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CSIR-Net 2015 June

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127) consider two planets ‘A’, ‘B’ with the following characteristics. The Relation between T_{1}  and T_{2} is

 

Distance from the sun

Radius of the planet

Incident solar flux density

Equivalent temperature

Planet A

d_{1}

r_{1}

F_{1}

T_{1}

Planet B

d_{2}=4d_{1}

r_{2}=2r_{1}

F_{2}

T_{2}

 

(Special Thanks to Chandrasekhar, ANU) 


Solution:

 Luminosity is the total amount of energy emitted by the star per unit volume

Luminosity is mainly depends on two things

 

1.       Radius of the star

 

2.     Surface temperature

Luminosity(L)= (4 \pi R^{2})(\sigma T^{4})

 

 When we are calculating the Luminosity for the planets the constant’s are negligible and then the Luminosity

Luminosity(L)= R^{2}T^{4}

 

From the above equation we can say that T^{4}\propto \frac{1}{R^{2}}

 

 For the Planet A

T^{4}_{1}\propto \frac{1}{R^{2}_{1}}

For the Planet B

 

T^{4}_{2}\propto \frac{1}{R^{2}_{2}}

From the above two equations we can write like this

\frac{T^{4}_{1}}{T^{4}_{2}}=\frac{R^{2}_{2}}{R^{2}_{1}}

 

\frac{T^{4}_{1}}{T^{4}_{2}}=\frac{(2R^{2}_{1})}{R^{2}_{1}}

 

\frac{T^{4}_{1}}{T^{4}_{2}}=\frac{4R^{2}_{1}}{R^{2}_{1}}

 

\frac{T^{4}_{1}}{T^{4}_{2}}=4

 

(\frac{T_{1}}{T_{2}})^{4}=4

 

\frac{T_{1}}{T_{2}}=(4)^{\frac{1}{4}}

 

\frac{T_{1}}{T_{2}}=(2^{2})^{\frac{1}{4}}

 

\frac{T_{1}}{T_{2}}=(2)^{\frac{1}{2}}

 

\frac{T_{1}}{T_{2}}=\sqrt{2}

 

T_{1}=\sqrt{2}T_{2}

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