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THEORETICAL TEST
Long problems
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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
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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 !
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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.
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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:
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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.
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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
;
RH
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sin 1 
R  H 2  R 2
2 RH  H 2
2 RH
2H


 1 ;
RH
RH
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


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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 
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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  h2  R 2
R
;
RH h
2 R  H  h   H  h 
2 R H  h 
2H  h


  2 ;
RH h
RH 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;
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- 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:
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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  .
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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      '.
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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 ;
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 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.
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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
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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
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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  m2l  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
2r  r 
m1 M
mM
mM 1
1
E  E  E0   K
 K 1  K 1 
 ;
2r  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  2r  Ft ; Ft  F sin  ;
K
m1M  r
 2r  Ft
2r 2
The variation of the altitude due to the action of the satellite will be:
4r 3 Ft 4r 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  2rFt  2  4rFt ;
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 
 ;
2r  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  2rFt   4rF   2rFt  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  2rFt ;
2
2rFt
M
v 
; v0  K ;
m1 v 0
r
Ec 
v 
2rFt
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
1b
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 4rPS
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
2rPL
  L  Soare, rPS
RL2

.
2
2rPL
Symilarly
proiectil, observator 
Freflectat, proiectil
  proiectil  Soare,rPS 
4rD,2 proiectil
2
Rproiectil
.
4rD,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
 0,4mLuna Plina  mproiectil ;
 proiectil,observator
Luna,observator
 log
proiectil,observator
RL2
RL2


L
2
2
rPL
2rPL

log
;
2
2
Rproiectil
Rproiectil

 proiectil  2
4rD,2 proiectil
2rD,proiectil
 L  Soare,r 
PS
 proiectil  Soare,r
PS
L 
log
 proiectil 
RL2
2
rPL
 0,4mL  mproiectil ;
2
Rproiectil
2rD,2 proiectil
2
 RL 
r

  2   D,proiectil   0,4mL  mproiectil ;
log

 r

 proiectil  Rproiectil 
 PL 
 L  0,12;  proiectil  1;
L
2
RL  1738 km; Rproiectil  400 mm;
rD,proiectil  rmax, observatorproiectil  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
 0,4mL  mproiectil ;
1738000 m
8700 km
 log 2  2 log
 0,4mL  mproiectil ;
0,400 m
384400 km
log 0,12  2 log
1738000
8700
 log 2  2 log
 0,4mL  mproiectil ;
0,400
384400
 0,920818754  13,27597956  0,301029995  3,290528253  0,4mL  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