Download Problem 1. Marking scheme Lagrange Point

THEORETICAL TEST
15 short problems
page 1 from 30
Content
Indications ................................................................................................................................................................ 2
Problem 1.
Lagrange Points .................................................................................................................................... 3
Problem 1. Marking scheme Lagrange Point ........................................................................................................... 3
Problem 2.
Sun gravitational catastrophe! .............................................................................................................. 4
Problem 2. Marking scheme Sun gravitational catastrophe! .................................................................................... 4
Problem 3.
Cosmic radiation ................................................................................................................................... 6
Problem 3. Marking scheme Cosmic radiation......................................................................................................... 7
Problem 4.
Mass function of a visual binary stellar system .................................................................................... 8
Problem 5.
The Astronaut saved by … ice from a can! .......................................................................................... 9
Problem 5. Marking scheme The Astronaut saved by … ice from a can! ................................................................ 9
Problem 6.
The life –time of a star from the main sequence................................................................................. 11
Problem 7. Marking scheme The life –time of a star from the main sequence ...................................................... 12
Problem 7.
The effective temperature on the surface of a star .............................................................................. 13
Problem 8. Marking scheme The effective temperature on the surface of a star.................................................... 14
Problem 8.
The transit of an exoplanet ................................................................. Error! Bookmark not defined.
Problem 9.
Gradient temperatures ......................................................................................................................... 15
Problem 11. Marking scheme . Gradient temperatures ......................................................................................... 16
Problem 10.
Pressure of light .................................................................................................................................. 17
Problem 12. Marking scheme . Pressure of light ................................................................................................... 17
Problem 11.
The density of the star......................................................................................................................... 18
Problem 15. Marking scheme The density of the star ............................................................................................ 19
Problem 12.
Space – ship orbiting the Sun ............................................................................................................. 21
Problem 16. Marking scheme Space – ship orbiting the Sun ................................................................................. 21
Problem 13.
The Vega star in the mirror ................................................................................................................. 22
Problem 14.
Stars with Romanian names................................................................................................................ 24
Problem 15.
Apparent magnitude of the Moon ....................................................................................................... 27
THEORETICAL TEST
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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:
c. Answer sheets
d. Draft sheets
e. 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 – 15 for short
problems, 16 – 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
15 short problems
page 3 from 30
Problem 1.
Lagrange Points
The Lagrange points are the five positions in an orbital configuration, where a small object is stationary
relative to two big bodies, only gravitationally interacting with them. For example, an artificial satellite relative to
Earth and Moon, or relative to Earth and Sun. In the Figure 1 are sketched two possible orbits of Earth relative to
1
Sun and of a small satellite relative to the Sun. Find out which of the two points L3 and L23 could be the real
Lagrange point relative to the system Earth – Sun, and calculate its position relative to Sun. You know the
following data: the Earth - Sun distance d E S  15 10 7 km and the Earth – Sun mass ratio M E / M S  1 / 332946
L13
S
E
L23
S
E
Figure 1
Problem 1. Marking scheme Lagrange Point
1.
2.
3.
4.
Correct derivation of forces’ equilibrium
Correct identification of the Lagrange point
Correct calculation of the position of Lagrange point
Deduction for incorrect value
According to the notations in fig.1.1 and fig. 2.1

FS
L

FP
m
z
"
3
SM S 
rPS
Figure 1a
PM P 
6 points
2 points
2 points
1 point
THEORETICAL TEST
Short problems
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m 
SM S 
L'3 FP
w
 


FS  FP  Fcp  ma cp

FS
PM P 
rPS
FS  FP  ma cp  m2 rPS  w ;
2 points
The sign “+” for position L'3 and „– “for L"3
K
mM S
mM P
K
 m2 rP S  w ;
2
rP S  w 
2rP S  w 2
w  rP S
Using the assumption that
Figure 1 b

w
1 

 rP S 
2points
2
w
1 2
rP S

w 
1 

 2rPS 
2
 1
w
;
rPS
p
The rotation speed
2 
KMS
3
rPS
.............................................................................................................................................. 2p
The final relation
w
M P rPS
12M S  M P 
The value has to be positive, thus the L"3 is the position of one Lagrange point ........................ 2p
w  37,54 Km ............................................................................................................................ 2p
Problem 2.
Sun gravitational catastrophe!
In a gravitational catastrophe, the mass of the Sun mass decrease instantly to half of its actual value. If you
consider that the actual Earth orbit is elliptical, its orbital period is T0  1 year and the eccentricity of the Earth
orbit is e0  0,0167.
Find the period of the Earth`s orbital motion, after the gravitational catastrophe, if it occurs on: a) 3rd of
July b) 3rd of January.
Problem 2. Marking scheme Sun gravitational catastrophe!
-
Correct analyze of the initial conditions when the catastrophe occurs ( A)
Correct calculations (B)
o Correct use of laws of conservation
o Finding out that in the first case the orbit will be elliptic, relations (1)
5 points
5 points
2 points
THEORETICAL TEST
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and (2)
o Correct conduct of calculations
2 points
1 point
Detailed solution
(A) The orbit of Earth is elliptical, so the shape of the orbit after the solar catastrophe will depend on
the moment when the decrees of the mass of the Sun will occur.
Initial analysis of the problem
a)
In 3rd July the Earth is at the aphelion. The speed of the Earth is smaller than the speed
of Earth on a circular orbit with radius r0,max  a0 1  e0 . .
b)
In 3rd January the Earth is at perihelion. The speed of the Earth is bigger than the speed
of Earth on a circular orbit with radius r0,max  a 0 1  e 0 . .
Conclusion (A) the period should be calculated only for situation a). The expected trajectory in this case
is an elliptic one. The possibility that Earth hit the Sun is available too.
(B) Calculations:
S0 ; M 0
rd
In 3 July the distance from Sun is
maximum: fig. 2.1,

