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THEORETICAL TEST Long problems page 1 from 31 Content Indications 2 16. LONG PROBLEM 1. EAGLES ON THE CARAIMAN CROSS ! 3 17. LONG PROBLEM 2. COSMIC PENDULUM 14 18. LONG PROBLEM 3. FROM ROMANIA …. TO ANTIPOD! … A BALLISTIC MESSENGER 22 THEORETICAL TEST Long problems page 2 from 31 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 15 for short problems, 16 to 19 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 leters 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 ! THEORETICAL TEST Long problems page 3 from 31 16. Long problem 1. Eagles on the Caraiman Cross ! In the Bucegi mountains, part of the Carpathian mountains, after the end of the First World War an iron cross was built by the former King of Romania called Ferdinand the I-st and his wife Queen Maria. The cross is an unique monument in Europe. The monument is an impressive iron cross called „The Heros’ Cross” which in 2013 entered in the Guiness Book as the cross build on the highest altitude mountain peek. The cross was built on the plane plateau situated on the top of the peek called Caraiman, at the altitude H 2300 m from sea level. Its height, including the base-support is h 39,3 m . The horizontal arms of the cross are oriented on the N-S direction. The latitude of the Cross is 45. A. In the evening of 21st of March 2014, the summer equinox day, two eagles stop from their flight, first near the monument, and the second, on the top of the Cross as seen in figure 1. The two eagles are on the same N S h Figure 1 vertical direction. The sky was very clear, so the eagles could see the horizon and observe the Sun set. Each eagle start to fly right in the moment that each of it observes that the Sun completely disappears. In the same time, an astronomer located at the sea-level, at the base of the Bucegi Mountains. Assume that he is on the same vertical with the two eagles. Assuming negligible the atmospheric refraction, solve the following questions: 1) Calculate the duration of the sunset, measured by the astronomer. THEORETICAL TEST Long problems page 4 from 31 2) Calculate the durations of sunset measured by each of the two eagles and indicate which of the eagles leaves first the Cross. What is the time interval between the leaving moments of the two eagles. The following information is necessary: The duration of the sunset measurement starts when the solar disc is tangent to the horizon line and stops when the solar disc completely disappears. 5 The Earth’s rotation period is TE 24 h , the radius of the Sun RS 6,96 10 km , Earth – Sun distance d E S 15 10 7 km , the local latitude of the Heroes Cross 45 . B) At a certain moment of the next day, 22nd March 2014, the two eagles come back to the Heroes Cross. One of the eagles lands on the top of the vertical pillar of the Cross and the other one land on the horizontal plateau, just in the end point of the shadow of the vertical pillar of the Cross. 1) Calculate the distance between the two eagles, if this distance has the minimum possible value. 