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At a distance of \(7.00 \times 10^{12} \mathrm{m}\) from a star, the intensity of the radiation from the star is 15.4 \(\mathrm{W} / \mathrm{m}^{2} .\) Assuming that the star radiates uniformly in all directions, what is the total power output of the star?

Short Answer

Expert verified
The total power output is approximately \(2.98 \times 10^{27} \mathrm{W}\).

Step by step solution

01

Understand the Relationship Between Intensity and Power

The intensity of radiation \( I \) at a distance \( r \) from a star is related to the total power \( P \) emitted by the star by the formula \( I = \frac{P}{A} \), where \( A \) is the surface area of a sphere with radius \( r \).
02

Expression for the Surface Area of a Sphere

The surface area \( A \) of a sphere with radius \( r \) is given by the formula \( A = 4\pi r^2 \).
03

Substitute for Intensity Formula

Substitute the surface area expression into the intensity formula: \[ I = \frac{P}{4\pi r^2} \] which can be rearranged to solve for power: \[ P = I \times 4\pi r^2 \]
04

Plug Values Into Power Formula

Substitute the given values into the power formula: \[ P = 15.4 \, \mathrm{W/m^2} \times 4\pi (7.00 \times 10^{12} \, \mathrm{m})^2 \]
05

Calculate the Total Power

Evaluate the expression to find total power \( P \): \[ P = 15.4 \, \mathrm{W/m^2} \times 4 \times 3.14159 \times (7.00 \times 10^{12} \, \mathrm{m})^2 \] \[ P \approx 2.98 \times 10^{27} \, \mathrm{W} \]
06

Conclusion

The total power output of the star is calculated to be approximately \( 2.98 \times 10^{27} \, \mathrm{W} \).

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Key Concepts

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

Power Output Calculation
Calculating the power output of a celestial object, like a star, requires understanding how intensity ties into total energy emitted. Intensity is defined as power per unit area. So, if we know the intensity at a certain distance, we can figure out the total power output by using an equation based on how much area that intensity spreads over.

The relationship between intensity, power, and surface area comes down to the formula:
  • \( I = \frac{P}{A} \)
where \( I \) is intensity, \( P \) is the power, and \( A \) is the surface area through which the power is distributed.
To calculate total power, rearrange this formula to solve for \( P \):
  • \( P = I \times A \)
This expression allows you to multiply the known intensity by the area to find how much total energy, or power, a star emits.
Spherical Surface Area
The intensity of radiation emitted by objects like stars usually diminishes as you move further away. This decrease relates directly to the geometry of how that light spreads out over space. Specifically, it spreads out over a spherical shape radiating outward from the star.

The surface area of this sphere is essential to our calculations. It's given by the formula:
  • \( A = 4\pi r^2 \)
Here, \( r \) is the radius of the sphere. In this context, \( r \) represents the distance from the star to the point where the intensity is being measured.
If you increase \( r \), the surface area grows significantly because it scales with the square of \( r \). That's why the intensity lowers with distance; the same amount of power spreads over a larger area. This understanding is crucial for deriving power from intensity.
Radiation Formula
The radiation formula connects the physical concept of how radiation spreads and the mathematical description used for calculations. When we talk about a star radiating energy uniformly in all directions, it means that the energy is equally distributed over the surface area of a virtual sphere centered on the star.

This uniform distribution is expressed in the formula:
  • \( I = \frac{P}{4\pi r^2} \)
Here, \( I \) is the intensity, \( r \) is the distance from the star, and \( P \) is the total power output. By rearranging the formula, we derive:
  • \( P = I \times 4\pi r^2 \)
This calculates the total power, showing how it depends on intensity and the sphere's radius. Understanding this formula not only helps in astrophysics but also in many engineering applications where energy distribution is important.

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Most popular questions from this chapter

A light wire is tightly stretched with tension \(F .\) Trans- verse traveling waves of amplitude \(A\) and wavelength \(\lambda_{1}\) carry average power \(P_{\mathrm{av}, 1}=0.400 \mathrm{W}\) . If the wavelength of the waves is doubled, so \(\lambda_{2}=2 \lambda_{1},\) while the tension \(F\) and amplitude \(A\) are not altered, what then is the average power \(P_{\mathrm{av}, 2}\) carried by the waves?

\mathrm{A} jet plane at takeoff can produce sound of intensity 10.0 \(\mathrm{W} / \mathrm{m}^{2} at 30.0 \)\mathrm{m} away. But you prefer the tranquil sound of normal conversation, which is 1.0$\mu \mathrm{W} / \mathrm{m}^{2} . Assume that the plane behaves like a point source of sound. (a) What is the closest distance you should live from the airport runway to preserve your peace of mind? (b) What intensity from the jet does your friend experience if she lives twice as far from the runway as you do (c) What power of sound does the jet produce at takeoff?

A piano tuner stretches a steel piano wire with a tension of 800 \(\mathrm{N}\) . The steel wire is 0.400 \(\mathrm{m}\) long and has a mass of 3.00 \(\mathrm{g}\) . (a) What is the frequency of its fundamental mode of vibration? (b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to \(10,000 \mathrm{Hz} ?\)

A 1.50 -m string of weight 0.0125 \(\mathrm{N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W\) . Neglect the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t)=(8.50 \mathrm{mm}) \cos \left(172 \mathrm{m}^{-1} x-4830 \mathrm{s}^{-1} t\right)$$ Assume that the tension of the string is constant and equal to \(W\) . (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W ?\) (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for wayes traveling down the string?

The speed of sound in air at \(20^{\circ} \mathrm{C}\) is 344 \(\mathrm{m} / \mathrm{s} .\) (a) What is the wavelength of a sound wave with a frequency of 784 \(\mathrm{Hz}\) , corresponding to the note \(\mathrm{G}_{5}\) on a piano, and how many milliseconds does each vibration take? (b) What is the wavelength of a sound wave one octave higher than the note in part (a)?

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