v 0,aph
r0,max  a0 1  e0 .
Before the catastrophe:
v 0,aph - the speed of Earth on aphelion,
Initial Earth orbit
F inal Earth orbit
r0,max  r0,aph
a 0 - big Earth’s elliptical orbit semi axis
v 0 - the speed of Earth if its orbit is
circular with radius r0  a 0
v 02,per
2
M 0 v 0,aph
M
K

K 0
2
r0,per
2
r0,ap h
r0, min  r0, per  a 0 1  e 0 
r0,max  r0,aph  a 0 1  e 0 
KM0  v 02 r0  v 02 a 0
M0
M
 K 0
r0
a0
v 0,per  v 0
1  e0
 v0
1  e0
v 0,per  v 0
v 0,aph  v 0
v 0,aph  v 0
r0,aph  rper
rmin  rper
According to Keppler’s second law and
the law of energy conservation (see figure 2.1) the
following relations can be written :
v 0,per r0,per  v 0,aph r0,aph ;
v0  K

v 0 ,per
(1)
1  e0
1  e0
(2)
S; M

v per
Figure 2.1
THEORETICAL TEST
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Conclusion – According to the relations (1) and (2) the new orbit of the Earth could be an elliptic one.
For the new elliptical Earth orbit:
rper  r0,aph ;
rmin  rper  a1  e ;
a0 1  e0   a1  e; a  a0
v per  v 0,aph ;
v per  v
1  e0
;
1 e
1 e
,
1 e
Where v este is the Earth’s speed on a circular orbit with the radius r  a, when the mass of the Sun becomes
M  M 0 / 2;
v
1  e0
1 e
 v0
;
1  e0
1 e
v K
M
M
M
 K 0  K 0
r
2a
a0
e  1  2e0 ; a  a0
a0
a0
 v0
;
2a
2a
1  e0
.
2e0
Conclusion
2r0 2a 0
T0 

v0
v0
2r 2a
T

;
v
v
T
a v 0 1  e0


T0 a0 v
2e0
T  T0
 1  e0 

2 
 2e 0 
1  e0
2a 1  e0

2
;
a0
2e0
2e0
3/ 2
 230 years
b) In 3rd of January the Earth is at perihelion. In that moment the Erath speed is larger than the speed
necessary for an Earth’s circular orbit. Thus the trajectory of the Earth after the catastrophe will be an open
trajectory, i.e. an hyperbolic or parabolic orbit.
Conclusion it is not necessary to calculate the period of revolution or could be issued as infinite
Problem 3.
Cosmic radiation
During studies concerning cosmic radiation, a neutral unstable particle – the  0 meson was identified. The
rest-mass of meson  0 is much larger than the rest-mass of the electron. The studies reveal that during its flight, the
meson  0 disintegrates into 2 photons.
THEORETICAL TEST
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Find an expression for the initial velocity of the meson  0 , if after its disintegration, one of the photons has
the maximum possible energy Emax and, consequently, the other photon has the minimum possible energy Emin .
You may use as known c - the speed of light.
Problem 3. Marking scheme Cosmic radiation
-
Correct use of general laws of conservation (A)
Correct applying of the laws of conservation for the conditions stated
in the problem (B)
Correct conduct of calculations and final solution (C)
5 points
4 points
1 point
Detailed solution
(A)
In the disintegration process the laws of energy conservation and the law of the conservation of momentum
are both obeyed.
In the general case the law of conservation of the momentum is represented in the down below figure.

p1

p
me sone π 0
1
f1
f2

p2

p
2
the total initial energy of the  0 meson is
E 2  p 2 c 2  m 02 c 4
And its kinetic energy is
E c  E  m0c2
The expressions of the 2 conservation laws written after the disintegration are:
  
p  p1  p2 ;
E  m0 c 2  E1  E2 ,
The energy of the photon 1 can be calculated using the notations in the figure
E  Ec  m0 c 2 ;
p1 sin 1  p2 sin  2 ;
E1
E
sin 1  2 sin  2 ;
c
c
E 2  p 2 c 2  m02 c 4 ;
p 2 c 2  E 2  m02 c 4 ;
E  E1  E2 ;
THEORETICAL TEST
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

E2  E  E1  Ec  m0 c 2  E1 ;
2
m c4
1
E1  0
.
2 Ec  m0 c 2  cos1 Ec Ec  m0 c 2


Similar the second photon energy is:
2
m c4
1
E2  0
2 Ec  m0 c 2  cos  2 Ec Ec  m0 c 2


