Math Problem Statement

The Sun has a surface temperature of 5800 K and a radius of 1 Solar Radius. Another star might have a surface temperature of 5000 K and a radius of 2 Solar Radii. Using Stefan-Boltzmann's Law, how many times more or less luminous is this star than the Sun? Also, find the wavelength of the radio waves produced by KISS-FM 99.5 MHz radio station.

Solution

Let's break down the two parts of this problem:

(a) Stefan-Boltzmann Law Problem

We are given two stars: the Sun and another star. We need to compare their luminosities using Stefan-Boltzmann's Law, which states:

L=4πR2σT4L = 4\pi R^2 \sigma T^4

Where:

  • LL is luminosity,
  • RR is the radius of the star,
  • TT is the surface temperature, and
  • σ\sigma is the Stefan-Boltzmann constant.

The question asks to find how many times more or less luminous the second star is compared to the Sun. To solve this, we need to compute the ratio of luminosities:

LstarL=4πRstar2σTstar44πR2σT4=(RstarR)2(TstarT)4\frac{L_{\text{star}}}{L_{\odot}} = \frac{4 \pi R_{\text{star}}^2 \sigma T_{\text{star}}^4}{4 \pi R_{\odot}^2 \sigma T_{\odot}^4} = \left(\frac{R_{\text{star}}}{R_{\odot}}\right)^2 \left(\frac{T_{\text{star}}}{T_{\odot}}\right)^4

We are given:

  • R=1RR_{\odot} = 1 R_{\odot}, T=5800KT_{\odot} = 5800 \, \text{K},
  • Rstar=2RR_{\text{star}} = 2 R_{\odot}, Tstar=5000KT_{\text{star}} = 5000 \, \text{K}.

Substituting these into the equation:

= 2^2 \times \left(\frac{5000}{5800}\right)^4 = 4 \times \left(0.8621\right)^4$$ Now calculating the fourth power of $$0.8621$$: $$0.8621^4 \approx 0.552$$ Thus, the luminosity ratio is: $$\frac{L_{\text{star}}}{L_{\odot}} = 4 \times 0.552 = 2.208$$ So, the star is about **2.21 times more luminous** than the Sun. --- ### (b) Wavelength of the Radio Waves We are asked to find the wavelength of the radio waves produced by **KISS-FM 99.5**, which broadcasts at a frequency of 99.5 MHz. We use the equation: $$c = \lambda \times f$$ Where: - $$c$$ is the speed of light ($$3 \times 10^8 \, \text{m/s}$$), - $$\lambda$$ is the wavelength, and - $$f$$ is the frequency (99.5 MHz = $$99.5 \times 10^6 \, \text{Hz}$$). Rearranging for $$\lambda$$: $$\lambda = \frac{c}{f} = \frac{3 \times 10^8}{99.5 \times 10^6} = \frac{3 \times 10^8}{9.95 \times 10^7} \approx 3.015 \, \text{meters}$$ Thus, the wavelength of the radio waves is approximately **3.015 meters**. --- Do you have any questions or would you like more details on any part of this solution? Here are some follow-up questions: 1. How would the luminosity change if the temperature of the second star was higher than the Sun's? 2. What would happen to the luminosity if the radius of the star was halved? 3. Can Stefan-Boltzmann’s law be used for objects other than stars? 4. How does frequency relate to energy in electromagnetic waves? 5. Why does the wavelength decrease as the frequency increases? **Tip:** For stars, small changes in temperature can cause large changes in luminosity due to the $$T^4$$ dependence in Stefan-Boltzmann's law.

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Math Problem Analysis

Mathematical Concepts

Thermodynamics
Electromagnetic Waves

Formulas

Stefan-Boltzmann Law: L = 4πR²σT⁴
Speed of Light: c = λ × f

Theorems

Stefan-Boltzmann Law

Suitable Grade Level

College-level Physics