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Chapter 12: Problem 88

When the earth is closest to the sun, we have winter in the northern hemisphere. Explain why. Also explain why we have summer in the northern hemisphere when the earth is farthest away from the sun.

Short Answer

Expert verified
Answer: The seasons in the Northern Hemisphere are caused by a combination of Earth's axial tilt and elliptical orbit around the Sun. Although Earth is closer to the Sun during winter (perihelion) and farthest during summer (aphelion), the more significant factor is the axial tilt. This tilt affects the distribution of sunlight, with the Northern Hemisphere being tilted away from the Sun during winter and towards the Sun during summer. This results in cooler temperatures and winter conditions when Earth is closest to the Sun and warmer temperatures and summer conditions when Earth is farthest from the Sun.

Step by step solution

01

Understand Earth's axial tilt

Earth rotates around an imaginary line called the axis. The axis is tilted at an angle of approximately 23.5 degrees with respect to the plane of Earth's orbit around the Sun. Due to this tilt, different parts of Earth receive different amounts of sunlight at different times of the year, resulting in the change of seasons.
02

Learn about Earth's elliptical orbit

Earth's orbit around the Sun is not a perfect circle, but slightly elliptical. This means that Earth's distance from the Sun varies throughout the year. The part of the orbit where Earth is closest to the Sun is called perihelion, and the part where it is farthest away is called aphelion.
03

Realize the effect of axial tilt on sunlight distribution

Due to Earth's axial tilt, the Northern Hemisphere is tilted toward the Sun during the summer months, which causes more direct sunlight to reach the surface, leading to warmer temperatures. Conversely, during the winter months, the Northern Hemisphere is tilted away from the Sun, causing sunlight to spread over a larger area and arrive at a shallower angle, which results in cooler temperatures.
04

Connect perihelion with winter in the Northern Hemisphere

Perihelion occurs around January 3rd each year, which corresponds with winter in the Northern Hemisphere. Though Earth is closer to the Sun at this point, the axial tilt's effect on sunlight distribution is more significant. Since the Northern Hemisphere is tilted away from the Sun during this time, the sunlight is more spread out and arrives at a shallower angle, causing cooler temperatures and winter conditions.
05

Connect aphelion with summer in the Northern Hemisphere

Aphelion occurs around July 4th each year, coinciding with summer in the Northern Hemisphere. Although Earth is farthest from the Sun at this point, the axial tilt makes the Northern Hemisphere tilted towards the Sun. As a result, sunlight is more concentrated and arrives at a steeper angle, leading to warmer temperatures and summer conditions. In conclusion, the distance between Earth and the Sun does not determine the seasons directly; instead, it is the combination of Earth's axial tilt and elliptical orbit that causes the variation in sunlight distribution and temperature, leading to the different seasons in the Northern Hemisphere.

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Most popular questions from this chapter

The reflectivity of aluminum coated with lead sulfate is \(0.35\) for radiation at wavelengths less than \(3 \mu \mathrm{m}\) and \(0.95\) for radiation greater than \(3 \mu \mathrm{m}\). Determine the average reflectivity of this surface for solar radiation \((T \approx 5800 \mathrm{~K})\) and radiation coming from surfaces at room temperature \((T \approx 300 \mathrm{~K})\). Also, determine the emissivity and absorptivity of this surface at both temperatures. Do you think this material is suitable for use in solar collectors?

The spectral emissivity function of an opaque surface at \(1000 \mathrm{~K}\) is approximated as $$ \varepsilon_{\lambda}= \begin{cases}\varepsilon_{1}=0.4, & 0 \leq \lambda<2 \mu \mathrm{m} \\ \varepsilon_{2}=0.7, & 2 \mu \mathrm{m} \leq \lambda<6 \mu \mathrm{m} \\ \varepsilon_{3}=0.3, & 6 \mu \mathrm{m} \leq \lambda<\infty\end{cases} $$ Determine the average emissivity of the surface and the rate of radiation emission from the surface, in \(\mathrm{W} / \mathrm{m}^{2}\).

A surface is exposed to solar radiation. The direct and diffuse components of solar radiation are 350 and \(250 \mathrm{~W} / \mathrm{m}^{2}\), and the direct radiation makes a \(35^{\circ}\) angle with the normal of the surface. The solar absorptivity and the emissivity of the surface are \(0.24\) and \(0.41\), respectively. If the surface is observed to be at \(315 \mathrm{~K}\) and the effective sky temperature is \(256 \mathrm{~K}\), the net rate of radiation heat transfer to the surface is (a) \(-129 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(-44 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(129 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(537 \mathrm{~W} / \mathrm{m}^{2}\)

You have probably noticed warning signs on the highways stating that bridges may be icy even when the roads are not. Explain how this can happen.

The variation of the spectral absorptivity of a surface is as given in Fig. P12-78. Determine the average absorptivity and reflectivity of the surface for radiation that originates from a source at \(T=2500 \mathrm{~K}\). Also, determine the average emissivity of this surface at \(3000 \mathrm{~K}\).
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