Answer:
the surface temperature of Regulus A is 11724.13 K
Explanation:
Given the data in the question;
Sun's surface temperature T = 5800 K
total radiated power, relative to the sun is; P/P[tex]_{sun[/tex] = [tex]([/tex] 1.5(M/M[tex]_{sun[/tex] [tex])^{3.5[/tex]
The star Regulus A is a bluish main-sequence star with mass 3.8M[tex]_{sun[/tex] and radius 3.1R[tex]_{sun[/tex] .
First, we determine the value power emitted by the sun or sun as follows;
P = eσAT⁴
where P is the power, e is surface emissivity, σ is Stefan Boltzmann, A is area and T is temperature.
so, lets assume emissivity of star and sun is same;
let p be power related to star and p[tex]_{sun[/tex] be power related sun.
Ratio of power radiated by star and power radiated by sun;
P/P[tex]_{sun[/tex] = eσAT⁴ / eσA[tex]_{sun[/tex]T[tex]_{sun[/tex]⁴
we know that AREA A = πR²
we input the formula for area
P/P[tex]_{sun[/tex] = eσ(πR²)T⁴ / eσ(π(R[tex]_{sun[/tex])²)T[tex]_{sun[/tex]⁴
such that we now have;
P/P[tex]_{sun[/tex] = R²T⁴ / R[tex]_{sun[/tex]²T[tex]_{sun[/tex]⁴
given that P/P[tex]_{sun[/tex] = [tex]([/tex] 1.5(M/M[tex]_{sun[/tex] [tex])^{3.5[/tex], we substitute
[tex]([/tex] 1.5(M/M[tex]_{sun[/tex] [tex])^{3.5[/tex] = R²T⁴ / R[tex]_{sun[/tex]²T[tex]_{sun[/tex]⁴
we find temperature of the star T
T = 5800 × [tex][[/tex] 1.5[tex]([/tex]M/M[tex]_{sun[/tex][tex])^{3.5[/tex] (R[tex]_{sun[/tex]²/R²)[tex]]^{1/4[/tex]
Given that; mass M is 3.8M[tex]_{sun[/tex] and radius R is 3.1R[tex]_{sun[/tex] .
we substitute
T = 5800 × [tex][[/tex] 1.5[tex]([/tex]3.8M[tex]_{sun[/tex]/M[tex]_{sun[/tex][tex])^{3.5[/tex] (R[tex]_{sun[/tex]²/( 3.1R[tex]_{sun[/tex] )²)[tex]]^{1/4[/tex]
T = 5800 × [tex][[/tex] 1.5[tex]([/tex]3.8[tex])^{3.5[/tex] ( 1/( 3.1)²)[tex]]^{1/4[/tex]
T = 5800 × [tex][[/tex] 1.5( 106.9652 ) ( 1/(9.61) [tex]]^{1/4[/tex]
T = 5800 × [tex][[/tex] 16.69592 [tex]]^{1/4[/tex]
T = 5800 × 2.02140152
T = 11724.13 K
Therefore, the surface temperature of Regulus A is 11724.13 K