2) Calculate the length of the horizontal arms of the Cross l b , if the shadow on the plateau of one of the arm of the cross has the length ub 7 m C) At midnight, the astronomer visit the cross and, from the top of it, he identifies a bright star at the limit of the circumpolarity. He named this star „Eagles Star”. Knowing that due to the atmospheric refraction the horizon lowering is 34' , calculate: 1) The “Eagles star” declination; 2) The “Eagles star” maximum height above the horizon. Long problem 1. Marking scheme - Eagles on the Caraiman Cross 1) 2) B1) B2) C1) C2) 10 10 10 10 5 5 A. 1 The following notations are used: DS the diameter of the Sun, d E S Earth-Sun distance, angular diameter of the Sun as seen from the Earth: THEORETICAL TEST Long problems page 5 from 31 Fig. 1 According the fig. 1 the angular diameter of the Sun can be calculated as follows sin 2 RS ; d PS 2 2 RS DS 2 6,96 105 km 0,00928 rad. d PS d PS 15 10 7 km The figure 2 presents the Sun’s evolution during sunset as seen by the astronomer. In an equinox day the Sun moves retrograde along the celestial equator. There are marked the following 3 positions of the Sun: Tdos - The solar disc is tangent to the equatorial plane above the standard horizon – the start of the sunset; Sdos - The center of the solar disc on the celestial equator in the moment of the sunset starts; Tsos - The solar disc is tangent to the equatorial plane bellow the standard horizon – the end of the sunset S s os - The center of the solar disc on the celestial equator in the moment of the sunset ends; Earth . During this time the center of the Sun moves along the equator from Sdos to Ssos . The vector-radius of the Sun rotates in equatorial plane with angle and in vertical plane with angle . i.e. The duration of the sunset is the angular diameter of the Sun as seen from the Earth. THEORETICAL TEST Long problems page 6 from 31 time written: Considering that the Sun travels the distance 2x along the equatorial path with merely constant i.e. during and that the spherical right triangle SdosTdosV can be considered a plane one the following relations can be 2 RS D R RS ; x S ; 2x S ; x sin sin sin DS DS d PS TP 2x 2x sin ; 2 v d PS 2 d 2 sin sin PS TP TP sin sin sin 90 cos; TP ; 2 cos 0,00928 rad 24 h 0,22272 60 min 3 min . 2 3,14 rad cos 45 21' 2 3,14 0,707 2) If the atmospheric refraction is negligible the eagle on the top of the cross V 1 on figure 3 is on the same latitude , as the astronomer but at the altitude H. Thus from the point of view of the V1 the horizon line is below the standard horizon line with an angle 1 , Standard horizon standard Lowered horizon Fig. 3 cos 1 R ; RH THEORETICAL TEST Long problems page 7 from 31 sin 1 R H 2 R 2 2 RH H 2 2 RH 2H 1 ; RH RH R R 2 2,3 km `1 0,02685 rad 1,54. 6380 km For the observer V1 the Sun will go below the lowered horizon after moving down under the standard horizon with angle 1 and moving along the equator with an angle 1 , as seen in fig. 4 Fig. In the right spherical triangle ABV by using the sinus formula : sin 90 sin 90 ; sin 1 sin 1 cos 1 1 ; 1 ; 1 1 cos 1 1 1 2 1 ; TP 1 TP 1,54 24 60 min 8,71 minute, cos 2 cos 45 21' 360 THEORETICAL TEST Long problems page 8 from 31 Which represents the delay of the start of the sunset from the point of view of V 1 regardless to the astronomer due to the V1 observer’s altitude. The altitude effect on the total time of sunset can be calculated by using the fig. 5 Equatorial plane Lowered horizon’s plane Pământul Fig.5 2 RS DS RS RS ; ; 2y ;y sin 1 sin 1 sin 1 y DS DS d PS sin 1 TP 2y 2y 1 ; 2 2 v d PS 2 sin 1 d PS sin 1 TP TP sin 1 90 ; sin 1 sin 90 1 sin 90 1 cos 1 ; TP 1 ; 2 cos 1 THEORETICAL TEST Long problems page 9 from 31 1 0,00928 rad 24 h 0,22272 60 