(B)
If one of the photon has the maximum possible energy E max and consequently the other photon has the
minimum possible energy E min the law of momentum conservation is sketched:
.
Thus the relations become very simple:
mv 
v2
1 2
c
m0c2
 v
1   ;
v2  c 
2 1 2
c
m0c2  v 

1   ;
v2  c 
2 1 2
c
E  E min
.
v  c max
E max  E min
E max 
E min
m0 v
Problem 4.
(C)
Mass function of a visual binary stellar system
For a visual binary stellar system consisted of the stars
mass function of the system:
f M 1 ; M 2  
 1 and  2 , the following relation represents the
M 23 sin 3 i
,
M 1  M 2 2
where M 1 is the mass of star  1 , M 2 is the mass of star  2 and i is the angle between the plane of the stars’
orbits and a plane perpendicular on the direction of observation.
The recorded spectrum of radiations emitted by the star 1 , during several months, reveals a sinusoidal
variation of radiation wavelength, with the period T = 7 days and a shift factor z   /   0,001.
a. Prove that the mass function of the system is:
M 23 sin 3 i
T
v1  sin i 3 ,
f M 1 ; M 2  

2
M 1  M 2  2K
THEORETICAL TEST
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Where: v1  sin i is the maximum speed of star 1 relatively to the observer; K  the gravitational constant , i is
the angle between the plane of the orbits and the plane normal to the observation direction.
Assumptions: The orbits of the stars are circular,
b. Derive an expression for the mass function of the system. The following values are known:
c  3 x108 m/s; K  6,67 x10 11 Nm2 kg 2 .
Problem 5.
The Astronaut saved by … ice from a can!
An astronaut, with mass M  100 kg, get out of the space ship for a repairing mission. He has to repair a
satellite standing still relatively to shuttle, at about d  90 m distance away from the shuttle. After he finished his
job he realizes that the systems designated to assure his come-back to shuttle were broken. He also observed that
he has air only for 3 minutes. He also noticed that he possessed a hermetically closed cylindrical can (base section
S  30 cm 2 ) firmly attached to its glove, with m  200 g of ice inside. The ice did not completely fill the can.
Determine if the astronaut is able to arrive safely to the shuttle, before his air reserve is empty. Briefly
explain your calculations. Note that he cannot throw away anything of its equipment, or touch the satellite.
You may use the following data: T  272 K / the temperature of the ice in the can, p s  550 Pa - the
pressure of the saturated water vapors at the temperature T  272 K ; R  8300 J/(kmol  K) - the constant of
perfect gas;   18 kg/kmol - the molar mass of the water.
Problem 5. Marking scheme The Astronaut saved by … ice from a can!
-
A. For the use with an adequate justify of one of the relationships (4)
B. Reasoning The student describe correctly the processes before and after
the can is opened.
C. Calculations according to the reasoning, and/or as support for reasoning
D. Correct result
3 points
4points
2 points
1 point
Detailed solution
Theoretical considerations:
The incident particle flux on a wall (i.e. a certain direction on a surface) is:
1
 m0  n  S  v
(3)
6
N m N m
m0  n  m0  3  03    ,
V
a
a
where: m0 is the mass of one molecule; m  mass of the gas in the cube ; V  volume of the cube ;   the density
  m0   
of the gas ;

p
RT
,
THEORETICAL TEST
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where p  the pressure of the gas in the cube; The relation (3) – the mass flux relation becomes:
A 

1
1 p
3RT 1
3
 S v  
S 
  pS 
.
6
6 RT

6
RT
(4)
1
1 p
3RT 1
3
 S v  
S 
  pS 
.
6
6 RT

6
RT
(5)
3points
B. Reasoning
Because the cylindrical can is not full of ice, in the empty part of it there are saturated vapors, i.e the mass flux of
the molecule which sublimate is equal with mass flux of gass which transform into ice. Thus the pressure in the
can is the saturated vapor pressure ps and it has the corresponding maximum density  s See figure 6.2
vapori saturați
2
ps
1
ps / 2
1
gheață
a
b
Fig. 6.2
1
1
  s  S  v   ps  S 
6
6
1
1
 2   solidification    s  S  v   ps  S 
6
6
1   2 .
1   sublimation 
3
;
RT
3
;
RT
After the can was opened, there be no molecules which sublimate thus the mass flux of the molecules
which gather the ice become null. So the pressure becomes  ps / 2.
Thus the force acting on the astronaut will be
C. Calculations according to the reasoning, and/or as support for reasoning
F
ps
 S,
2
Opening the can the astronaut will be accelearated with:
F ps  S 550 Nm2  30 10 4 m 2
a


 0,00825 ms 2 .
2
M
2M
2 10 kg
The total time of the acceleration movement will be the total time of ice sublimation:
2 points
THEORETICAL TEST
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
m
m
6m
RT



 150 s .
1 1
ps  S 3
3
 ps  S 
6
RT
D. Correct result
The travel distance in this time will be :
L
a 2 0,00825 ms 2  225 10 2 s 2

