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At a distance of \(7.00 \times 10^{12}\) m from a star, the intensity of the radiation from the star is 15.4 W/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 of the star is approximately \(9.49 \times 10^{27}\) Watts.

Step by step solution

01

Understanding the Relationship

To find the total power output of the star, we need to use the concept of intensity, which is the power per unit area. The equation relating intensity (I), power (P), and radius (r) is given by: \[ I = \frac{P}{A} \]where A is the surface area of a sphere of radius r. This gives the equation \[ A = 4 \pi r^2 \].
02

Substituting Known Values

Substitute the values for intensity (I = 15.4 \text{ W/m}^2), radius (r = 7.00 \times 10^{12} \text{ m}), and the equation for surface area into the equation: \[ I = \frac{P}{4 \pi r^2} \] which becomes \[ 15.4 = \frac{P}{4 \pi (7.00 \times 10^{12})^2} \].
03

Rearranging the Equation

Rearrange the equation to solve for the total power output (P): \[ P = I \times 4 \pi r^2 \].
04

Calculating the Total Power

Now calculate the total power by substituting the intensity and radius into the rearranged formula: \[ P = 15.4 \times 4 \pi \times (7.00 \times 10^{12})^2 \].
05

Performing the Calculation

Perform the calculation: - First calculate the surface area: \[ 4 \pi (7.00 \times 10^{12})^2 = 4 \pi \times 4.9 \times 10^{25} \approx 6.16 \times 10^{26} \text{ m}^2\] if approximated \[ 15.4 \times 6.16 \times 10^{26} \approx 9.49 \times 10^{27} \text{ Watts} \].

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Key Concepts

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

Intensity of radiation
Intensity of radiation is a fundamental concept that describes how much energy in the form of light or radiation hits a surface per unit area. This measurement, denoted by "I", is expressed in watts per square meter (W/m\(^2\)). It specifically measures the brightness or strength of radiation as perceived or measured at a given distance from a source.
For example, in the context of stars, the intensity of radiation diminishes as one moves further away from the source. This is because the same amount of energy spreads out over a larger surface area. In the exercise provided, we learn that the star's radiation at a certain distance has an intensity of 15.4 W/m\(^2\), which is a critical data point for determining the total power output. By understanding intensity, we can infer details about the power a star emits just by observing how bright it appears from our location.
In calculations, intensity is used along with the formula \( I = \frac{P}{A} \), where "P" is the total power and "A" is the area over which this power is distributed, illustrating the concept quantitatively.
Surface area of a sphere
The surface area of a sphere is calculated using the formula \( A = 4 \pi r^2 \), where "r" is the radius of the sphere. This formula is essential when analyzing celestial bodies or any spherical object emitting energy uniformly across its surface.
Understanding the surface area is crucial in astronomy for evaluating how the intensity of radiation from celestial objects, like stars, decreases with distance. As radiation moves outward equally in all directions, it covers a spherical surface that grows larger the further it gets from the source.
In the provided problem, the surface area of a sphere at a distance of the star is calculated to help transform the intensity into total power output. The method involves substituting the radius of the sphere (given as \(7.00 \times 10^{12}\) m in the exercise) into the area formula, hence allowing us to find out how widely the star's energy is distributed.
Calculating total power
Calculating the total power output of a star involves using both the intensity of radiation and the surface area of the sphere through which it spreads. The equation \( P = I \times A \) comes from rearranging the original intensity formula \( I = \frac{P}{A} \), emphasizing the purpose of each term: \( P \) is the power, \( I \) is the intensity, and \( A \) is the area over which the power is dispersed.
In practical terms, knowing a star's intensity at a certain distance and the distance itself allows astronomers to calculate the star's total power or luminosity. This data is crucial for understanding the star's energy and influence in its surrounding space.
Using the values given in the original problem—intensity (15.4 W/m\(^2\)) and the calculated surface area \((6.16 \times 10^{26} \text{ m}^2)\)—one can easily compute the total power output of the star: \( P = 15.4 \times 6.16 \times 10^{26} \approx 9.49 \times 10^{27} \text{ Watts} \). This represents the star's energy output, illustrating the fundamental application of these concepts in real astronomy.

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Most popular questions from this chapter

(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wavelength \(\lambda\). Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) \(x = \lambda/2\), (ii) \(x = \lambda\)/4, and (iii) \(x = \lambda\)8, from the left-hand end of the string. (b) At each of the points in part (a), what is the amplitude of the motion? (c) At each of the points in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?

(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wavelength \(\lambda\). Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) \(x = \lambda/2\), (ii) \(x = \lambda /4\), and (iii) \(x = \lambda /8\), from the left-hand end of the string. (b) At each of the points in part (a), what is the amplitude of the motion? (c) At each of the points in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?

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