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Chapter 33: Problem 28

The average intensity of the solar radiation that strikes normally on a surface just outside Earth's atmosphere is \(1.4 \mathrm{~kW} / \mathrm{m}^{2}\). (a) What radiation pressure \(p_{r}\) is exerted on this surface, assuming complete absorption? (b) For comparison, find the ratio of \(p_{r}\) to Earth's sea-level atmospheric pressure, which is \(1.0 \times 10^{5} \mathrm{~Pa}\).

Short Answer

Expert verified
(a) \(4.67 \times 10^{-6} \mathrm{~Pa}\); (b) Ratio is \(4.67 \times 10^{-11}\).

Step by step solution

01

Understand the concept

The radiation pressure (\(p_{r}\)) on a surface is the pressure exerted when radiation (light) is absorbed by the surface. The formula for radiation pressure is \(p_{r} = \frac{I}{c}\), where \(I\) is the intensity of the radiation, and \(c\) is the speed of light in vacuum, approximately \(3.00 \times 10^8\) m/s.
02

Calculate the radiation pressure

Using the formula for radiation pressure, we substitute the given intensity \(I = 1.4\, \mathrm{kW/m^2} = 1400 \, \mathrm{W/m^2}\) and the speed of light \(c = 3.00 \times 10^8 \) m/s into the equation: \[p_{r} = \frac{1400 \, \mathrm{W/m^2}}{3.00 \times 10^8 \, \mathrm{m/s}}\]This simplifies to \[p_{r} = 4.67 \times 10^{-6} \, \mathrm{Pa}\]
03

Calculate the ratio of radiation pressure to atmospheric pressure

The ratio of the radiation pressure to the atmospheric pressure can be calculated by dividing the radiation pressure by Earth's sea-level atmospheric pressure. The atmospheric pressure \(p_a = 1.0 \times 10^{5} \, \mathrm{Pa}\). Thus, the ratio is:\[\frac{p_{r}}{p_{a}} = \frac{4.67 \times 10^{-6} \, \mathrm{Pa}}{1.0 \times 10^{5} \, \mathrm{Pa}}\]This gives a ratio of \[4.67 \times 10^{-11}\]
04

Conclusion

The radiation pressure exerted is \(4.67 \times 10^{-6} \mathrm{~Pa}\), and it is extremely small compared to Earth's sea-level atmospheric pressure, with a ratio of \(4.67 \times 10^{-11}\).

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Key Concepts

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

Radiation Intensity
Radiation intensity is a vital concept when studying the effect of solar radiation on Earth. It refers to the power received per unit area from the sun. In simpler terms, it measures how much energy from the sun hits a square meter on the Earth's surface per second.

The standard unit for measuring radiation intensity is watts per square meter (W/m虏). For instance, just outside Earth's atmosphere, the average solar radiation intensity is about 1.4 kW/m虏 or 1400 W/m虏. This amount varies slightly depending on the Earth's position relative to the sun.

Understanding radiation intensity helps in determining how much solar power can be harnessed, and how this impacts Earth's climate and weather patterns. It's like measuring how intense the sunlight is that reaches us, just before entering our protective atmospheric layer, influencing various systems:
  • Solar energy absorption by surfaces
  • Impact on temperature and climate
  • Influence on atmospheric processes
Speed of Light
The speed of light is a fundamental constant crucial to physics and our understanding of the universe. This constant, denoted by "c", represents the speed at which light travels through a vacuum. It is approximately 299,792,458 meters per second, often rounded to 3.00 脳 10鈦� m/s for simplicity in calculations.

Light's speed is not just significant in scientific equations; it also defines the foundation for the relationship between energy, mass, and space-time. In the context of radiation pressure, the speed of light is a key variable required to calculate how much force light exerts on surfaces it strikes. This relationship is expressed in the formula for radiation pressure:

\[ p_{r} = \frac{I}{c} \]
Here, "I" represents the radiation intensity, and "c" is the speed of light. This formula helps determine the pressure exerted on any surface due to light absorption, highlighting light's momentum transfer properties.
  • Fundamental constant in physics
  • Integral to calculating radiation pressure
  • Underpins theories on relativity and quantum mechanics
Atmospheric Pressure
Atmospheric pressure is the force per unit area exerted against a surface by the weight of air above that surface in the Earth's atmosphere. It plays a critical role in determining weather patterns and how natural and artificial systems behave on Earth.

At sea level, the standard atmospheric pressure is approximately 1.0 脳 10鈦� pascals (Pa), which can vary with altitude and weather conditions. This pressure comes from the gravitational pull of Earth, compressing atmospheric gases, like squeezing air in a balloon.

In comparison to radiation pressure, atmospheric pressure is significantly larger, as seen in exercises comparing values. By comparing radiation pressure with atmospheric pressure, you often arrive at very small ratios, as the Earth's atmosphere exerts a much greater force due to its mass and gravity:
  • Effects on weather and climate
  • Guides our understanding of thermodynamics
  • Influences biological processes on Earth

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Most popular questions from this chapter

An isotropic point source emits light at wavelength \(500 \mathrm{~nm}\), at the rate of \(200 \mathrm{~W}\). A light detector is positioned \(400 \mathrm{~m}\) from the source. What is the maximum rate \(\partial B / \partial t\) at which the magnetic component of the light changes with time at the detector's location?

Frank D. Drake, an investigator in the SETI (Search for Extra-Terrestrial Intelligence) program, once said that the large radio telescope in Arecibo, Puerto Rico (Fig. 33-36), "can detect a signal which lays down on the entire surface of the earth a power of only one picowatt." (a) What is the power that would be received by the Arecibo antenna for such a signal? The antenna diameter is \(300 \mathrm{~m}\). (b) What would be the power of an isotropic source at the center of our galaxy that could provide such a signal? The galactic center is \(2.2 \times 10^{4}\) ly away. A light-year is the distance light travels in one year.

Calculate the (a) upper and (b) lower limit of the Brewster angle for white light incident on fused quartz. Assume that the wavelength limits of the light are 400 and \(700 \mathrm{~nm}\).

(a) How long does it take a radio signal to travel \(150 \mathrm{~km}\) from a transmitter to a receiving antenna? (b) We see a full Moon by reflected sunlight. How much earlier did the light that enters our eye leave the Sun? The Earth-Moon and Earth-Sun distances are \(3.8 \times 10^{5} \mathrm{~km}\) and \(1.5 \times 10^{8} \mathrm{~km}\), respectively. (c) What is the round-trip travel time for light between Earth and a spaceship orbiting Saturn, \(1.3 \times 10^{9} \mathrm{~km}\) distant? (d) The Crab nebula, which is about 6500 light-years (ly) distant, is thought to be the result of a supernova explosion recorded by Chinese astronomers in A.D. 1054 . In approximately what year did the explosion actually occur? (When we look into the night sky, we are effectively looking back in time.)

At a beach the light is generally partially polarized due to reflections off sand and water. At a particular beach on a particular day near sundown, the horizontal component of the electric field vector is \(2.3\) times the vertical component. A standing sunbather puts on polarizing sunglasses; the glasses eliminate the horizontal field component. (a) What fraction of the light intensity received before the glasses were put on now reaches the sunbather's eyes? (b) The sunbather, still wearing the glasses, lies on his side. What fraction of the light intensity received before the glasses were put on now reaches his eyes?
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