 93 m,
2
2
The astronaut could arrive safely in at the shuttle if he didn’t lose to much time by solving the problem.
Problem 6.
The life –time of a star from the main sequence
The plot of the function log L/LS   f log M/M S  for data collected from a large number of stars is
represented in figure 3. The symbols represents: L and M the luminosity respectively the mass and of a star and LS
and respectively M S the luminosity and the respectively the mass of the Sun.
4
3
2
log
L
LS
1
0
1
 0,5
0
0,5
log
1,0
M
MS
Figure 6
Find an expression for the life- time for each star in the Main Sequence from Hertzprung – Rossell diagram
if the time spent by Sun in the same Main Sequence is  S . Consider the following assumptions: for each star the
THEORETICAL TEST
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percentage of its mass which changed into energy is  , the percent of the mass of Sun which changes into energy
is S , the mass of each star is M  nM S and the luminosity of each star remains constant, during its entire life
time.
Problem 6. Marking scheme The life –time of a star from the main sequence
-
A. The analysis of the graph
o Obtaining the formula (1) from the linearity of the graph
o Correct use of the luminosity formula (2) for finding out the final formula
6 points
4 points
Detailed solution
A. The analysis of the graph :
The graph is linear:
y  ax  b  ax;
From the graph it can be obtain
the following data:
4
3
L
M
log
 a  log
;
LS
MS
y 3,5
a  tan  

 3,5;
x
1
1,0
y
0,5

1,0
L ~ M 3,5 . (1)
The total energy of the star is:
1
E  Mc 2 ,
So the emitted energy due to the
mass variation of the star is:
E  c 2 M ,
According to the text
M  M ;
E  c 2M .
By using the definition of the luminosity :
E
 L; (2)
t
t   ;
c 2M

 L;
x
0,5
0
0,5
1,0
THEORETICAL TEST
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
c 2M
, (2)
L
Which represents the life-time of the star.
By using the results from the graph analysis
LS
 M 3, 5 ,
3, 5
MS
L
Thus :

c 2M S3,5
 M 2,5 .
LS
If use the same calculations for the Sun it can be obtain
ES  M S c 2 ;
S 
c 2S M S
,
LS
Which is the life-time of the Sun


  S  M S2,5  M 2,5 ;
S
2 , 5

 M 

  
 S  S  M S 

  n 2,5 S .
S
; M  nM S ;
Problem 7.
The effective temperature on the surface of a star
A star emits radiation with wavelength values in a narrow range    , i.e. the wavelength have values
between  and
 and    . According to Planck's relationship (for an absolute black body), the following
relation define, the energy emitted by star in the unit of time, through the unit of area of its surface, per length-unit
of the wavelength range:
r
2πhc 2
.
5 e hc/kλc  1


The spectral intensities of two radiations with wavelengths 1 and respectively  2 , both in the range 
measured on Earth are I1 1  and, respectively I 2  2 .
a. Establish the equation which, in a general case, allows determining the effective temperature
on the surface of the star using only spectral measurements.
b. Find out the approximate value of the effective temperature on the star surface if hc  kT .
c. Find out the relation between wavelength  1 and  2 , if I1 1   2I 2  2 , when hc  kT.
THEORETICAL TEST
Short problems
Page 14 from 30
You know: h  Planck’s constant; k  Boltzmann’s constant; c  speed of light in vacuum.
Problem 7. Marking scheme The effective temperature on the surface of a star
a. Identifying the expression of spectral intensity and correct use of the given relation for obtaining the
relation (1)
4 points
b. correct use of the assumption hc  kT and find out relation (2)
3 points
c. correct use of the assumption hc  kT. and find out relation (3)
3 points
Detailed solution
a. We start from the definition of r:
r
E
E
where R is the radius of the star

t  Sstea   t  4R 2  
2hc 2
r  5 hc / kT
 e
1


Considering d as the distance from the star to the Earth, the definition- relation of the spectral intensity can
be written as follows:
E
I   t  2
4d
2hc 2 R 2
I   2 5 hc / kT
;
d  e
1


Particularly for each wavelength:
I1 1  
2hc 2 R 2
2hc 2 R 2


;
I


;
2
2
d 2 15 e hc / 1kT  1
d 2 52 e hc / 2kT  1




The ratio of the 2 above relations
I1 1    2  e hc /  2kT  1
  
I 2  2   1  e hc / 1kT  1
5
(1)
Represents an equation which allow to find out the temperature of star’s surface T by using spectral
measurements
b. If we consider that
hc  kT, then:
e hc / 1kT  1  e hc / 1kT and e hc /  2kT  1  e hc /  2kT
The relation (1) becomes
I1 1    2

I 2  2   1
5
5
 e hc /  2 kT   2 
  hc /  kT     e hc /  2 kT hc / 1kT
1
 e
 1 
THEORETICAL TEST
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I1 1   1 
    e hc /  2 kT hc / 1kT
I 2  2    2 
5
T
hc  1   2 
 I    
k 1 2  ln  1 1   1
 I 2  2    2
hc
k1T
hc
k2T