min 2,9350 min, 2 3,14 0,725 2 3,14 rad cos 45 1,54 Which represents the total duration of sunset for V1 at altitude H. Similarly for eagle V2 at the same latitude , but altitude H h (the top of the cross), the lowering effect on the horizon is measured by angle 2 thus cos 2 sin 2 R H h2 R 2 R ; RH h 2 R H h H h 2 R H h 2H h 2 ; RH h RH h R R 2 2,3 0,0393 km `2 0,02707 rad 1,55 ; 6380 km sin 90 sin 90 ; sin 2 sin 2 2 cos 1 2 ; 2 ; 2 2 cos 2 2 2 2 ; TP 2 TP 1,55 24 60 min 8,77 minute, cos 2 cos 45 21' 360 Which represents the delay of the start moment of the sunset for V2 due to the altitude H h . 2 Similar the total duration of the sunset for the observer V2 : 2 2 TP ; 2 cos 2 0,00928 rad 24 h 0,22272 60 min 2,9309 min, 2 3,14 rad cos 45 1,55 2 3,14 0,726 We may note the following: - the horizon- lowering is increased by the increase of the altitude; H H h 1 2 ; H . - the delay of the moment of sunset start is increased by the increase of the altitude: H H h 1 2 ; H . - the total duration of sunset is reduced by the increase of the altitude: 0 H H h 1 2 ; H . Conclusions: If we consider t 0 the moment of sunset star for the astronomer - for V1 the sunset starts at t 0 8,71 min and ends at t0 8,71 min 2,9350 min t0 11,6450 min - for V2 the sunset starts at t 0 8,77 min and ends at t0 8,77 min 2,9309 min t0 11,7009 min - Thus eagle from the plateau leaves first the cross; THEORETICAL TEST Long problems page 10 from 31 - The time between the leaving moments is: t t0 11,7009 min t0 11,6450 min 0,0559 min 3,354 s. b) As seen in fig. 6 the length of the cross on the plateau will be minimum when the Sun passes the local meridian, i.e. the height of the Sun above the horizon will be maximum: h max 90 . Thus the shadow of the horizontal arms of the cross is superposed on the shadow of the vertical pillow. Local Meridian locului Fig. 6 In this conditions : h ; 90 ; d h h h h 39,3 m 39,3 d m 55,58 m; sin sin sin 90 cos cos 45 0,707 sin The distance between the two eagles is umin h cot h cot h cot 90 h tan 39,3 m. 2) In the above mentioned conditions the shadow of the arm oriented toward South is on the vertical pillow of the cross, as seen in fig. 7: THEORETICAL TEST Long problems page 11 from 31 Fig. 7 tan lb ; lb u b tan 7 m tan 45 7 m, ub Which represents the length of the cross arm. C) 1) For an observer situated in the center O of the celestial topocentric sphere, at latitude , at sea level, al the stars are circumpolar ones see fig. 8. Their diurnal parallels, parallel with the equatorial parallel, are above the real local horizon N 0S0 . The star 0 is at the circumpolar limit because its parallel touches the real local horizon in point N 0 but still remains above it. Thus 0 is a marginal circumpolar star. Without taking into account the atmospheric refraction: From the isosceles triangle O 0 N 0 results the 0 declination: 0,min 90 min 180 ; 0,min 90 . THEORETICAL TEST Long problems page 12 from 31 Fig. 8 By taking into account the atmospheric refraction the horizon line changes to line N'S' , with angle ' 34' , as seen in fig. 9. The star 0 remains a circumpolar one but above the limit. In this conditions the ' star ' gather the limit conditions its declination been min min . In this conditions for an observer situated in the center of the topocentric celestial sphere, at latitude and altitude zero, the star ' , with declination ' min 0 min is on the limit