Star A is twice as luminous as star B but has half the temperature of star B.How large is the radius of star A compared to star B?
A) 23/2 as large
B) 22 as large
C) 25/2 as large
D) 23 as large

Question

Quizplus · Accepted Answer

We can use the Stefan-Boltzmann law which states that the luminosity is proportional to the fourth power of the temperature and the square of the radius.

Let L_a be the luminosity of star A, L_b be the luminosity of star B, T_a be the temperature of star A, and T_b be the temperature of star B. Let R_a be the radius of star A and R_b be the radius of star B.

We have:

L_a = 2 L_b          (star A is twice as luminous as star B)
T_a = 1/2 T_b       (star A has half the temperature of star B)

From the Stefan-Boltzmann law, we know that:

L_a/L_b = (T_a/T_b)^4 (R_a/R_b)^2

Substituting the given values, we get:

2 = (1/2)^4 (R_a/R_b)^2

Simplifying, we get:

R_a/R_b = sqrt(2)

R_a is (sqrt(2)) times R_b.

Converting this to a ratio, we get:

R_a/R_b = 2^(1/2) / 1

R_a is 2^(1/2) times R_b.

Simplifying this, we get:

R_a = (sqrt(2)) R_b

Therefore, the radius of star A is 25/2 (or 12.5) times as large as the radius of star B.

Hence, the answer is C) 25/2 as large.

Star a Is Twice as Luminous as Star B but Has