5



c. If hc  kT , then:
hc
hc
1 
 1; e hc / 1kT  1  1 
k1T
k1T
hc
hc
1 
 1; e hc /  2 kT  1  1 
k 2 T
k 2 T
The relation (1) becomes:
hc
I1 1    2  k  T   2
  
 
hc
I 2  2   1 
 1
k  T
5



4
4
 
I1  2I 2 ;  2   2; 2  1  4 2  1,2  1 .
 1 
Problem 8.
Gradient temperatures
The spectra of two stars with different temperatures T1 and respectively T2 were compared. In the
spectrum of each star, two very close spectral lines corresponding to the wavelength with values 1 and
respectively 2 were found. For each line of this spectral lines, the difference between the corresponding visual
apparent magnitudes of the stars are m 1  m1,1  m 2,1 and m  2  m1, 2  m 2, 2 . m1,1 is the apparent
magnitude of the star 1 for the wavelength 1 , m1, 2 is the apparent magnitude of the star 1 for the wavelength  2
, m 2 ,1 is the apparent magnitude of the star 2 for the wavelength 1 , m 2 , 2 is the apparent magnitude of the star 2
for the wavelength  2 .
Determine the temperature T1 of one of the two stars, if the temperature T2 of the other star is already
known, by using the Plank expression of black body radiation:
THEORETICAL TEST
Short problems
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r λ  


1
2πhc 2 hc/kλc
e
1 ,
5
λ
where: h  Planck’s constant; k  Boltzmann’s constant; c  speed of light in vacuum. You will consider that
hc  kT .
Problem 8. Marking scheme . Gradient temperatures
-
Finding out the relations (1) and (2) by the correct using of the
approximation hc  kT .
Correct using of the Pogson’s formula
For correct conduct of calculations and obtaining the final formula (5)
4 points
3 points
3points
Detailed solution
By using the Plank expression the spectral intensities for the two wavelengths in the doublet emitted by the
star number 1 are:
2hc 2
2hc 2
r1 1   5 hc / 1kT1
; r1 2   5 hc / 2kT1
;
1 e
1
2 e
1
And by considering hc  kT;

2hc 2
r1 1  

5
1
e
 hc / 1kT1
; r1 2  

2hc 2

5
2


e hc / 2kT1 .
(1)
Respectively, the spectral intensities for the two wavelengths in the doublet emitted by the star number 2
are:
r2 1  
r2 1  
2hc 2

5
1
e hc / 1kT2 ; r2 2  
2hc 2
2hc 2


;
r


;
2
2
15 e hc / 1kT2  1
52 e hc / 2kT2  1
And by considering hc  kT ;


2hc 2

5
2


e hc / 2kT 2 .
(2)
Using the Pogson formula for star 1 and 2 and for 1 , result:
log
r1 1 
 0,4 m1,1  m2,1  0,4  m1 ;
r2 1 


log ehc / 1k 1/ T2 1/ T1   0,4  m1
Similar the Pogson formula for 2 , :
log
r1  2 
 0,4 m1, 2  m 2, 2  0,4  m  2
r2  2 


log ehc / 2k 1/ T2 1/ T1   0,4  m2
Using the relations (3) and (4)
(3)
(4)
hc /  k 1/ T 1/ T 
log e hc / 2k 1/ T2 1/ T1   log e 1 2 1  0,4  m2  0,4  m1 ;
THEORETICAL TEST
Short problems
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 hc / 1k 1 / T2 1 / T1 
 0,4  m2  0,4  m1 ;
log e hc / 2k 1/ T2 1/ T1   log e
 hc  1 1   1 1 
         log e  0,4 m1  m2 ;
 k  T2 T1   2 1 
0,4
Because log e  0,43; log e  0,93;
m1  m2
k12 m1  m2
1 1
  0,93 
 0,93 
;
T2 T1
hc1  2 
hc  1 1 
  
k  2 1 


T1  T2 
Problem 9.




hc1  2 
. (5)
hc1  2   0,93  k12T2 m1  m2


Pressure of light
One particle of star dust is in static equilibrium at a certain distance from Sun. Assuming that the particle is
spherical and its density is  , calculate the diameter of the particle.
The following assumption may be useful for solving the problem:
The pressure of electromagnetic radiation is equal with the volume density of the electromagnetic radiations
Problem 9. Marking scheme . Pressure of light
-
Correct use of the formula (1) for the pressure of light
Correct identify of the equilibrium condition
Correct solution and reasoning
The pressure of the emitted radiation is
prad  w 
p rad 
4
2
T planet
R planet
cD 2
3 points
3 points
4 points
 planet,D
c
(1)
As seen in the below image, the pressure due to the solar radiation is effectively acting on an equivalent
plane disc with the diameter d of the spherical star dust particle
THEORETICAL TEST
Short problems
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
Fg
Pemis

Frad,S
d
SUN
Pabsorbed

Frad,S
d
SUN
p rad,S
Fig.12 b
Thus the force acting by the Sun radiation on the star-dust particle is:
Frad,S  prad,S  r 2  prad,S 
d 2
4

TS4 RS2 d 2
cL2S

4
,
The equilibrium condition is
Frad,S  Fg ,
Where Fg is the gravitational attraction force between Sun and the star-dust particle.
TS4 RS2 d 2
2
S
cD