of the circumpolarity. From the isosceles triangle N' O ' the declination ' will be : Fig. 9 ' ' 2 min min 90 ' 180 ; ' min 90 '. THEORETICAL TEST Long problems page 13 from 31 For the observer at latitude , but at height h, taking into account the effect of lowering of the horizon the star " will meet the problem requirements see figure 10. The new horizon is N"S" and declination of " 0 min . Star 0 remains a circumpolar one but above the limit. " is min From the isosceles triangle N"O " the declination of star " will be: Fig. 10 " " 2 min min 90 " 180 ; " min 90 ". By taking into account the refraction effect and the altitude effect, from triangle NO in figure 11, the declination will be Fig. 11 2 min min 90 180 ; ' " ; ' 34': " 2 1,55 ; THEORETICAL TEST Long problems page 14 from 31 min 90 ' " 90 ' 2 ; min 90 45 0,56 1,55 42,9. 2) The maximum height above the horizon will be hmax 90 min 90 42,9 45 87,9 . 17. Long problem 2. Cosmic Pendulum A space shuttle (N) orbits the Earth in the equatorial plane on a circular trajectory with radius r. From the spaceship. The shuttle has an arm designated to place satellites on the orbit. The arm is a metallic rod (negligible mass) with length l r . The arm is connected to the shuttle with frictionless mobile articulation. A satellite S is attached to the arm and let out from the shuttle. At a certain moment the angle between the rod and the shuttle’s orbit radius is , see figure 1. You know the mass of the Earth –M , and the gravitational constant k. Space shuttle (N) Satelite (S) Earth Ecuatorial plane Fig.2 a) Find out the values of angle for which the configuration of the system shuttle – rod –satellite remains unchanged regardless to Earth (the system is in equilibrium), during orbiting the Earth. For each found value of angle , specify the type of the system equilibrium i.e. stable or unstable. THEORETICAL TEST Long problems page 15 from 31 You will take in to account the following assumptions: the initial orbit of the shuttle is not affected by the presence of the satellite S, all the external friction-type interactions are negligible, the satellite – shuttle gravitational interaction is negligible too. The following data are known: , m1 - the mass of the shuttle, the mass of the satellite m2 m1 . b) In the moment of one stable equilibrium configuration, the rod with the satellite attached is slightly rotated with a very small angle in the orbital plane and then released. Demonstrate that the small oscillations of the satellite S relative to the shuttle are harmonically ones. Express the period T0 of this cosmic pendulum as a function of the orbiting period T of the shuttle around the Earth. It is known the linear harmonic oscillator equation: 2 d2 , 02 0; 0 2 T0 dt Where : the instantaneous angular deviation; T0 the period of the linear harmonic oscillator. c) If we consider that the mass of the satellite S , m2 is not negligible by comparison with the shuttle’s one m1 in the conditions from point a) the evolution on the orbit of the shuttle would be influenced by the presence of the satellite S rigid attached to the shuttle by the rod. Identify and determine the consequences on the shuttle’s movement after one complete rotation around the Earth. d) Propose a special technical maneuver which can cancels the influence of the non negligible mass satellite S on the shuttles movement. a) b) c) d) Long problem 