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Chapter 16: Problem 71

Suppose a spherical particle of mass \(m\) and radius \(R\) in space absorbs light of intensity \(I\) for time \(t\). (a) How much work does the radiation pressure do to accelerate the particle from rest in the given time it absorbs the light? (b) How much energy carried by the electromagnetic waves is absorbed by the particle over this time based on the radiant energy incident on the particle?

Short Answer

Expert verified
(a) The work done by the radiation pressure to accelerate the particle from rest in the given time is: \(W = \frac{I}{3 \times 10^8 m/s} \times \pi R^2 \times \frac{1}{2}\frac{I}{3 \times 10^8 m/s}t^2\) (b) The energy absorbed by the particle from the electromagnetic waves over this time is: \(E_r = I \times \pi R^2 \times t\)

Step by step solution

01

Calculate the radiation pressure exerted on the spherical particle

Radiation pressure, \(P_r\), is the force exerted by electromagnetic radiation on a surface. It can be calculated as: \(P_r = \frac{I}{c}\), where \(I\) is the intensity of the light and \(c\) is the speed of light. Given the intensity of light \(I\), we can find the radiation pressure as: \(P_r = \frac{I}{3 \times 10^8 m/s}\)
02

Calculate the force exerted on the particle due to radiation pressure

According to Newton's second law, the force acting on a sphere due to radiation pressure can be calculated as: \(F = P_r \times A\), where \(A\) is the cross-sectional area of the sphere, which can be calculated as: \(A = \pi R^2\), where \(R\) is the radius of the spherical particle. Now we can find the force exerted on the particle: \(F = \frac{I}{3 \times 10^8 m/s} \times \pi R^2\)
03

Calculate the work done by the radiation pressure on the particle

The work done by the radiation pressure can be calculated as: \(W = F \times d\), where \(d\) is the displacement caused by the force exerted by the radiation pressure. Since the particle starts from rest, we can use the equation: \(d = \frac{1}{2}at^2\), where \(a\) is the acceleration and \(t\) is the time. From Newton's second law, \(F = ma\), and hence \(a = \frac{F}{m}\). So the displacement can be written as: \(d = \frac{1}{2}\frac{F}{m}t^2\) Now we find the work done: \(W = F \times d = F \times \frac{1}{2}\frac{F}{m}t^2\)
04

Calculate the radiant energy incident on the spherical particle

The radiant energy incident on the particle, \(E_r\), can be calculated as: \(E_r = I \times A \times t\) We already have the values for \(I\), \(A\), and \(t\), so we can substitute them into the equation: \(E_r = I \times \pi R^2 \times t\) (a) The work done by the radiation pressure to accelerate the particle from rest in the given time is: \(W = F \times \frac{1}{2}\frac{F}{m}t^2 = \frac{I}{3 \times 10^8 m/s} \times \pi R^2 \times \frac{1}{2}\frac{I}{3 \times 10^8 m/s}t^2\) (b) The energy absorbed by the particle from the electromagnetic waves over this time is: \(E_r = I \times \pi R^2 \times t\)

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Key Concepts

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

Newton's Second Law
When exploring the behavior of objects in motion, Newton's Second Law is essential. It's elegantly summed up in the formula:
\[ F = ma \]
This principal formula tells us that the force (\( F \) exerted on an object is equal to the mass (\( m \) of the object multiplied by its acceleration (\( a \) ). In the context of radiation pressure, imagine that a particle in space is struck by light (which, surprisingly, exerts a force). The resulting force causes the particle to accelerate. By calculating the force due to radiation pressure (\( F = P_r \times A \) ) on the particle, and knowing the mass (\( m \) ), we figure out the particle's acceleration.

Applying Newton's Law to Radiation Pressure

As the step-by-step solution demonstrates, after determining the force with the radiation pressure and the area of the particle, the acceleration is derived, allowing us to establish the distance (\( d \) ) the particle travels. This distance helps calculate the work done by the radiation, tying Newton's second law to the concept of energy transfer via radiation.
Electromagnetic Waves
Electromagnetic waves are ripples of electromagnetic fields that transport energy through space or through a medium. Light is the most familiar form of electromagnetic waves. These waves are characterized by their frequency and wavelength and travel at the universal constant speed of light, denoted by \( c \) in equations, which is approximately \( 3 \times 10^8 \text{m/s} \).

Illuminating Radiation Pressure

When electromagnetic waves strike an object, they exert pressure, known as radiation pressure. This is precisely what happens when light hits our spherical particle in space 鈥� it pushes on it. The light鈥檚 intensity (\( I \) ), which is the power per unit area, directly influences this pressure. It's this relationship between electromagnetic waves and mechanical force that allows us to analyze phenomena like solar sails on spacecraft or the behavior of dust particles in space.
Radiant Energy
Radiant energy is the energy carried by electromagnetic waves. Whether it's sunlight warming your face or microwaves heating your food, it's all about the transfer of energy. The intensity of this energy is a measure of how much power these waves transfer per unit area, and it directly affects the amount of pressure these waves can exert on an object.

Energy Absorption of a Particle

When dealing with a particle absorbing light, the radiant energy incident on it is defined by the product of the light's intensity, the particle's cross-sectional area, and the duration of exposure. Radiant energy isn't just a theoretical concept; it has practical implications like powering solar cells or influencing the motion of particles in space, as shown in the exercise. Calculating the radiant energy (\( E_r \) ) gives us a way to quantify the amount of energy a particle can absorb over a certain period, which is directly related to the work done by radiation pressure.

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Most popular questions from this chapter

An outdoor WiFi unit for a picnic area has a \(100-\mathrm{mW}\) output and a range of about \(30 \mathrm{m}\). What output power would reduce its range to \(12 \mathrm{m}\) for use with the same devices as before? Assume there are no obstacles in the way and that microwaves into the ground are simply absorbed.

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