1 Answer

Nishant Nagwani · Answer 1 · 2022-03-14T21:43:05+0000

Answer:

the surface temperature of Regulus A is 11724.13 K

Step-by-step explanation:

Given the data in the question;

Sun's surface temperature T = 5800 K

total radiated power, relative to the sun is; P/P
$_{sun$ =
1.5(M/M
$_{sun$
$)^{3.5$

The star Regulus A is a bluish main-sequence star with mass 3.8M
$_{sun$ and radius 3.1R
$_{sun$ .

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
$_{sun$ be power related sun.

Ratio of power radiated by star and power radiated by sun;

P/P
$_{sun$ = eσAT⁴ / eσA
$_{sun$ T
$_{sun$ ⁴

we know that AREA A = πR²

we input the formula for area

P/P
$_{sun$ = eσ(πR²)T⁴ / eσ(π(R
$_{sun$ )²)T
$_{sun$ ⁴

such that we now have;

P/P
$_{sun$ = R²T⁴ / R
$_{sun$ ²T
$_{sun$ ⁴

given that P/P
$_{sun$ =
1.5(M/M
$_{sun$
$)^{3.5$ , we substitute

1.5(M/M
$_{sun$
$)^{3.5$ = R²T⁴ / R
$_{sun$ ²T
$_{sun$ ⁴

we find temperature of the star T

T = 5800 ×
1.5
M/M
$_{sun$
$)^{3.5$ (R
$_{sun$ ²/R²)
$]^{1/4$

Given that; mass M is 3.8M
$_{sun$ and radius R is 3.1R
$_{sun$ .

we substitute

T = 5800 ×
1.5
3.8M
$_{sun$ /M
$_{sun$
$)^{3.5$ (R
$_{sun$ ²/( 3.1R
$_{sun$ )²)
$]^{1/4$

T = 5800 ×
1.5
3.8
$)^{3.5$ ( 1/( 3.1)²)
$]^{1/4$

T = 5800 ×
1.5( 106.9652 ) ( 1/(9.61)
$]^{1/4$

T = 5800 ×
16.69592
$]^{1/4$

T = 5800 × 2.02140152

T = 11724.13 K

Therefore, the surface temperature of Regulus A is 11724.13 K

Please log in or register to add a comment.