Problem Solution Ieso2010 r7

Problem & Solution

International Earth Science Olympiad (IESO)

Yogyakarta, 19-28 September 2010

By

Suryadi Siregar

Chatief Kunjaya

Hesti Retno Tri Wulandari

Dhani Herdiwijaya

Confidential Round-7

1

A. Multiple Choices

1. (Solar System (E)) Suppose you see a new planet in the sky. Based on observation you

conclude that the planet is close to the Sun with maximum elongation of 30 degrees. As

comparison the maximum elongation of Venus is 46 degrees, meanwhile the maximum

elongation of Mercury is 23 degrees. According to these data you consider:

a. the planet is nearer to the Sun than Mercury

b. the planet is between Mercury and Venus

c. you do not know the position of planet

d. all answers are incorrect

e. the planet is between Earth and Venus

A distant planet has a bigger elongation than a near planet. In this diagram 1 is the maximum

elongation of Mercury, 2 is the maximum elongation of planet, and 3 represents the

maximum elongation of Venus.

2. (Solar System, (E)) For an astronaut who is walking in the surface of the Moon :

a. The Earth will always appear full circle

b. The length of one day and one night is equal to the sinodic period of the Moon seen by an

observer in the Earth

c. The length of the day is half of the sidereal period of the Moon orbiting the Earth

Earth

Sun


2

d. The duration since the Earth rises until it sets is the same with the duration between New

Moon and Full Moon seen from the Earth

e. The surface of the Earth facing the Moon is always the same so that only one side of the Earth

is visible from the moon

Answer : b

The rotational period of the Moon is the same with its revolution period, so that only one side of

the moon always faces the Earth. The length of one day and one night in the Moon is from the

Sunrise until the nexts sunrise that is from the new moon until the next new moon for an

observer in the Earth, or the sinodic period.

3. (Solar System (M)) How would the length of the solar day change if the direction of the

Earths rotation suddenly changed, opposite to its initial direction?

a. It would be 4 minutes longer than before

b. It would be 4 minutes shorter than before

c. It would be 8 minutes longer than before

d. It would be 8 minutes shorter than before

e. It would not change, but remains the same as before

Answer : d

Solar day is determined from the daily motion of the Sun. Local noon in solar day is defined as

the time when the Sun is in its culmination (highest) point on the sky. After the Earth has

completed one rotation (known as one sidereal day), the Earth has moved (360/365) ~ 1

= 4

along its orbit around the Sun. Therefore, the Earth still has to rotate a bit further until the Sun

reaches the culmination point. It means that one solar day is 4 minutes longer than one sidereal

day. On the contrary, if the direction of the Earths rotation is opposite to this direction, one solar

day would be 4 minutes shorter than one sidereal day. Consequently, one solar day would be 8

minutes shorter if the direction of the Earths rotation changed to the opposite direction.

4. (Sun (D)) According to stellar evolution theory, the Sun will evolve into red giant stage in

few billion years. How is the average temperature in the surface of the Earth compared to present

temperature, in the time when the sun become a red giant with radius 11.2 million km and

temperature drop to 2900 K. Assume that currently the radius of the Sun is 700 000 km and its

average surface temperature is 5800 K. How the temperature of the Earth would change ?


3

a. Become four times of the present temperature

b. Become twice of the present temperature

c. Become half of the present temperature

d. Become a quarter of the present temperature

e. No change

Solution :

In equilibrium, the amount of Suns radiation energy absorbed is equal the radiated by the Earth.

22 4

2

14

4

R ( A)LR T

d

Where A is albedo of the Earth and d is the distance of the Sun from the Earth

14 4

2

1( A)LT T L

16 d

The temperature of the Earth is proportional to L

The radius of the Sun become 16 times and the temperature become half (2900/5800). Then

4

2 4 2 14 16 162

L R T L L

4 16 2T

Then the temperature of the Earth will become twice.

5. (Stars (M)) Parallax of a star measured on the Earth is 0.05 arc- seconds. Determine the

parallax of this star if we measure from Jupiter where heliosentric distance of Jupiter is 5.2 AU

a. 0.25 arc-seconds

b. 0.33 arc-seconds

c. 0.5 arc-seconds

d. 0.26 arc-seconds

e. 1.0 arc-seconds


4

Answer: d (0.26 arc- seconds)

r-distance of the star from the Earth

d-distance of the star from the Sun

From the Sun (Earth) distance of the star is

120r

p pc

From the Jupiter parallax of star is

(heliocentric distance of Jupiter)

5 20 26

20

Jd .p .r

6. (Celestial Mechanics (M)) If mass of the Sun increases two times than the present mass, and

the planets remains in present orbit, and one year is 365.25 days, then period revolution of the

Earth is (rounded):

a. 423 days

b. 365 days

c. 321 days

d. 253 days

e. 147 days

Answer: d (253 days)

2

1

22

1

2 2

rG( M )

P rrP

GM

r

P2= 0.707 P1 = 253 days

7. (Celestial Mechanics (M)) The perihelion of Halley comet is 8.9 1010

meters, its period is

76 years, then the eccentricity of Halley is:


5

a. 0.567

b. 0.667

c. 0.767

d. 0.867

e. 0.967

Answer: e (0.967)

2 3 1 1/r

a P r a( e ) ea

0.9669

8. (Stars, D) The typical spectrals line of star was observed at 4999 A. According to laboratory

expriment this spectral appear at 5000 A. How is the velocity of this star ?:

a. 60 km/s approach observer b. 60 km/s recede from observer c. 75 km/s approach to observer d. 75 km/s recede from observer e. The star does not move

Answer: a

Solution:

Suppose 1 = 5000 and 2 = 4999

2 1 = 4999 5000 = -1

From Dopplers equation 1

rV

c

where Vr is the radial velocity and c =3 x 10

10 cm/s is the

velocity of light

101

13 10 6000000

5000rV c cm/s -60 km/s

, negative sign represents the star

aproach to observer.