2. Marking scheme - Cosmic Pendulum a) In figure 1 are represented the forces in the system. F1 the gravitational attraction force acting on the shuttle due to the Earth; F2 the gravitational attraction force acting on the satelite due to the Earth; F the tension force in the suspention rod. 15 15 10 10 THEORETICAL TEST Long problems page 16 from 31 For a value of 0 the movement equations are: For the shuttle on circular orbit around the Earth: m1 2 r K m1 M F cos ; r2 For the satelite on circular orbit around the Earth: m2 2 r l cos K m2 M F . 2 r l cos cos Spre centrul Pământului Fig.1 If the gravitational interaction between the shuttle and the satellite is negligible results: m1 M ; r2 KM 4 2 mM m1 2 r K 1 2 ; 2 3 2 ; r r T F cos K T 2 r3 , KM The revolution time of the shuttle around the Earth: m2 KM r l cos K m2 M 2 F ; 3 r r l cos cos THEORETICAL TEST Long problems page 17 from 31 m2 KM r l cos 3 Km2 M F r l cos 2 ; 3 cos r 3 2 KM l F 2 l m2 3 r 3 1 cos Km2 M r 1 cos ; cos r r r 3 2 F 2 l l m2 KM 1 cos Km2 M r 1 cos ; cos r r l F 2 l m2 KM 1 3 cos Km2 M r 1 2 cos ; r cos r l F 2 3Km2 M cos r ; r cos l F 3Km2 M 3 cos 2 , r The tension force in the rod. The satellite movement regardless to the shuttle is non uniform circular one, described by the equation: m2 at F"; d2 l m2 at m2l m2 2 l F " F tan 3Km2 M 3 sin cos ; dt r 2 d KM 3 3 sin cos 0, 2 dt r The solutions of this equation is t . If during the system evolution the configuration of the system remains the same results: d2 constant; 2 0; dt sin cos 0; sin 0; 1 0; 2 ; (echilibru stabil); cos 0; 3 / 2; 4 3 / 2; (echilibru instabil), As seen in the pictures . THEORETICAL TEST Long problems page 18 from 31 Stabile eq Stabile eq unstabile eq unstabile eq Fig. b) Coresponding to the position with 0, the movement equation of the satellite displaced from equilibrium position with a very small angle : d2 KM 3 3 sin cos 0; 2 dt r sin ; cos 1; d2 KM 3 3 0, 2 dt r Which represents the linear harmonic oscillator; d2 02 0; dt2 Thus the period of the small oscillations will be 02 3 1 r3 T KM 4 2 T 2 , ; 0 3 2 3 KM r T0 3 c) m1M F cos ; r2 m2 M F m2 2 r l cos K ; 2 r l cos cos KM F cos 2 3 ; m1r r m1 2 r K THEORETICAL TEST Long problems page 19 from 31 m2 2 r l cos Km2 M 3 F r l cos 2 ; cos 3 2 F 2 l l m2 r 1 cos Km2 M r 1 cos ; cos r r l F 2 l m2 2 r 3 1 3 cos Km2 M r 1 2 cos ; r cos r KM F cos 3 l F 2 l r 1 3 cos Km2 M m2 3 r 1 2 cos ; m1r r cos r r 2 3 F cos 2 l F 2 l m2 KM r 1 3 cos Km2 M r 1 2 cos ; m1 r cos r l 3Km2 M cos r F ; m2 m2 1 2 2 r rl cos cos 1 cos m1 m1 KM F cos ; m1r r3 m l 3K 2 M 2 cos 2 m1 r KM 2 3 . r m2 m2 1 2 2 r rl cos cos 1 cos m1 m1 2 Observation: If m2 m1 and l r , the results are already found : 2 KM 02 ; 3 r l 3Km2 M cos l r F 3Km2 M 3 cos 2 . r 1 2 r cos Corresponding to the initial circulae orbit with radius r, the total mechanical energy of the system shuttle – Earth is: E0 m1 v 02 mM mM K 1 K 1 . 2 r 2r After one complete rotation, the tangential component of the tension in the rod Ft , acting on the shuttle will determine the change of the radius of the orbit i.e. r r . The total energy of the system will be: m1 v 2 mM m1M K 1 K . 