Problem 29 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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 \( r \).

Step 2: Expression for the Surface Area of a Sphere

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

Step 3: 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: \[ P = I \times 4\pi r^2 \]

Step 4: Plug Values Into Power Formula

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

Step 5: Calculate the Total Power

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

Step 6: Conclusion

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

Key Concepts

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.