vest},

Somewhere South EEst from Tasmania (South from Australia).

Original maps of Romania and the world, with the launch and antipodal landing locations marked.

b)

The schetch of the trajectory

Earth as a shaded circle with centre F1(O), launch point Romania at the top and Antipod at the bottom; a dashed elliptical trajectory with the second focus F2, Apogeu on the left with velocity v_min, Perigeu inside the Earth, points B, C, D; initial velocity v0 at launch angle α to the horizon and at the Antipod; angles α and β, radius R, height h_max; labels Tangent to the trajectory, Horison at launch point, Bysector of F1AF2.

In order to hit the point the trajectory of the missile has to be an elypse with the Earth center in the center of the Earth. Se știe că:

\mathrm{F_2B} = 2\cdot\mathrm{F_1B};\ \mathrm{F_1F_2} = 3\cdot\mathrm{F_1B}.

Rezultă:

\tan 2\alpha = \frac{\mathrm{F_1F_2}}{R};\ \mathrm{F_1F_2} = R\cdot\tan 2\alpha;

\tan\alpha = \frac{\mathrm{F_1B}}{R};\ \mathrm{F_1B} = R\cdot\tan\alpha;

R\cdot\tan 2\alpha = \mathrm{F_1B} = R\cdot\tan\alpha;

\tan 2\alpha = 3\cdot\tan\alpha;

\frac{\sin 2\alpha}{\cos 2\alpha} = 3\frac{\sin\alpha}{\cos\alpha};

\frac{2\sin\alpha\cdot\cos\alpha}{\cos 2\alpha} = 3\frac{\sin\alpha}{\cos\alpha};

2\cos^2\alpha = 3\cos 2\alpha;\ 2\cos^2\alpha = 3(\cos^2\alpha-\sin^2\alpha);

3\sin^2\alpha = \cos^2\alpha;\ \tan^2\alpha = \frac{1}{3};

\tan\alpha = \frac{\sqrt{3}}{3};\ \alpha = 30^{\circ};\ 2\alpha = 60^{\circ};\ \beta = 90^{\circ}-2\alpha = 30^{\circ};\ 2\beta = 60^{\circ};

\Delta(\mathrm{RF_2A}) \to \text{triunghi echilateral};

\mathrm{RF_2} = \mathrm{AF_2} = \mathrm{RA} = 2R;

\mathrm{RF_2} + \mathrm{RF_1} = 2a = 3R;

a = \frac{3}{2}R;

\mathrm{v}_0 = \sqrt{KM\left(\frac{2}{r}-\frac{1}{a}\right)};\ r = R;\ g_0 = K\frac{M}{R^2};

\mathrm{v}_0 = \sqrt{\frac{KM}{R^2}\cdot R^2\left(\frac{2}{R}-\frac{2}{3R}\right)} = 2\sqrt{\frac{g_0R}{3}}.

c)

\mathrm{v}_{\mathrm{Antipod}} = \mathrm{v}_0.

d)

\mathrm{F_1F_2} = R\cdot\tan 2\alpha = 2c;\ c = \frac{R}{2}\cdot\tan 2\alpha = \frac{R}{2}\cdot\tan 60^{\circ} = \frac{\sqrt{3}}{2}R;

b = \sqrt{a^2-c^2} = \sqrt{\frac{3}{2}}R;

2a = 2r_{\min}+2c;\ r_{\min} = a-c = \frac{1}{2}(3-\sqrt{3})R;

r_{\max} = 2a-r_{\min} = \frac{1}{2}(3+\sqrt{3})R;

\mathrm{v}_{\min} = \sqrt{KM\left(\frac{2}{r_{\max}}-\frac{1}{a}\right)};\ r_{\max} = \frac{1}{2}(3+\sqrt{3})R;

\mathrm{v}_{\min} = \sqrt{\frac{KM}{R^2}\cdot R^2\left(\frac{4}{(3+\sqrt{3})R}-\frac{2}{3R}\right)} = \sqrt{\frac{2g_0R}{3}\cdot\frac{3-\sqrt{3}}{3+\sqrt{3}}}.

e)

Accordin to Kepplers laws:

Two shaded ellipses with the foci marked: left, area S0 and period T with launch point R, vertical line to A and a dashed circle around the focus; right, the ellipse divided by dashed axes into S0/4, S0/4 and two areas S1 near the focus, with Δt, and points R and A.