4
K
mM S
;
DS2
4r 3
4 d 3
d 3


;
3
3 8
6
TS4 RS2 d 2
d 3 M S


K

;
4
6 DS2
cDS2
m  V  
3  TS4 RS2
d 

.
2 K M S
Problem 10. The density of the star
In a very simple model, a star is assumed to be a sphere of gas in a state of equilibrium in its on
gravitational field. The stellar gas is consisted of plasma, i.e. hydrogen and helium atoms, completely ionized. Find
an expression for the value of the mass of the star if you know: r  radius of the star; T  the temperature of the
star; n  the relative proportion of hydrogen in the mass of the star;
 H  molar mass of the hydrogen;  He 
molar mass of the helium; R  universal gas constant; K  gravitation constant. You may use the formula of the
pressure of radiation inside the star
negligible.
1
p rad  aT 4 , where a is a known constant. The rotation of the star is
3
THEORETICAL TEST
Short problems
Page 19 from 30
Problem 10. Marking scheme The density of the star
-
Correct reasoning for finding the expression of the inner stellar gas (2)
Correct expression for the equilibrium condition
The correct conduct of calculations, and obtain the final correct result
Correct solution and reasoning
2 points
2 puncte
2 puncte
4 points
Detailed solution
The hydrostatic equilibrium inside the star means that in each point of the inside of the star the
gravitational forces are compensated by the hydrostatic pressure forces. That means that the mater of the
star remains localized in a region of space.
The total pressure of the stellar gas has two components: the pressure due to the movement of the
stellar –gas particles  p gaz  and the pressure due to the emitted radiation by the stellar-gas particles  prad  ,
thus:
ptotal  pgaz  prad ;
pgaz  p H  p He ;
1
p rad  aT ;
3
n  1  n  H
1
p total   He
RT  aT . (1)
 H  He
3
In order to calculate the gravitational pressure of the stellar /gas let’s consider a narrow cylinder with
section area S ,along the radius of the star . See the figure bellow. If the total gravitational force acting on this

cylinder is Fg than the gravitational pressure exert by the gas –column is pgrav  Fg / S

Fn
prad,S
DS
r  r

F2

Frad,S
d

Frad,S

F1
prad,S

Fg
Fig. 15 a
d
THEORETICAL TEST
Short problems
Page 20 from 30

In order to calculate Fg let divide the cylinder in n identically small cylinders each of it with height
r and mass m Considering an homogenous star :
m  M 1
2
m  M 1
m  M 1  r 
F1  K
K
K
1   ;
2
2
r 2  2r 
r 
r 

2
r 1  
r  
2 

 2r 
M1
r  r   M  1  r  ;
M

; M1  M 


3
3
r 
r
r3
r  r 

m  M 
r 
F1  K
1  2 ;
2
r 
r

m  M 
r 
F2  K
1  3 ;
2
r 
r

3
3
………………………………………………..
m  M 
r 
1  n  1 ;
2
r 
r

Fg  F1  F2  .......  Fn ;
Fn  K
mM 
1 1  n n 
m  M  1  n  
n
; Fg  K
1
;
2 

n 2 
2n 
nr 
r 2 
m  M n 1 n  1 1
Fg  K 2
;
 ;
2n
2n
2
r
mM
Fg  K
. (2)
2r 2
Thus the gravitational pressure is
4r 3
  r  S   
Fg
1
3
pg 

2
S S
2r
2
pg  Kr 2  2
3
Using the relations (1) and (2) in the pressures equilibrium relationship
p total  pgravitational
Fg  K
Results :
n  1  n  H
2
1
Kr 2  2   He
RT  aT ;
3
 H  He
3
n  1  n  H
2
1
Kr 2   2  He
RT    aT  0;
3
 H  He
3
This is an second degree equation in 
A 2  B  C  0
Where the coefficients are
2
A  Kr 2
3
THEORETICAL TEST
Short problems
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B
n He  1  n  H
 H  He
RT
1
C  aT
3
The positive solution is the valid one i.e.
n He  1  n  H

B  B 2  4 AC

2A
 H  He
 n  1  n  H
 8
RT   He
RT   Kr 2 aT
 H  He

 9
4
Kr 2
3
2
Problem 11. Space – ship orbiting the Sun
A spherical space –ship orbits the Sun on a circular orbit, and spin around one of its axes. The temperature
on the exterior surface of the ship is TN . Find out the apparent magnitude of the Sun and the angular diameter of
the Sun as seen by the astronaut on board of the space – ship. The following values are known:, TS - the effective
temperature of the Sun; R S - the radius of the Sun; d 0 - the Earth –Sun distance; m 0 - apparent magnitude of
Sun measured from Earth; R N - the radius of the space –ship.
Problem 11. Marking scheme Space – ship orbiting the Sun
-
Correct use of the formulas (1) for the apparent brightness
Correct use of the formula (2) by using Pogson formula
Correct solution and reasoning
3 points
3 points
4 points
Detailed solution
According to the Stefan – Boltzmann law, the luminosity of the Sun is:
Lsun  Qsun  4RS2un  TS4un  4RS2un ,
(1)
At distance d from the Sun , where the space ship is the energy which passes the unit of surface in an unit of time
is:
S un,d
LS
TS4  4RS2