2 r r 2r r m1 M mM mM 1 1 E E E0 K K 1 K 1 ; 2r r 2r 2 r r r m M r r r mM r E K 1 K 1 ; 2 r r r 2 r r r E THEORETICAL TEST Long problems page 20 from 31 r r r; The variation of the mechanical energy after a complete rotation E K Thus m1 M r 0, 2r 2 E Lt 2r Ft ; Ft F sin ; K m1M r 2r Ft 2r 2 The variation of the altitude due to the action of the satellite will be: 4r 3 Ft 4r 3 F sin r , Km1 M Km1 M The potentialenergy v m1 M mM 1 1 K 1 Km1 M ; r r r r r r r r r r Ep Km1M Km1 M ; r r r r r r r r r; m M r Ep K 1 2 0, r Ep K Thus m1M r 2 E 2 2rFt 2 4rFt ; 2r 2 m v2 m v2 Ec 1 1 0 ; 2 2 2 m1 v 0 mM mM K 1 2 ; m1 v 02 K 1 ; r r r 2 m1 v m1 M mM K ; m1 v 2 K 1 ; 2 r r r r r r Ep K m1 M mM mM 1 1 K 1 K 1 ; 2r r 2r 2 r r r mM 1 mM 1 1 1 Ec K 1 K 1 ; 2 r r r 2 r r r mM 1 m M r r r 1 Ec K 1 K 1 ; 2 r r r 2 r r r mM r Ec K 1 ; 2 r r r Ec K THEORETICAL TEST Long problems page 21 from 31 r r r; m M r Ec K 1 2 0, 2r reprezentând variația energiei cinetice a navei, după o rotație completă în jurul Pământului; Ec 0; v v 0 , rezultat care constituie “paradoxul sateliților”, adică, atunci când altitudinea orbitei navei scade, viteza navei crește; E Ec Ep ; Ec E Ep 2rFt 4rF 2rFt 0; m1 v 2 m1 v 02 m1 2 2 m v v 0 1 v v 0 v v 0 ; 2 2 2 2 v v 0 v; v v 0 v; v v 0 2 v 0 v 2 v 0 ; m Ec 1 2 v 0 v m1 v 0 v 2rFt ; 2 2rFt M v ; v0 K ; m1 v 0 r Ec v 2rFt r . m1 KM d) Manevra tehnică propusă este reprezentată în figura alăturată: o forță reactivă, egală în modul și de sens contrar cu tensiunea din tijă: Fr F , astfel încât forța rezultantă care acționează asupra navei să rămână forța de atracție gravitațională din partea Pământului: FNava F1 F Fr F1. THEORETICAL TEST Long problems page 22 from 31 Fig. 18. Long problem 3. From Romania …. to Antipod! … a ballistic messenger The 8th IOAA organizers plan to send to the antipode the official flag using a ballistic projectile. The projectile will be launched from Romania, and the rotation of the Erath will be neglected. a) Calculate the coordinates of the target-point if the launch-point coordinates are: Romania 44 North; Romania 30 Est . b) Determine the elements of the launching-speed vector, regardless to the center of the Earth, 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. e) Calculate the flying –time of the projectile, from the launch-moment to the impact one. You will use the value of the gravitational acceleration at Erath surface g 0 9,81 ms 2 ; the Earth radius R 6370 km. f) Evaluate the possibility that the projectile to be seen with the naked eye in the moment that it passes at the maximum distance from the Earth. You will use the following values: The Moon albedo M 0,12; The Moon radius RM 1738 Km ; the Earth –Moon distance rEL 384400 km; the apparent magnitude of the full THEORETICAL TEST Long problems page 23 from 31 moon mM 12,7 m. You assume that the projectile is perfectly metallic sphere with radius rprojectile 400 x 10 3 m and with perfectly reflective surface. THEORETICAL TEST Long problems page 24 from 31 Long problem 3. Marking scheme - From Romania …. to Antipod! … a ballistic messenger a) b) c) d) e) f) 10 10 10 10 10 10 a) The two places are represented in the figure. Greenwich Ecuatorul Fig. Y Y,Sud X, Nord X ; Y,Vest X, Eest 180 ; Y X 180. Romania 43 Nord; Romania 30 Est, The landing point is Antipod 43 Sud; Antipod 150 Vest, Somewhere South EEst from Tasmania (South from Australia). THEORETICAL TEST Long problems page 25 from 31 b) The schetch of the trajectory THEORETICAL TEST Long problems page 26 from 31 Bysector of Tangent to the trajectory Romania Horison at launch point Apogeu Perigeu Antipod 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: F2 B 2 F1B; F1F2 3 F1B. Rezultă: F1F2 ; F1F2 R tan 2 ; R FB tan 1 ; F1B R tan ; R R tan 2 F1B R tan ; tan 2 3 tan ; sin 2 sin 3 ; cos 2 cos 2 sin cos sin 3 ; cos 2 cos 2 cos 2 3 cos 2 ; 2 cos 2 3 cos 2 sin 2 ; 1 3sin 2 cos 2 ; tan 2 ; 3 3 tan ; 30 ; 2 60 ; 90 2 30 ; 2 60 ; 3 RF2 A triunghi echilateral; tan 2 RF2 AF2 RA 2R; RF2 RF1 2a 3R; THEORETICAL TEST Long problems page 27 from 31 a 3 R; 2 M 2 1 v 0 KM ; r R; g 0 K 2 ; R r a v0 g R KM 2 2 2 R 2 0 . 