Question

The Sun emits electromagnetic waves (including light) equally in all directions. The intensity of the waves at Earth's upper atmosphere is \(1.4 \mathrm{kW} / \mathrm{m}^{2} .\) At what rate does the Sun emit electromagnetic waves? (In other words, what is the power output?)

Short answer

The Sun emits electromagnetic waves at a power rate of approximately \(3.85 \times 10^{26}\) watts.

Step-by-step solution

Understanding the problem

The problem asks us to find the rate at which the Sun emits electromagnetic waves in terms of power. We know the intensity of the waves at Earth's atmosphere is given, and we need to find the power emitted by the Sun.

Apply the formula for intensity

Intensity \(I\) is defined as the power \(P\) per unit area \(A\). The formula for intensity is given by:\[ I = \frac{P}{A} \] We are given \(I = 1.4 \text{ kW/m}^2\). The power \(P\) is what we need to find, and the area \(A\) is the surface area over which this power is distributed.

Determine the area

The power emitted by the Sun radiates out in a sphere centered on the Sun. At the distance of the Earth from the Sun, this forms a sphere with a radius equal to the average distance between the Earth and the Sun, approximately \(1.496 \times 10^{11} \text{ m}\). The surface area \(A\) of a sphere is given by:\[ A = 4\pi r^2 \]Therefore, we substitute for \(r\):\[ A = 4\pi (1.496 \times 10^{11})^2 \]

Solve for power

Substitute the expression for area \(A\) back into the intensity formula and solve for \(P\):\[1.4 \times 10^3 = \frac{P}{4\pi (1.496 \times 10^{11})^2} \]Multiply both sides by \(4\pi (1.496 \times 10^{11})^2\) to solve for \(P\):\[ P = 1.4 \times 10^3 \times 4\pi (1.496 \times 10^{11})^2 \]Calculate to find the value of \(P\).

Final calculation

Perform the calculations:\[ P = 1.4 \times 10^3 \times 4 \times 3.1416 \times (1.496 \times 10^{11})^2 \]\[ P \approx 3.85 \times 10^{26} \text{ W} \]This is the rate at which the Sun emits electromagnetic waves.

Key concepts

Solar Power Output

The power output of the Sun refers to the total energy it emits in the form of electromagnetic radiation every second. This is a colossal amount of energy, and it's what powers life on Earth, driving processes like photosynthesis and the weather cycles. To put it in perspective, the Sun's power output, also known as its luminosity, is approximately \[3.85 \times 10^{26} \text{ watts}\]This incredible figure shows just how much energy the Sun pumps out, sustaining the solar system.

• **Explained in watts**: A watt is a unit energy flow, representing one joule per second.
• **Sustaining life**: This massive energy output is why Earth can support life as we know it, maintaining stable climate and weather patterns.

Intensity Formula

The intensity of solar radiation is a measure of how much power the Sun emits per unit area. This value decreases the farther you go from the Sun because the energy spreads over a larger area. The intensity is calculated using the formula:\[ I = \frac{P}{A} \]where \(P\) is the power output and \(A\) is the area over which the power is distributed. For Earth, the intensity of solar radiation at the upper atmosphere is about\[1.4 \text{ kW/m}^2\]

• **Units to remember**: Intensity is often expressed in kilowatts per square meter \(\text{kW/m}^2\).
• **Significance for Earth**: This intensity value is crucial for understanding how much solar energy the Earth receives and will influence climate and energy calculations.

Surface Area of a Sphere

When considering how sunlight spreads out in space, it's helpful to think of a sphere with the Sun at its center. As the Sun's energy radiates in all directions, it covers an ever-expanding spherical surface. The formula for the surface area of a sphere is:\[ A = 4\pi r^2 \]where \(r\) is the radius of the sphere.

This formula is key for finding out how intensity changes with distance.

• **Sphere basics**: Multiply \(4\pi\) by the radius squared to get the area.
• **Application in space**: As the distance from the Sun increases, the surface area increases, causing intensity to decrease.

Distance Earth to Sun

The average distance from the Earth to the Sun is a significant factor in calculating solar intensity on Earth. This distance is known as an astronomical unit (AU) and measures about\(1.496 \times 10^{11}\) meters. This distance serves as the radius in the sphere's surface area calculation around the Sun.

• **Impact on solar calculations**: The larger the radius, the lower the intensity for a given power output.
• **Seasonal variations**: Though the average distance is a constant, the Earth's elliptical orbit causes slight variations in the solar intensity received throughout the year.