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Measurements on two stars indicate that Star \(\mathrm{X}\) has a surface temperature of \(5727^{\circ} \mathrm{C}\) and Star \(\mathrm{Y}\) has a surface temperature of \(11727^{\circ} \mathrm{C}\). If both stars have the same radius, what is the ratio of the luminosity (total power output) of Star Y to the luminosity of Star X? Both stars can be considered to have an emissivity of \(\underline{1.0 .}\)

Short Answer

Expert verified
The ratio of the luminosity of Star Y to the luminosity of Star X is 16:1.

Step by step solution

01

Convert temperatures to Kelvin

First, convert the given temperatures from Celsius to Kelvin. The formula to convert Celsius to Kelvin is \(K = C + 273.15\). So, for Star X: \(K_x = 5727 + 273.15 = 6000 K\), and for Star Y: \(K_y = 11727 + 273.15 = 12000 K\).
02

Apply the Stefan-Boltzmann law

Stefan-Boltzmann law states that the power radiated per unit area of a black body is directly proportional to the fourth power of its temperature. Mathematically, it can be represented as \(L = 4\pi R^2 \sigma T^4\), where L is Luminosity, R is Radius, \(\sigma\) is Stefan-Boltzmann constant and T is temperature. Since both stars have the same radius and emissivity, when calculating the ratio, these terms will cancel out. Thus, the ratio can be simplified to \(L_{ratio} = \frac{T_{Y}^4}{T_{X}^4}\).
03

Calculate the ratio

Substitute the values of \(T_y\) and \(T_x\) obtained in Step 1 into the equation obtained in Step 2: \(L_{ratio} = \frac{(12000 K)^4}{(6000 K)^4} = \frac{2^4}{1} = 16\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Black Body Radiation
It's essential to understand the concept of black body radiation in astrophysics as it forms the basis for studying various stellar properties. A black body is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. It also reemits radiation at a characteristic rate, dependent on its temperature, which is given by the Stefan-Boltzmann law.

To comprehend how this relates to stars, envision them as black bodies radiating energy generated from nuclear reactions occurring within their cores. The energy output is emitted across a spectrum of wavelengths, but the peak of this emission shifts according to the star's surface temperature. This peak output can be calculated precisely using Planck's law, and overall, the Stefan-Boltzmann law enables us to relate the amount of energy emitted to the temperature of a star's surface.
Temperature Conversion

When dealing with temperatures in astronomical contexts, it's common to use the Kelvin scale. It's imperative because the Kelvin scale starts at absolute zero, which is the theoretical point at which particles have the least amount of energy possible. In contrast, Celsius is a scale often used for everyday temperatures but is not appropriate for the high temperatures characteristic of stars.

To convert from Celsius to Kelvin, as the step in the exercise informed, one must add 273.15 to the Celsius temperature. This conversion is crucial for observations, calculations, and simulations in astrophysics, including those involving luminosity or thermal radiation.

Luminosity of Stars
In astronomy, luminosity refers to the total amount of energy emitted by a star, galaxy, or other astronomical object per unit time. It's a measure of a star's power output, and it's directly influenced by both the star's radius and its surface temperature. The Stefan-Boltzmann law serves as a pivotal formula for determining the luminosity of a star, assuming we already know the star's radius and its effective surface temperature. As illustrated in the textbook solution, when comparing the luminosity of two stars with identical radii and emissivities, one can isolate the variable of temperature to find the luminosity ratio; the temperatures, raised to the fourth power, indicate how much more luminous one star is over another.

The relation between luminosity and temperature is highly non-linear due to the exponent in the law's formula, which demonstrates why a star with double the surface temperature of another radiates much more than twice the energy. This concept is critical for astrophysicists to understand the various stellar classifications and evolutionary stages in a star’s lifecycle.

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Most popular questions from this chapter

ecp A sprinter of mass \(m\) accelerates uniformly from rest to velocity vin \(t\) seconds. (a) Write a symbolic expression for the instantaneous mechanical power \(\mathcal{P}\) required by the sprinter in terms of force \(F\) and velocity \(u\). (b) Use Newton's second law and a kinematics equation for the velocity at any time to obtain an expression for the instantaneous power in terms of \(m, a\), and \(t\) only. (c) If a \(75.0-\mathrm{kg}\) sprinter reaches a speed of \(11.0 \mathrm{~m} / \mathrm{s}\) in \(5.00 \mathrm{~s}\), calculate the sprinter's acceleration, assuming it to be constant. (d) Calculate the \(75.0\) -kg sprinter's instantaneous mechanical power as a function of time \(\ell\) and \((\mathrm{e})\) give the maximum rate at which he burns Calories during the sprint, assuming \(25 \%\) efficiency of conversion form food energy to mechanical energy.

In a showdown on the streets of Laredo, the good guy drops a \(5.0-g\) silver bullet at a temperature of \(20^{\circ} \mathrm{C}\) into a \(100-\mathrm{cm}^{3}\) cup of water at \(90^{\circ} \mathrm{C}\). Simultaneously, the bad guy drops a \(5.0-\mathrm{g}\) copper bullet at the same initial temperature into an identical cup of water. Which one ends the showdown with the coolest cup of water in the West? Neglect any energy transfer into or away from the container.

B. For bacteriological testing of water supplies and in medical clinics, samples must routinely be incubated for \(24 \mathrm{~h}\) at \(37^{\circ} \mathrm{C}\). A standard constant-temperature bath with electric heating and thermostatic control is not suitable in developing nations without continuously operating electric power lines, Peace Corps volunteer and MIT engineenAmy Smith invented a low-cost, low-maintenance incubator to fill the need. The device consists of a foaminsulated box containing several packets of a waxy material that melts at \(37.0^{\circ} \mathrm{C}\), interspersed among tubes, dishes, or bottles containing the test samples and growth medium (food for bacteria). Outside the box, the waxy material is first melted by a stove or solar energy collector. Then it is put into the box to keep the test samples warm as it solidifies. The heat of fusion of the phasechange material is \(205 \mathrm{~kJ} / \mathrm{kg}\). Model the insulation as a panel with surface area \(0.490 \mathrm{~m}^{2}\), thickness \(9.50 \mathrm{~cm}\), and conductivity \(0.0120 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\). Assume the exterior temperature is \(23.0^{\circ} \mathrm{C}\) for \(12.0 \mathrm{~h}\) and \(16.0^{\circ} \mathrm{C}\) for \(12.0 \mathrm{~h}\). (a) What mass of the waxy material is required to conduct the bacteriological test? (b) Explain why your calculation can be done without knowing the mass of the test samples or of the insulation.

A \(50-g\) ice cube at \(0^{\circ} \mathrm{C}\) is heated until \(45 \mathrm{~g}\) has become water at \(100^{\circ} \mathrm{C}\) and \(5.0 \mathrm{~g}\) has become steam at \(100^{\circ} \mathrm{C}\). How much energy was added to accomplish the transformation?

A \(100 \mathrm{~g}\) aluminum calorimeter contains \(250 \mathrm{~g}\) of water. The two substances are in thermal equilibrium at \(10^{\circ} \mathrm{C}\). Two metallic blocks are placed in the water. One is a \(50-\mathrm{g}\) piece of copper at \(80^{\circ} \mathrm{C}\). The other sample has a mass of \(70 \mathrm{~g}\) and is originally at a temperature of \(100^{\circ} \mathrm{C}\). The entire system stabilizes at a final temperature of \(20^{\circ} \mathrm{C}\). Determine the specific heat of the unknown second sample.

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