Question

Betelgeuse is a red supergiant star in the constellation Orion. It radiates heat at the rate of 2.70 × 1030 W and has a surface temperature of 3000 K. Assuming that it is a perfect emitter, what is the radius of Betelgeuse? The Stefan-Boltzmann constant is 5.670 × 10⁻⁸ W/m²⋅K⁴.

A) 2.30 × 10¹¹ m
B) 1.40 × 10¹¹ m
C) 1.90 × 10¹¹ m
D) 7.80 × 10¹⁰ m
E) 8.70 × 10¹⁰ m

Answer 1

To find the radius of Betelgeuse, we use the Stefan-Boltzmann law and calculate the result to be approximately 1.40 × 10^11 meters.

Step-by-step explanation:

The radius of Betelgeuse, a red supergiant star in the constellation Orion, can be calculated using the given heat radiation rate and surface temperature, along with the Stefan-Boltzmann constant. Using the formula for luminosity L = 4πR^2σT^4 where L is the luminosity, R is the radius, σ is the Stefan-Boltzmann constant, and T is the surface temperature in kelvins, we can solve for R. Plugging in the values for L (2.70 × 10^30 W), σ (5.670 × 10^\u22128 W/m²•K^4), and T (3000 K) and rearranging the formula to solve for R, we get the expression R = √(L / (4πσT^4)). When we do the calculation, we find that the radius of Betelgeuse is approximately 1.40 × 10^11 meters.