.
4d 2
4d 2
(2)
The space ship receive through its entire surface, in the unit of time, the energy:
Preceived 
TS4un  4RS2un
2
 Rship
.
2
4d
Corresponding to its temperature, TN , according the Stefan - Boltzmann law, the emitted energy by
starship through its hole surface in the unit of time :
Pemis, N  TN4  4RN2 .
When the temperature stabilized at thermic equilibrium :
Preceived , N  Pemis, N
THEORETICAL TEST
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TS4  4RS2
 RN2  TN4  4RN2
2
4d
the distance of orbiting the Sun of the space ship is:
T 2R
d  S 2S ,
2TN
The angular diameter of the Sun as seen from the space ship :
 /2 
RS
d
T 
2R
  S  4 N 
d
 TS 
2
According the Pogson formula written for Sun seen from Earth and space ship the following relation
occurs:
lg
ES, Nava
ES, P
 0,4m  m0 
(2)
d 
2  lg  0   0,4m  m0 
d 
The apparent magnitude of the Sun as seen from the space ship
2d T 2
m  m0  5 m  lg 0 2N
RSTS
Problem 12. The Vega star in the mirror
Inside a photo camera a plane mirror is placed along the optical axis of the lens of the objective (as seen in
figure 13). The length of the mirror is half of the focal distance of the lens of the objective. The photo camera is
oriented as on the photographic plate situated in the focal plane of the photo camera are captured two images with
different illuminations of the Vega star. Find out the difference between the apparent photographical magnitudes of
the two images of the Vega stars.
VEGA
star
Photographical
plate
Lens
objective
f /2
2
r/2
Plane mirror
r
f
Figure 12
1
THEORETICAL TEST
Short problems
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Problem 12. Marking scheme The Vega star in the mirror
The light beam arriving from Vega Star can be considered paraxial, due to the distance from it to the
observer on Earth. The explanation for the existence of two distinct images of the star is that the optical axis of the
objective is not parallel with the light beam from the star.
The images on the camera plate are symmetrical placed relative to the principal optical axis.
P
V
E
2
C
N
G
O
M
r/2
f /2
A
1
r
Q
f
Fig. 12
Each of the point images of the Vega Star 1 and  2 didn’t concentrate the same light fluxes. In the
down below figure it can be seen the sections of the lens which correspond to each image. The sector APBC is
passed by the light which concentrates in the image  2 and the light passing the sector ACBQ concentrates into
the point image 1 See the picture in figure 13 .
THEORETICAL TEST
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Page 24 from 30
P
S2
B
C
A
30
60
r
O
S1
Q
Fig.13 b
The ratio between the light fluxes concentrated into the two image points will directly depend on the ratio
of the two sectors areas.
From the geometry of the figure 2 results :
r
MN  OM; N 1  OC  ;
2

CBO  30 ; BOC  60 ; AOB  120 ;
S1 8  3 3

 4.
S 2 4  3 3
Using the Pogson formula :
TV4  4RV2
 S1
2
4d PV
E1
log
 log
 0,4m1  m2 ;
E2
TV4  4RV2
 S2
2
4d PV
log
S1
 0,4m1  m2 ;
S2
m2  m1  1,5m.
Problem 13. Stars with Romanian names
Two Romanian astronomers Ovidiu Tercu and Alex Dumitriu from The Astronomical Observatory of the
Museum Complex of Natural Sciences in Galati Romania, recently discovered – in September 2013- two variable
stars. They used for that a telescope with the main mirror diameter of 40 cm and a SBIG STL-6303e CCD camera.
THEORETICAL TEST
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With the accord of the AAVSO (American Association of Variable Stars Observers), the two stars have
now Romanian names: Galati V1 and respectively Galati V2. The two stars are circumpolar, located in Cassiopeia
and respectively in Andromeda constellation. The two stars are visible above the horizon, form the territory of
Romania,
all
over
the

year.
The
galactic
coordinates
of


the
two
stars
are:

Galati V 1 G1  114.371 ; g1  11.35 and Galati V 2 G2  113.266 ; g 2  16.177  .
Another star, discovered by the Romanian astronomer Nicolas Sanduleak, has also a Romanian name –
Sanduleak -69° 202; it explodes as the supernova SN 1987. This star was localized in the Dorado constellation
from the Large Magellan Cloud, by the coordinates:
  5h 35min 28,03s ;   6916' 11,79"; G  279,7  ; g  31,9.
Estimate the angular distance between the stars Galati V1 and Galati V2
Problem 13. Marking scheme Stars with Romanian names
1. Correct geometrical calculations
2. Correct calculations
5 points
5 points
In the figure bellow the two stars  1 and  2 , are located using the galactic coordintates G1; g1  and
respectively G2 ; g 2 . on the geocentric celestial sphere. The spherical triangles  1 A 2 G1; g1  and respectively
may be considered rectangular plane triangles because the angles G  G2  G1
g  g 2  g1 are very small
Thus:
 1 2 
 1 A2   2 A2 ,
 1 2 
1 B2   2 B2 ,
or:
G1; g1 
and respectively
THEORETICAL TEST
Short problems
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Galactic North Pole
J
O2
r2
2
A
g
r
O1
g
r
r1
B
1

r
galactic
equator
g1
O
F'
Nodul Galactic
ascendending node
of the North Pole
Polului Nord

Galactic
g1
g2
F
G2
G
G1
Fig.
 1 A  r  g;
 2 A  r2  G  r  cos g 2  G;
 1 2  r   ,
Where  is the angular distance between two stars
r  g 2  r  cos g2  G2 ;
2
2
  g   cos g 2  G  ;
r   
1 B  r1  G  r  cos g1  G;
G
1
 
cos g1  G2  g 2 ;

 
 
 4,827   0,96  1,105 
0,98   1,105    4,827 
 2
2
The angular distance between Galați V1 and Galați V2.
.

