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ivanzaharov [21]

3 years ago

11

Most stars are main-sequence stars, a group of stars for which size, mass, surface temperature, and radiated power are closely r

elated. The sun, for instance, is a yellow main sequence star with a surface temperature of 5800 K. For a main-sequence star whose mass M is more than twice that of the sun, the total radiated power, relative to the sun, is approximately P/Psun = 1.5(M/Msun)3.5. The star Regulus A is a bluish main-sequence star with mass 3.8Msun and radius 3.1Rsun. What is the surface temperature of Regulus A?

1 answer:

quester [9]3 years ago

4 0

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 $_{sun$ = $($ 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

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