Problem 27 - University Physics with Modern Physics

Question

If the intensity of direct sunlight at a point on the earth's surface is \(0.78 \mathrm{kW} / \mathrm{m}^{2},\) find \((\mathrm{a})\) the average momentum density (momentum per unit volume) in the sunlight and (b) the average momentum flow rate in the sunlight.

Step-by-Step Solution

Answer

(a) Average momentum density is \( \frac{13}{3 \times 10^7} \, \text{kg/m}^2 \cdot \text{s} \). (b) Average momentum flow rate is \( 1.3 \times 10^{-5} \, \text{kg/s}^2 \).

Step 1: Understand Relationship between Intensity and Energy Density

The intensity of light is related to the energy density (u) of the light by the equation: \( I = c u \), where \( I \) is the intensity and \( c \) is the speed of light in vacuum. We can rearrange this equation to find the energy density: \( u = \frac{I}{c} \).

Step 2: Calculate Energy Density

Given that the intensity \( I = 0.78 \, \text{kW/m}^2 = 780 \, \text{W/m}^2 \), and knowing \( c = 3 \times 10^8 \, \text{m/s} \), substitute these values into the formula: \( u = \frac{780}{3 \times 10^8} \, \text{J/m}^3 \).

Step 3: Calculate Average Momentum Density

The average momentum density \( p \) is related to the energy density via: \( p = \frac{u}{c} \). Use the energy density calculated in the previous step: \( p = \frac{780}{(3 \times 10^8)^2} \). This simplifies to \( p = \frac{780}{9 \times 10^{16}} = \frac{13}{3 \times 10^7} \, \text{kg/m}^2 \cdot \text{s} \).

Step 4: Calculate Average Momentum Flow Rate

The average momentum flow rate (momentum per unit time) is equivalent to the radiation pressure times the speed of light. For electromagnetic radiation, this can be written as \( P = \frac{I}{c} \). Use the given intensity \( I = 780 \, \text{W/m}^2 \): \( P = \frac{780}{3 \times 10^8} = \frac{13}{5 \times 10^7} \, \text{N/m}^2 = 1.3 \times 10^{-5} \, \text{N/m}^2 \). Since this is already in units of momentum flow rate, it represents \( 1.3 \times 10^{-5} \, \text{kg/s}^2 \) through each square meter.

Key Concepts

Energy Density

In discussing electromagnetic radiation, energy density is a core component that allows us to understand how much energy is contained within a specific volume. We use the relationship between intensity and energy density, which is described by the equation: \( I = c u \). Here, \( I \) represents the intensity of the light in watts per square meter, \( c \) is the speed of light, and \( u \) is the energy density in joules per cubic meter.

To find the energy density, the equation can be rearranged to become \( u = \frac{I}{c} \). Given sunlight intensity at \( 780 \, \text{W/m}^2 \) and the speed of light as \( 3 \times 10^8 \, \text{m/s} \), we substitute these values into the formula to get \( u = \frac{780}{3 \times 10^8} \, \text{J/m}^3 \).

• Energy density provides insight into how intense the light is, effectively measuring the energy available within that space.
• A low energy density implies less energy in the radiation, impacting how much can be absorbed or used for processes like photosynthesis.

Intensity of Light

The intensity of light is central to understanding how much power is transferred through a given area. Defined as power per unit area, it is measured in watts per square meter \( \text{W/m}^2 \).

Consider the intensity of sunlight calculated in the original problem at \( 780 \, \text{W/m}^2 \). This value indicates the concentration of energy hitting the earth's surface directly from the sun.

• Higher intensity means more energy per unit area and can significantly affect temperature and energy dynamics in the environment.
• Intensity is also essential in determining luminosity and overall energy output in astrological settings.

Knowing the intensity helps in calculating the energy density and ultimately the momentum carried by photons in light, key in understanding how light interacts with matter and contributes to solar power devices.

Radiation Pressure

Radiation pressure is exerted upon a surface by light or other electromagnetic radiation. It arises because photons, which make up light, possess momentum. When light strikes a surface, it can transfer this momentum, resulting in a small but measurable force.

• In the case of sunlight, the radiation pressure is calculated by \( P = \frac{I}{c} \), where \( I \) is the intensity and \( c \) is the speed of light.
• For the intensity of \( 780 \, \text{W/m}^2 \), the radiation pressure is \( 1.3 \times 10^{-5} \, \text{N/m}^2 \).

This pressure is minute, yet it is critical in understanding solar sails in spacecraft technologies, where radiation pressure from the sun's light is used to propel spacecraft without fuel. Such a concept leverages the multiplication of this tiny force over an extended area to generate forward momentum and illustrates the practical applications of radiation pressure in technology.