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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