Question

The average intensity of sunlight at the Earth's surface is approximately \(1400 \mathrm{~W} / \mathrm{m}^{2}\), if the Sun is directly overhead. The average distance between the Earth and the Sun is \(1.50 \cdot 10^{11} \mathrm{~m}\). What is the average power emitted by the Sun? a) \(99.9 \cdot 10^{25} \mathrm{~W}\) c) \(6.3 \cdot 10^{27} \mathrm{~W}\) e) \(5.9 \cdot 10^{29} \mathrm{~W}\) b) \(4.0 \cdot 10^{26} \mathrm{~W}\) d) \(4.3 \cdot 10^{28} \mathrm{~W}\)

Short answer

a) \(2.0 \times 10^{26} \mathrm{~W}\) b) \(4.0 \times 10^{26} \mathrm{~W}\) c) \(6.0 \times 10^{26} \mathrm{~W}\) d) \(8.0 \times 10^{26} \mathrm{~W}\) The closest value to the average power emitted by the Sun is: b) \(4.0 \times 10^{26} \mathrm{~W}\)

Step-by-step solution

Write down the given intensity and average distance from the Sun to Earth

We are provided with the average solar intensity at Earth's surface when the Sun is directly overhead: \(I = 1400 \mathrm{~W}/\mathrm{m}^2\) The average distance from the Earth to the Sun is given as: \(D = 1.5 \times 10^{11} \mathrm{~m}\)

Calculate the area of the sphere with Earth's distance from the Sun as a radius

Since the intensity is distributed uniformly on a sphere around the Sun, we can calculate the area of this sphere with radius 'D' by using the formula: \(A = 4 \pi D^2\) \(A = 4 \pi (1.5 \times 10^{11})^2\) \(A = 2.83 \times 10^{23} \mathrm{~m}^2\)

Rearrange the intensity formula to find the power

Now, let's use the intensity formula to find the power 'P': \(I = \frac{P}{A}\) Rearranging for 'P': \(P = I \times A\)

Substitute the known values and calculate the average power emitted by the Sun

Substitute the known values 'I' and 'A' into the equation: \(P = (1400 \mathrm{~W}/\mathrm{m}^2) \times (2.83 \times 10^{23} \mathrm{~m}^2)\) \(P = 3.96 \times 10^{26} \mathrm{~W}\)

Compare the calculated power to the given options

By comparing the calculated power with the given options, the closest answer is: b) \(4.0 \times 10^{26} \mathrm{~W}\)

Key concepts

Solar Intensity

Understanding solar intensity is akin to grasping the strength of sunlight. In practical terms, it refers to how much solar energy hits a given area. Think of this as the sun’s power reaching us here on Earth, and it's measured in watts per square meter (W/m²).

When the sun is directly overhead, the intensity at the Earth's surface is around 1400 W/m². This value changes based on time of day, location, and atmospheric conditions. However, for the sake of calculations in physics, we often use average or standard values, such as the 'solar constant' — which is an average measure of solar intensity outside of Earth’s atmosphere.

Intensity Formula Physics

In the realm of physics, the intensity formula is the bedrock for understanding how energy is distributed across different areas. It's presented as:

\[\begin{equation}I = \frac{P}{A}\end{equation}\]

where 'I' is intensity, 'P' is power, and 'A' is the area over which the power is spread. You can rearrange this equation to find any of the three variables, provided you have the other two. In the context of our solar intensity problem, we rearrange it to solve for the power emitted by the Sun, making it:

\[\begin{equation}P = I \times A\end{equation}\]
Simply put, if you know how intense the sunlight is over a certain area, you can use this relationship to determine the total power output. It's how we quantify the Sun's might!

Sphere Surface Area Calculation

Let's dive into the sphere surface area calculation, which is an essential concept in many physics problems involving spherical shapes. The formula to calculate the surface area 'A' of a sphere is given by:

\[\begin{equation}A = 4\pi r^2\end{equation}\]
where 'r' is the radius of the sphere. In astronomical terms, when we talk about the power of the Sun reaching Earth, we envision an imaginary sphere around the Sun with a radius equal to the distance from the Sun to the Earth. This sphere's surface is where all the sunlight spreads out through space.
Calculating the surface area of this sphere helps us understand the total area that the Sun's light touches, which is crucial when determining the solar power emitted by using the intensity formula. By calculating this area, we ended up with a vast sphere around the Sun, reaffirming not only the immense power of our star but also the expansive scale of our solar system.