GATE-2018 (24) :Wyllie – Time average equation

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

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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

 

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