\Omega = \frac{\mathrm{d}S}{\mathrm{d}t} = \text{constant};

\frac{S_0}{T} = \frac{2\dfrac{S_0}{4}+2S_1}{\Delta t};\ \frac{S_0}{T} = \frac{\dfrac{S_0}{2}+2S_1}{\Delta t};\ \frac{S_0}{T} = \frac{S_0+4S_1}{2\cdot\Delta t};

S_0 = \pi ab;\ S_1 = \frac{ab}{2}\left[\sqrt{1-\frac{b^2}{a^2}}\cdot\frac{b}{a}+\arcsin\sqrt{1-\frac{b^2}{a^2}}\right];

\Delta t = \frac{S_0+4S_1}{2S_0}\cdot T = \left(\frac{1}{2}+2\frac{S_1}{S_0}\right)\cdot T;

T = 2\pi\sqrt{\frac{a^3}{KM}};\ T = \frac{2\pi}{R}\sqrt{\frac{a^3}{g_0}};

\frac{2S_1}{S_0} = \frac{1}{\pi}\left(\frac{b}{a}\cdot\sqrt{1-\frac{b^2}{a^2}}+\arcsin\sqrt{1-\frac{b^2}{a^2}}\right);

\sqrt{1-\frac{b^2}{a^2}} = e;\ \frac{2S_1}{S_0} = \frac{1}{\pi}\left(\frac{b}{a}\cdot e+\arcsin e\right);

\Delta t = \left(\frac{1}{2}+\frac{eb}{\pi a}+\frac{\arcsin e}{\pi}\right)\cdot T.

f)

The integral luminosity of Sun:

L_{\mathrm{S}} = \frac{E_{\mathrm{emis,Soare}}}{t} = 3{,}86\cdot 10^{26}\ \mathrm{W},

Dacă For a circumsolar surface $\Sigma$ with radius $r_{\mathrm{PS}}$ , see picture bellow the solar radiation energy passing through the surface in one second is $L_{\mathrm{S}}$ .

Sun (radius R_S) at the centre of a circumsolar sphere of radius r_LS = r_PS, surface Σ; 4πr_PS²; L_S, with radial arrows; an inner surface Σ0; L_S; parallel rays towards the Moon disc πR_L² labelled Luna; caption text in Romanian: Sfera circumsolară a cărei rază este egală cu distanța de la Soare la Pământ (Lună).

Density of solar flux

\phi_{\mathrm{Soare},r_{\mathrm{PS}}} = \frac{E_{\mathrm{emis,Soare}}}{St} = \frac{\dfrac{E_{\mathrm{emis,Soare}}}{t}}{S} = \frac{L_{\mathrm{S}}}{S} = \frac{L_{\mathrm{S}}}{4\pi r_{\mathrm{PS}}^2} = \text{constant}.

F_{\mathrm{incident},FullMoon} = \phi_{\mathrm{S}un,r_{\mathrm{PS}}}\cdot\pi R_{\mathrm{L}}^2.

Dacă $\alpha_{\mathrm{L}}$ este albedoul Lunii, rezultă:

\alpha_{\mathrm{L}} = \frac{F_{\mathrm{reflectat},\,FullMoon}}{F_{\mathrm{incident},FullMoon}},

unde $F_{\mathrm{reflectat,\,Luna\,Plina}}$ – fluxul energetic al radiațiilor reflectate de Luna Plină spre observatorul de pe Pământ;

F_{\mathrm{reflectat},\,FullMoon} = \alpha_{\mathrm{L}}\cdot F_{\mathrm{incident},FullMoon} = \alpha_{\mathrm{L}}\cdot\phi_{\mathrm{Soa,,re},r_{\mathrm{PS}}}\cdot\pi R_{\mathrm{L}}^2.

În consecință, densitatea fluxului energetic ajuns la observator, după reflexia pe suprafața Lunii, este:

\phi_{moon,\ \mathrm{observator}} = \frac{F_{\mathrm{reflectat},\ FullMoon}}{2\pi r_{\mathrm{PL}}^2} = \alpha_{\mathrm{L}}\cdot\phi_{\mathrm{Soare},r_{\mathrm{PS}}}\cdot\frac{\pi R_{\mathrm{L}}^2}{2\pi r_{\mathrm{PL}}^2}.

Symilarly

\phi_{\mathrm{proiectil,\ observator}} = \frac{F_{\mathrm{reflectat,\ proiectil}}}{4\pi r_{\mathrm{D,proiectil}}^2} = \alpha_{\mathrm{proiectil}}\cdot\phi_{\mathrm{Soare},r_{\mathrm{PS}}}\cdot\frac{\pi R_{\mathrm{proiectil}}^2}{4\pi r_{\mathrm{D,proiectil}}^2}.

În expresia anterioară s-a avut în vedere faptul că densitatea fluxului energetic al proiectilului la observator rezultă din distribuirea prin suprafața sferei cu raza $r_{\mathrm{P,proiectil}}$ .