6

B. Essay

1. (Solar System (D)) Some time ago gossip media reported that Mars will be seen as big as the

Moon. Determine the maximum angular diamater of Mars, given that the semimajor axis and

eccentricity of Earth and Mars are (a=1 AU, e = 0.017) and (a=1.5 AU, e = 0.093) respectively,

and the radius of Mars is R= 3393.4 kilometres

To answer these equations you have to

a. Make a sketch of this situation

b. Show the formula(s) that will be used

c. Show the calculation and final results

Solution

The closest position of Mars to the Earth is at opposition. Suppose, r, represents the nearest

distance of Mars to the Earth at opposition (Mars at perihelion and Earth at aphelion). See

configuration of Mars-Earth-Sun at opposition and the angular diameter of Mars seen from Earth

in following diagram.

(30%)

1 1M Br a ( e ) a ( e ) =1.5(1-0.093) (1+0.017)=0.3435 AU = 51.525 106 km (20%)

Linier diameter of Mars, D =6786.8 km

So the angular diameter of Mars is, =6

6786 8 3600 0075 27 1689

51 525 10 2

o. . " ..

(20%)

Earth

Sun


7

Compared to the angular diameter of the Moon, =30=1800, we can conclude that the ratio of

Mars diameter is only 0.15 times the angular diameter of the Moon.

Ratio of surface area of Mars to that of the Moon seen from the Earth is:

2 224

2

27 16892 278 10

1800

M M

B B

R R " .f .

R R "

=0.02% (30%)

Thefrore, the conclusion is WRONG, Mars is never be seen as big as the Moon.

2. (Celestial Mechanics (M)) On January 15, 2010 there was an annular eclipse, where at

maximum 97% of Solar disk is covered by the Moon. At that time Earth was very close to its

perihelion. If the semi major axis of Earth orbit is 150 million km, Solar radius is 700 000 km,

eccentricity is 0.017 and the radius of the Moon is 1738 km, what is the distance of the Moon

from the observer on the Earth surface ?

(Show the formula(s), calculation and final results)

Solution :

Annular eclipse is maximum if the Earth is at perihelion

Solar distance : d = a(1-e) = 147.6 million km. (20%)

If 97% of Solar disk was covered, b is the angular radius of the Moon , M is the angular radius

of the Sun, then

M =(20.7/147.6) rad = 32.61 arc minutes (30%)

2 20 97b M.

b = 32.12 arc minutes = 0.009342 rad (30%)

Distance of the Moon : 1738/0.009342 = 372083 km (20%)

3. (Celestial mechanics (M)) Assume that the Moon moves around the Earth in a circular orbit. Imagine you stand on the Moons surface, on the side facing the Earth. How would you see the phase of the Earth compared to the phase of the Moon as seen by an observer


8

standing on the Earths surface, facing the Moon? Give your answer by completing the following table and show a sketch that describes the situations.

You may draw the following symbols in the table :

New Full Gibbous Half Cresent

Answer :

(75%)

Moons phase as seen from the Earth

Earths phase as seen from the Moon

New Moon Full Earth

Full Moon New Earth

Gibbous Moon Crescent Earth

Half Moon Half Earth

Crescent Moon Gibbous Earth

(25%)


9

Practical Test

Plan A (Clear night)

Problem:

Time: 15 minutes

Night observation using telescope with eye piece 1. Find and look carefully Jupiter planet (RA2000: 23h 56m 32s and Dec2000: -02

0 06 59) 2. How many satellites of Jupiter and draw with the proper orientation on the provided answer

sheet 3. You can and select or use different (provided) eye-pieces 4. Give marking of the N-E directions on the answer sheet

Plan B (cloudy night)

Problem:

Time: 15 minutes

1. Show the position of the bright stars and a planet as listed below on the answer sheet

a. Antares (alpha Scorpii) (RA2000: 16h 29m 24.461s Dec2000: -260 25 55.209)

b. Vega (alpha Lyra) (RA2000: 18h 36m 56.336s Dec2000: +380 47 01.290)

c. Arcturus (alpha Aquilae) (RA2000: 14h 15m 39.672s Dec2000 : +190 10 56.67)

d. Mars (RA2000: 14h 05m 07s Dec2000: -120 50 53)

2. Pointing the telescope to Jupiter planet. Show to the jury


10

Table of constants and units

Constants Symbols Values

Solar luminosity L

3.86 x 1026 Js-1 = 3.86 x 1026 watt

Solar constant F

1.368 x 103 Jm-2

Universal gravitational constant G 6.67 x 10-11 Nm2kg-2

Earths gravitational acceleration g 9.8 ms-2

Earth mass M 5.98 x 1024kg

Lunar mass M

7.34 x 1022kg

Solar mass M 1.99 x 1030kg

Stefan-Boltzmann constant 5.68 x 10-8 Js-1m-2K-4

Astronomical Unit AU 1.496 x 1011 m

Moon-Earth average distance d 3.84 x 108 m

Earth radius R 6.37 x 106 m

Solar radius R

6.96 x 108 m

Sidereal year 365.256 days = 3.16 x 107 s

Solar effective temperature T

5880 K

Light year Ly 9.5 x 1015 minutes

Parsec pc 3.26 Ly

Documents

Problem Solution Ieso2010 r7