In April 2, 2008 an active region 0987 appeared at South 8o and West 75o

Soal tentang matahari

1. In April 2, 2008 one of telescope (10 cm diameter, f/10) at Bosscha Observatory observed an active region 0987 (following to the NOAA number) appeared at South 8o and West 40o of solar disk. It was recorded with CCD SBIG ST-8 (1600x1200 pixels, 9 m). It sized 5x4 pixels. According to almanac the solar diameter is 32 arc min. How large the corrected area of sunspot in unit of millionth of solar hemisphere (msh)?a. 15

b. 35

c. 55

d. 75 (optional for multiple choices)Answer:Pixel size () = 206265 x pixel size (m) / (1000 x focal length (mm)) 5

= 206265 x 9 / (1000 x 1000) = 1.86 arc sec

= 1.86 x 720 km = 1339.2 km

Solar diameter = 32 x 60 arc sec / 1.86 arc sec/pixel = 1034 pixelsRadius = 1034/2 = 517 pixel = 700000 km

( 1 pixel ~ 700000/517 ~ 1354 km

(estimate 40 deg W; R=517 pix = 90 deg ( 40 deg = 40x517/90=229.7 = 230 (330) ( 8 deg = 8x517/90 = 45.95 = 46 (100)Area should be corrected to the foreshorteningsDisk area

= R2Hemispheric area = 2 R2Am = As . 106 / (2 R2 cos B cos (L))

5

= 20 pixel2 x 106/ (2 x 3.14 x 5172 x cos 8 x cos 40)

( cos 40 = 0.76604444311897803520239265055542

( cos 8 = 0.99026806874157031508377486734485

= 15.69 msh

The result is the same if pixel converts to mThe area of a flare is commonly measured in units of one millionth of the visible solar hemisphere. One millionth of the hemisphere is approximately 3 million square kilometers (3.04x1016 cm2 ~ 2 R2 / 106; R= 700 000 km).

Am = As . 106 / (2 R2 cos B cos L)As = 2 R2 cos B cos (L) / 106 = 2 x 3.14 x 5172 x cos 8 x cos 40 / 106 = 1.27 pixel2 = 1.27 x 1.862 arc sec2 = 4.39 (5.86) arc sec2 x 7202 km

= 2.27 million square km.In the past, the unit of heliocentric square degrees was also used. This is the square area which subtends, at the solar centre, an angle of one degree of arc on each side. One square degree so defined is equivalent to an area of 48.5 hemispheric millionths. An even more obscure unit is the solid angle subtended at the Earth by one square degree on the Sun at the centre of the solar disc. This solid angle is (16.73)2 square seconds of arc.

(Short question; tingkat kesulitan: mudah)

2. Assuming that the Sun and the Earth is a perfect blackbody, estimate the mean temperature (K) and energy per area (watt/m2) of the outer Earths atmosphere that determined by the solar radiation? Answer:

Lsun = Tsun4 . 4 Rsun2

2Fsun = Lsun/4d2 (where d = 1 AU)

Total flux received by the Earth:

Fearth = Fsun . Rearth2Learth= Tearth4 . 4 Rearth2 Learth = Fearth ( Tearth4 . 4 Rearth2 = Fsun . Rearth2 = Lsun/4d2 . Rearth2 = Tsun4 . 4 Rsun2 / 4d2 . Rearth2( Tearth4 = Tsun4 . Rsun2/4d2

4With Tsun = 5780 K; Rsun=6.9x105 km; Rearth=6400 km; d= 1.49x108 km, then

Tearth = 278.77 K ~ 300 K

Eearth = Lsun / (4 d2) = 4 Rsun2 Tsun 4/ 4 d2

4= (6,96x1010 cm)2 5,67x10-5 erg cm-2K-1s-1 x (5780 K)4 / (1,49597892x1013 cm)2= 1369,8 watt/m2

Eearth = 4 Tearth4(Short question; Tingkat kesulitan: mudah)

Soal tentang binary star and instrumentasi

3. A binary system consists of two stars A and B with a brightness ratio of 2. However, they are not unresolved and we see them as a point of magnitude +5.0. What are the magnitude of the star A and B, respectively?

a. 5.44 and 6.19b. 7.13 and 8.0

c. 6.89 and 4.78d. 7.39 and 4.01

(optional for multiple choices)Answer:

Assume A and B have flux FA and FBFormula: m1 m2 = -2.5 log F1/F2 ( 5 = -2.5 log ((FA+FB)/FA)

5FA = 2 FBThen, 5 = -2.5 log 3/2 + MA ( MA = 5.44

5

5 = -2.5 log 3 + MB( MB= 6.19

(Short question; Tingkat kesulitan: mudah)

