ACTPOHOM~~ECKOE EUR0-ASIAN Round I Thea I

06~ECTBO ASTRONOMICAL SOCIETY

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XIV International Astronomy Olympiad
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KIIT8" , XSII'n"o,'
. 8 -·16. XI. 2009 flangzholl, China
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Theoretical round. Sketches for solutions

Note for jury and team leaders. The proposed sketches are not fuJI; the team leaders have to give more detailed
explanations for students. But the correct solutions in the students' papers (enough for 8 pts) may be shorter.

af3-l. Sirius. Sirius is the brightest star (in historical-classical meaning of a star) on the Earth's sky. Therefore in
a first approximation it will be the brightest star in those districts on the Earth where it is visible, that is, at
least sometimes appears over horizon. As the declination of Sirius is 6 = -16°43', it will be visible in all
southern hemisphere, and also in northern hemisphere at latitudes not nearer 16°43' to the pole, that is, at
latitudes not higher 73°17' North. Theoretically, in the second approximation refraction may be taken into
account (35' at horizon). With the refraction Sirius can be above horizon at districts till latitudes of 73°52'
North. But refraction doesn't matter for our goal. There is a talk about real (not theoretical) sky in the text.
In the third approximation light absorption should be taken into account. It is obvious that Vega or
Arcturus in high sky essentially are more bright than Sirius at horizon. HeIghts on which Sirius becomes
weaker than Vega or Arcturus may be roughly estimated as -5+8°. Thus, Sirius is the brightest star in the
sky for districts to the south from the latitudes of 65°+ 69° North.
Note for jury: the solution without a correct mention of refraction (that is, without mentions that it not effected here)
should be estimated with a deduction of 0.5 p ·nts. Values -5+8° are approximate, it is important the understanding of
the effect and a correct order ofthis height.

af3-2. Number of molecules. The pressure of an atmosphere is the total weight of all number (N) of its
molecules, distributed over the whole surface of a planet, that is,
P=N'mg/S,
where m is the rtlean value of mass of molecules in atmosphere. Taking into account that the surface of the
Earth is
S = 47tRi
and mass of molecule is
m=~A,
•
where N A is Avogadro number, we can write
•
2
N = N A 'P-47tRE / ~'g
Since the Earth atmosphere consists mostly on N2 (~N2 = 28 glmol) and O 2 ' (J.!02 = 32 glmol) in
2
approximate proportion 3: 1, the mean value of ~ is 29 glmol or in SI: 2.9.10- kg/mol. Calculations:
l 2 2 1 I
N = 6.022'10 mor 'l OSN·m- ·12.6·(6.37 m·l 06i / 2.9'10- kg-mor '9.81 N'kg- = 1.08·1 0 ~ 1.1'10 .
23 44 44

a-3. Efficiency of eye. For detecting a star the human retina should uninterruptedly remember the image of the
star. That is the eye should receive from this star at least 7 photons per second. Human eyepiece is about
d = 6 mm in the very dark, so the limit of illumination is:
2
7 photons / 7td /4 ~ 250 000 photons/m (very rougWy, of course)
2

It is about 10000000000/250000 = 40 000 less than the illumination from Om star. In stellar magnitudes
the difference is:
m
ilin = 2 .5 19 40 000 ~ 11 m.5.

So the theoretical limit for human eye in the very dark is to see stars of 11 m.5. Of course, it is quite far
m
from reality. Nobody reported about possibility to see stars fainter than 8 even in the absolute dark.
 6
FEd/FSa "" 1.91.10- .

&n =
m
-2m.5·lg(FEd/Fsa) "" 14 .3,
and the magnitude of Eris in the "Great opposition" is
mEd = mS a + ~m = m m
Om.7 + 14 .3 "" 15 .O.
And we should repeat that this way with comparing the Eris and Saturn it is not the only possible correct
way for solution.

ap-4. Catastrophe. At first we should realise that "suddenly decreasing of solar mass to half its original value"
is a hypothetic process and many physical laws of conservation cannot be used to compare parameters
before and after since the system is not closed. Since no any other changing done, we should assume that
at the moment of the mass decreasing other parameters of the SW1 and the Earth have not changed:
Positions of the SW1 and the Earth, velocity of the Earth. Taking into account this postulate, a first
approximation, if mass of a central body is reduced to half its original value, the circular speed becomes
parabolic one. That is the Earth will move on a parabolic orbit and will never return. The answer to the
problem is meaningless (the period is equal to infinity).
Nevertheless, it is known, that on July 5 the Earth is near the aphelion of its orbit. Its speed in this case is
less than circular. Thus, in this case if mass of a central body to reduce twice, the speed of the Earth
already is less than parabolic, the Earth will be rotate around the Sun on a prolate elliptic orbit.
For the beginning let us find a relation between the speed of the Earth in aphelions Vaph and circular speed
Vo for a motion on a circular orbit with the same semi-axis.
Using the II Kepler law one may write

using the law of conservation of energy

2 2
V per/2 - GMJRper = V aph/2 - GMJRaph .,
and also taking into account, that ,
2
R per = ao (l-e), Raph = ao (He), GM = Vo ·Ro,
it is possible to fmd
Vper = Vo·{(l +e)/(l-e)} 1/2 ,
V.ph = Vo·{(l-e)/(l+e)}l/2.
The orbit parameters of the Earth after that the mass of Sun has decreased to half its present value are
designate as:
aN - semi-axis of an orbit,
TN - period of revolution,
eN - eccentricity,
VN- speed for a motion on a circular orbit with a semi-axis aN. •
Let us write three additional equations: •
distance Raph ., former distance in apWion, became now the distance in perihelion
ao (l+e) = Raph = Rper-~ = aN'(l-eN), •

speed Vaph ., former speed in aphlion, will now become the speed in perihelion
'" Vaph = Vper-:'I/ = VN' {(l +eN)/(l-eN)} 1/2,
and the relation
VN= 21ta:'l/ / TN.
By solving all equations together it is possible to receive a beautiful expression
a:'l/ = ao(l +e)/2e.
According to the general III Kepler law

one can find

•