Utilizând formula lui Pogson, vom compara magnitudinea aparentă vizuală a Lunii Pline cu magnitudinea aparentă vizuală a proiectilului balistic:

\log\frac{\phi_{\mathrm{Luna,observator}}}{\phi_{\mathrm{proiectil,observator}}} = -0{,}4(m_{\mathrm{Luna\,Plina}}-m_{\mathrm{proiectil}});

\log\frac{\phi_{\mathrm{Luna,observator}}}{\phi_{\mathrm{proiectil,observator}}} = \log\frac{\alpha_{\mathrm{L}}\cdot\phi_{\mathrm{Soare},r_{\mathrm{PS}}}\cdot\dfrac{\pi R_{\mathrm{L}}^2}{2\pi r_{\mathrm{PL}}^2}}{\alpha_{\mathrm{proiectil}}\cdot\phi_{\mathrm{Soare},r_{\mathrm{PS}}}\cdot\dfrac{\pi R_{\mathrm{proiectil}}^2}{4\pi r_{\mathrm{D,proiectil}}^2}} = \log\frac{\alpha_{\mathrm{L}}\cdot\dfrac{R_{\mathrm{L}}^2}{r_{\mathrm{PL}}^2}}{\alpha_{\mathrm{proiectil}}\cdot\dfrac{R_{\mathrm{proiectil}}^2}{2r_{\mathrm{D,proiectil}}^2}};

\log\frac{\alpha_{\mathrm{L}}\cdot\dfrac{R_{\mathrm{L}}^2}{r_{\mathrm{PL}}^2}}{\alpha_{\mathrm{proiectil}}\cdot\dfrac{R_{\mathrm{proiectil}}^2}{2r_{\mathrm{D,proiectil}}^2}} = -0{,}4(m_{\mathrm{L}}-m_{\mathrm{proiectil}});

\log\frac{\alpha_{\mathrm{L}}}{\alpha_{\mathrm{proiectil}}}\cdot\left(\frac{R_{\mathrm{L}}}{R_{\mathrm{proiectil}}}\right)^2\cdot 2\cdot\left(\frac{r_{\mathrm{D,proiectil}}}{r_{\mathrm{PL}}}\right)^2 = -0{,}4(m_{\mathrm{L}}-m_{\mathrm{proiectil}});

\alpha_{\mathrm{L}} = 0{,}12;\ \alpha_{\mathrm{proiectil}} = 1;

R_{\mathrm{L}} = 1738\,\mathrm{km};\ R_{\mathrm{proiectil}} = 400\,\mathrm{mm};

r_{\mathrm{D,proiectil}} = r_{\text{max,observator-proiectil}} = h_{\max} = r_{\max}-R;\ r_{\max} = \frac{1}{2}(3+\sqrt{3})R;

h_{\max} = \frac{1}{2}(3+\sqrt{3})R-R = \frac{1}{2}(1+\sqrt{3})R \approx 8700\,\mathrm{km};

r_{\mathrm{PL}} = r_{\mathrm{observator,Luna}} = 384400\,\mathrm{km};\ m_{\mathrm{L}} = -12{,}7^{\mathrm{m}};

\log\frac{\alpha_{\mathrm{L}}}{\alpha_{\mathrm{proiectil}}}+2\log\frac{R_{\mathrm{L}}}{R_{\mathrm{proiectil}}}+\log 2+2\log\frac{r_{\text{D-proiectil}}}{r_{\mathrm{PL}}} = -0{,}4(m_{\mathrm{L}}-m_{\mathrm{proiectil}});

\log(0{,}12)+2\log\frac{1738000\,\mathrm{m}}{0{,}400\,\mathrm{m}}+\log 2+2\log\frac{8700\,\mathrm{km}}{384400\,\mathrm{km}} = -0{,}4(m_{\mathrm{L}}-m_{\mathrm{proiectil}});

\log(0{,}12)+2\log\frac{1738000}{0{,}400}+\log 2+2\log\frac{8700}{384400} = -0{,}4(m_{\mathrm{L}}-m_{\mathrm{proiectil}});

-0{,}920818754+13{,}27597956+0{,}301029995-3{,}290528253 = -0{,}4(m_{\mathrm{L}}-m_{\mathrm{proiectil}});

23{,}4^{\mathrm{m}} = 12{,}7^{\mathrm{m}}+m_{\mathrm{proiectil}};

m_{\mathrm{proiectil}} = 10{,}7^{\mathrm{m}};

m_{\max} \approx 6^{\mathrm{m}};\ m_{\mathrm{proiectil}} > m_{\max},

The projectile wasn’t seen when it was at its apogee

Оригинал в Архива: long_problems.pdf · официални решения: long_final_sol.pdf · допълнителни данни: th_long_ver1_sol.pdf