 114.371 ; g1  11.35 ; G2  113.266 ; g 2  16.177  ;
G  G2  G1  1,105 ; g  g 2  g1  4,827  ;
cos g1  0,98; cos g 2  0,96;
2
 2
 2
 4,942 ;
 2
 4,946 ,
THEORETICAL TEST
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Problem 14. Apparent magnitude of the Moon
You know that the absolute magnitude of the Moon is M L  0,25 . Calculate the values of the apparent
m
magnitudes of the Moon corresponding to the following Moon –phases : full-moon and the first quarter. You
know: the Moon – Earth distance - d LP  385000 km , the Earth – Sun distance - d PS  1 A U , the Moon –Sun
distance, d LS  1 A U
Problem 14. Marking scheme Apparent magnitude of the Moon
1. General analysis of the problem
2. The analysis of the 2 particular situations
6 points
4 points
The apparent magnitude of a planet from the Solar System depends on the phase angle M  M  .
The apparent magnitude of the body is given by the relation:
m  M  2,5  log
unde: d B ,S
d C,2 S  d C,2 O
,
d 04  p 
 the distance between the body and the Sun; d B ,O  distance between the body and observer;
d 0  1 AU;   the phase angle ; p   the phase function :

2   
1
p    1   cos   sin  ,
3   


 as seen in the figure bellow is given by the cosine law.
Sun
d B ,S
The body fro
Solar System

d O,S
Observer
d B ,O
Fig.
2
d B2 O  d B2 S  d OS
cos  
.
2d B O  d B S
In particularaly for the Moon
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Moon
Sun
dM S
dES
Observer
d ME
Earth
cos  
2
d ME
 d M2 S  d E2 S
;
2d ME  d M S

2   
1
p    1   cos   sin  ;
3   


2
2
d d
mM  M M  2,5  log M4 ,S ME .
d 0  p 
Particular cases:
1) Full moon
  0;
cos   1; sin   0;
2
p    ;
3
d M S  1 AU; d ME  385000 km  0,00256 AU  256 10 5 SU; d 0  1 SU ;
mM  M M  12,5m  0,25m  12,5m  12,25m.
2) First Quarter
  90 ;
cos   0; sin   1;
2
p  
 0,2;
3
THEORETICAL TEST
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2
d M2 ,S  d ME
65536 10 10

 491520 10 10 ;
4
0,2
d 0  p 
mM  M M  10,75m  0,25m  10,75m  10,5m.
Problem 15. Absolute magnitude of a cepheide
The cepheides are cariable stars, whom luminosities and luminosities varies due to volume oscillations.
The period of the oscilations of a cepheide star is:
P  2R
R
,
KM
where: R  the radius of the cepheide; M  the mass of the cepheid (constant during oscillation);
R  Rt ; P  Pt .
M cef , depend on the period of cepheide’s
Demonstrate that the absolute magnitude of the cepheide
pulsation P according the following relation:
m
where k is constant; P  Pt ; M cef
 10 
M cef  2,5 m  log k     log P,
3
 M cef t .
Problem 15. Marking scheme Apparent magnitude of the Moon
P  2R
R
,
KM
rezults
4 2 R 3
KMP 2  KM 
P 
;R3
 2 
4 2
KM
 4 
1/ 3
 P2/3;
2
 KM 
R2   2 
 4 
2/3
 P4/ 3.
The absolute brightness is:
Lcef  Tcef4  4R 2 ,
And the apparent brightness :
Ecef 
Lcef
Tcef4  4R 2

,
4d P,2 cef
4d P,2 cef
d P ,cef is the distance between the observer on Erath and the cepheide
 KM 
T  4   2 
 4 

4d P,2 cef
4
cef
Ecef
Similarly for Sun
2/3
 P4/3
.
THEORETICAL TEST
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LS
TS4  4RS2
ES 

.
2
2
4d PS
4d PS
By using the Pogson formula:
log
Ecef
 0,4mcef  mS ;
ES
M cef  M S  5 m log
d P,cef
d PS
 2,5  log
Ecef
;
ES
3/ 2
 KM 
2
T   2   d PS
4



 k1  constant;
TS4  RS2  d P,2 cef
4
cef
M cef  M S  5 m log
M S  5m log
d P,cef
d P,cef
d PS
d PS
 2,5  log k1 
10
 log P;
3
 2,5  log k1  2,5  log k ;
k  constant;
M cef  2,5  log k 
10
 log P.
3