2 R 3 R 3R c) v Antipod v 0 . d) R R 3 tan 2 tan 60 R; 2 2 2 3 b a2 c2 R; 2 1 2a 2rmin 2c; rmin a c 3 3 R; 2 1 rmax 2a rmin 3 3 R; 2 F1F2 R tan 2 2c; c 2 1 1 v min KM ; rmax 3 3 R; 2 rmax a v min 2g0 R 3 3 KM 2 4 2 R . 2 3 3 3 R 3 3 R 3R e) Accordin to Kepplers laws: R dS constant; dt S S0 2 0 2S1 2 S1 S0 S0 S S 4S1 4 2 ; ; ; 0 0 T 2 t T t T t THEORETICAL TEST Long problems page 28 from 31 ab b2 b b2 S 0 ab; S1 1 2 arcsin 1 2 ; 2 a a a 1 S 4S1 S t 0 T 2 1 T ; 2S 0 S0 2 a3 2 ;T KM R T 2 a3 ; g0 2S1 1 b b2 b2 1 2 arcsin 1 2 ; S 0 a a a 2S 1b b2 1 2 e; 1 e arcsin e ; a S0 a 1 eb arcsin e t T . 2 a f) The integral luminosity of Sun: LS Eemis,Soare t 3,86 10 26 W, THEORETICAL TEST Long problems page 29 from 31 Dacă For a circumsolar surface with radius rPS , see picture bellow the solar radiation enegy passing through the surface in one second is LS . Sfera circumsolară a cărei rază este egală cu distanța de la Soare la Pământ (Lună) Density of solar flux E emis, Soare Soare, r PS E emis, Soare St t S LS L S 2 constant . S 4rPS Fincident,FullMoon S un,rPS RL2 . Dacă L este albedoul Lunii, rezultă: THEORETICAL TEST Long problems page 30 from 31 L Freflectat, FullMoon Fincident,FullMoon , unde Freflectat, Luna Plina fluxul energetic al radiațiilor reflectate de Luna Plină spre observatorul de pe Pământ; Freflectat, FullMoon L Fincident,FullMoon L Soa„ re,rPS RL2 . În consecință, densitatea fluxului energetic ajuns la observator, după reflexia pe suprafața Lunii, este: moon, observator Freflectat, FullMoon 2 2rPL L Soare, rPS RL2 . 2 2rPL Symilarly proiectil, observator Freflectat, proiectil proiectil Soare,rPS 4rD,2 proiectil 2 Rproiectil . 4rD,2 proiectil 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 rP,proiectil. Utilizând formula lui Pogson, vom compara magnitudinea aparentă vizuală a Lunii Pline cu magnitudinea aparentă vizuală a proiectilului balistic: log log Luna,observator 0,4mLuna Plina mproiectil ; proiectil,observator Luna,observator log proiectil,observator RL2 RL2 L 2 2 rPL 2rPL log ; 2 2 Rproiectil Rproiectil proiectil 2 4rD,2 proiectil 2rD,proiectil L Soare,r PS proiectil Soare,r PS L log proiectil RL2 2 rPL 0,4mL mproiectil ; 2 Rproiectil 2rD,2 proiectil 2 RL r 2 D,proiectil 0,4mL mproiectil ; log r proiectil Rproiectil PL L 0,12; proiectil 1; L 2 RL 1738 km; Rproiectil 400 mm; rD,proiectil rmax, observatorproiectil hmax rmax R; rmax hmax 1 3 3 R; 2 1 1 3 3 R R 1 3 R 8700 km; 2 2 THEORETICAL TEST Long problems page 31 from 31 rPL robservator,Luna 384400 km; mL 12,7 m ; log L proiectil 2 log log 0,12 2 log RL Rproiectil log 2 2 log rD proiectil rPL 0,4mL mproiectil ; 1738000 m 8700 km log 2 2 log 0,4mL mproiectil ; 0,400 m 384400 km log 0,12 2 log 1738000 8700 log 2 2 log 0,4mL mproiectil ; 0,400 384400 0,920818754 13,27597956 0,301029995 3,290528253 0,4mL mproiectil ; 23,4m 12,7 m mproiectil; mproiectil 10,7 m ; mmax 6 m ; mproiectil mmax , The projectile wasn’t seen when it was at its apogee