4. A variable star exhibits light variations of 60 sec and longer with amplitude of 0.02 magnitudes. In session A at 7 UT, the readout time is 2 sec and the read noise is 17 electrons. In session B at 8 UT, the readout time of CCD camera is 10 sec and the read noise is 3 electrons. In session C at 10 UT, the readout time is 17 sec and the read noise is 4 electrons. In the last session D, the readout time is 17 sec and the read noise is 30 electrons In 10 sec, the star produces 1500 counts. Assuming the star occupies 1 pixel and the gain is 3 electrons per count. The sky count can be neglected. Which is session could have the highest possibility to resolve the variability?a. A

b. B

c. C

d. D

(optional for multiple choices)Session A: For 2 sec readout, star produce 1500 counts every 10 sec or 150 count/sec.

( Nstar = 2 sec x 150 count/sec = 300 count and ( Nsky is neglected

2Formula Noise ( 2 = gain x Nstar + Npixel x Nnoise2

= 3 x 300 + 1 x 172 = 900 + 289 = 34.48

2Formula signal to noise ( S/N = gain x Nstar / = 900/34.48 = 26.10m = -2.5 log F + C ( m = -2.5 log F/F F = -2.5 F/(F ln 10)

F ~ gain x Nstar = 3 x 300 count = 900 count

F ~

=2.5 x / (gain x Nstar ln 10) = 2.5 x 34.48/ (3 x 300 x 2.3025) = 0.0416 2Session B: for 10 sec readout

( Nstar = 10 x 150 = 1500

( 2 = 3 x 1500 + 1 x 9 = 4509 ( = 67.15( S/N = 3 x 1500 / 67.15 = 67.01

m = 2.5 x / (gain x Nstar ln 10) = 2.5 x 67.15/ (3 x 1500 x 2.3025) = 0.016

2Session C: For 17 sec readout

( Nstar = 17 x 150 = 2550

2 = 3 x 2550 + 1 x 16 ( = 87.56( S/N = 3 x 2550 / 87.56 = 87.37 (highest S/N ratio)m = 2.5 x / (gain x Nstar ln 10) = 2.5 x 87.56 / (3 x 2550 x 2.3025) = 0.012

2Session D: For 17 sec readout

( Nstar = 17 x 150 = 2550

2 = 3 x 2550 + 1 x 900 = 8550 ( = 92.47( S/N = 3 x 2550 / 92.47 = 82.73m = 2.5 x / (gain x Nstar ln 10) = 2.5 x 82.73/ (3 x 2550 x 2.3025) = 0.012

2So, the session C could resolve the amplitude variable star due to the largest S/N (and smaller m)

(Long question; Tingkat kesulitan: sedang)5. One student carried out observations with a CCD camera of a star whose magnitude is known to be +10 visual mag. In the final CCD images, the star occupies 10 pixels of which data number are (30, 36, 37, 29, 40, 35, 38, 31, 39, 32). The nearby reference pixels are (4, 5, 7, 1, 3, 9, 2, 8). Assuming the gain is 1 electron per count.a. Determine the total flux of the star (with the background subtracted) with its error using the above data number

b. Now lets assume he observe a second star with the same CCD. The data numbers from the second star are (120, 130, 114, 124, 117, 116, 139, 122, 110, 101) and those from the nearby reference pixels are (7, 8, 6, 3, 5, 9, 8, 7, 4). Assuming exposure times for the two stars are identical. Whats the magnitude of this second star and its error? Answer:

a. Formula: Nstar = Nsum Nmed x Npixel

Nsky2 = (ni - )2 / N (for nearby reference pixels)2 = gain x Nstar + Npixel x Nsky2

4

Nsum = (30, 36, 37, 29, 40, 35, 38, 31, 39, 32) = 347

Nmed = median (4, 5, 7, 1, 3, 9, 2, 8) = 4.5

Npixel= 10

( Nstar1 = 347 4.5 x 10 = 302

Nsky12 = 58.875/8 = 7.359375 ( Nsky1 = sqrt (7.359375) = 2.713

( 2 = 1 x 302 + 10 x 7.359375 = 375.5938 ( 1 = sqrt(375.5938) = 19.38

2b. Second starNsum = 1193

Nmed = 7

( Nstar2 = 1193 7 x 10 = 1123

( Nsky2 = 1.66

( 2 = 33.9

Since Nstar ~ F; m1 = 10; F1=302; F2=1123

2m2 10 = -2.5 log 1123/302 ( m = 8.57

2(Short question; Tingkat kesulitan: sedang)