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Astro MarathonProblem-3

 

#Astro_Marathon
#Problem 3
Submission Deadline: 10.00AM, 27December
*Please write your full name, institution name and class before your solution.

একই open cluster এ অবস্থিত দুইটি নক্ষত্র এর আপাত উজ্জ্বলতার মধ্যে পার্থক্য 2 magnitude. নক্ষত্রগুলোর পৃষ্ঠের তাপমাত্রা যথাক্রমে 6000K এবং 5000K. নক্ষত্রদুটির ব্যাসার্ধের অনুপাত নির্ণয় কর।

The difference in brightness between two main sequence stars in an open cluster is 2 magnitudes. Their effective temperatures are 6000K and 5000K respectively. Estimate the ratio of their radii.

 

 

 

 

Turja Roy

Notre Dame College

Class : 11

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Ok

Abdullah Mohammad Mahi

NOTRE DAME COLLEGE

Class:11

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this could be wrong, but this is what i got

Nafi

Chittagong college,class 11

for m1-m2 be positive ,m1 must be greater,F1 must be less and as they are both in same cluster ,L1 must be less and thus T1 must be  less as well.

5000k=T1 ,6000k=T2

m1-m2=-2.5log(F1/F2) ,as in the same cluster distance from earth is same  , so F1/F2=L1/L2

we get, 2=-2.5log((R1^2 xT1^4)/(R2^2/T2^4)

T1=5000K and T2=6000k,

((10^-2)^(1/2.5))/0.482=R1^2/R2^2

(0.3288)^1/2=R1/R2

R1/R2=0.5731

so the hotter star is larger.

Meraj Hossain Promit

MC College, Sylhet

Class - 11

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Sayed Shafaat Mahmud

Class:10

Rajuk Uttara Model College

 

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Denote the stars by S_1, S_2 respectively.

Assume m_1 > m_2, meaning m_1 - m_2 = 2 and S_2 is brighter than S_1.

Now for main sequence stars, L \propto R^2T^4, where L, R, T denote the star's luminosity, radius and surface temperature respectively. So, \frac{L_1}{L_2} = \frac{{R_1}^2{T_1}^4}{{R_2}^2{T_2}^4}.

Now, from Pogson's law,

m_1 - m_2 = -2.5log\frac{L_1}{L_2}

\rightarrow 2 = -2.5log\frac{{R_1}^2{T_1}^4}{{R_2}^2{T_2}^4}

\rightarrow 2 = -5log\frac{{R_1}{T_1}^2}{{R_2}{T_2}^2}

\rightarrow 2 = -5log\frac{{R_1}}{{R_2}} + (-5log\frac{{T_1}^2}{{T_2}^2})

\rightarrow 2 = -5log\frac{{R_1}}{{R_2}} - 10log\frac{{T_1}}{{T_2}}

\rightarrow 5log\frac{{R_1}}{{R_2}} = -2 - 10log\frac{{T_1}}{{T_2}}

\rightarrow log\frac{{R_1}}{{R_2}} = -0.4 - 2log\frac{{T_1}}{{T_2}}

\rightarrow \frac{{R_1}}{{R_2}} = 10^{-0.4 - 2log\frac{{T_1}}{{T_2}}}

Now, the main problem here is whether we assume T_1 > T_2 or T_1 < T_2. However, as both stars are in the main sequence, we can assume that the brightest star is the hotter one, meaning T_1 = 5000 K, T_2 = 6000 K.

Putting in these values one can get \frac{{R_1}}{{R_2}} \approx 0.5733

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