Chapter 7.1, Problem 72AYU

In Problems 79-84, use the following discussion. The formula $D=24\left[1-\frac{{\cos }^{-1}\left(\tan i\tan \theta \right)}{\pi }\right]$ can be used to approximate the number of hours of daylight D when the declination of the Sun is $i^{\circ }$ at a location ${\theta }^{\circ }$ north latitude for any date between the vernal equinox and autumnal equinox. The declination of the Sun is defined as the angle i between the equatorial plane and any ray of light from the Sun. The latitude of a location is the angle $\theta$ between the Equator and the location on the surface of Earth, with the vertex of the angle located at the center of Earth. See the figure. To use the formula, ${\cos }^{-1}(\tan i\tan \theta )$ must be expressed in radians.

Approximate the number of hours of daylight in Anchorage, Alaska ( ${61}^{\circ }10\text{'}$ north latitude), for the following dates:

(a) Summer solstice $\left(i={23.5}^{\circ }\right)$

(b) Vernal equinox $\left(i=0^{\circ }\right)$

(c) July 4 $\left(i={22}^{\circ }48\text{'}\right)$

To determine

To calculate: The number of hours of daylight $D$ at the location $\theta$ north latitude Approximate the number of hours of daylight in anchorage, Alaska (${61}^{\circ }10\prime$ north latitude), for the following dates:

a. Summer solstice $i={23.5}^{\circ }$.

Answer to Problem 72AYU

Solution:

a. $12.0024$ hours.

Explanation of Solution

Given:

The declination of the Sun is defined as the angle $i$ between the equatorial plane and any ray of light from the Sun. The latitude of a location is the angle $\theta$ between the Equator and the location on the surface of Earth, with the vertex of the angle located at the center of Earth.

a. Summer solstice $i={23.5}^{\circ };\text{Θ }=\text{ 6}{1}^{\circ }10\prime$ north latitude $=\text{ 6}{1.16}^{\circ }$.

Formula used:

$D=24\left[1–\left({\text{cos}}^{-1}\left(\tan i\tan \theta \right)\right)/\pi \right]$

Calculation:

a. Summer solstice $i={23.5}^{\circ };\text{Θ }={61}^{\circ }10\prime$ north latitude $={61.16}^{\circ }$.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{23.5}^{\circ }\text{tan6}{1.16}^{\circ }\right)\right)/\pi \right]$

Convert degree into radians.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0.410}^c\text{tan}{1.0674}^c\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0.0071\times 0.0186\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0.000132\right)\right)/\pi \right]$

$D=24\left[1–\left(89.992\right)/\pi \right]$

$D=24\left[1–0.4999\right]$

$D=24\left[0.5001\right]$

$D=12.0024$ hours.

To determine

To calculate: The number of hours of daylight $D$ at the location $\theta$ north latitude Approximate the number of hours of daylight in anchorage, Alaska (${61}^{\circ }10\prime$ north latitude), for the following dates:

b. Vernal equinox $i=0^{\circ }$.

Answer to Problem 72AYU

Solution:

b. $23.86$ hours.

Explanation of Solution

Given:

The declination of the Sun is defined as the angle $i$ between the equatorial plane and any ray of light from the Sun. The latitude of a location is the angle $\theta$ between the Equator and the location on the surface of Earth, with the vertex of the angle located at the center of Earth.

b. Vernal equinox $i=0^{\circ }$; $\text{Θ }=\text{ 6}{1}^{\circ }10\prime$ north latitude $=\text{ 6}{1.16}^{\circ }$.

Formula used:

$D=24\left[1–\left({\text{cos}}^{-1}\left(\tan i\tan \theta \right)\right)/\pi \right]$

Calculation:

b. Vernal equinox $i=0^{\circ }$; $\text{Θ }={61}^{\circ }10\prime$ north latitude $={61.16}^{\circ }$.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0}^{\circ }\text{tan6}{1.16}^{\circ }\right)\right)/\pi \right]$

Convert degree into radians.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0}^c\text{tan}{1.0674}^c\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0\times 0.0186\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0\right)\right)/\pi \right]$

$D=24\left[1–\left(1\right)/\pi \right]$

$D=24\left[1–0.0055\right]$

$D=24\left[0.9944\right]$

$D=23.86$ hours.

To determine

To calculate: The number of hours of daylight $D$ at the location $\theta$ north latitude Approximate the number of hours of daylight in anchorage, Alaska (${61}^{\circ }10\prime$ north latitude), for the following dates:

c. July 4 $i={22}^{\circ }48’$.

Answer to Problem 72AYU

Solution:

c. $12.00$ hours.

Explanation of Solution

Given:

The declination of the Sun is defined as the angle $i$ between the equatorial plane and any ray of light from the Sun. The latitude of a location is the angle $\theta$ between the Equator and the location on the surface of Earth, with the vertex of the angle located at the center of Earth.

c. July 4 $i={22}^{\circ }48’\text{; }\text{Θ }=\text{ 6}{1}^{\circ }10\prime$ north latitude $=\text{ 6}{1.16}^{\circ }$.

Formula used:

$D=24\left[1–\left({\text{cos}}^{-1}\left(\tan i\tan \theta \right)\right)/\pi \right]$

Calculation:

c. July 4 $i={22}^{\circ }48’\text{ = 22}{\text{.8}}^{\circ };\text{Θ }={61}^{\circ }10\prime$ north latitude $={61.16}^{\circ }$.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan }{22.8}^{\circ }\text{tan}{61.16}^{\circ }\right)\right)/\pi \right]$

Convert degree into radians.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0.3979}^c\text{tan}{1.0674}^c\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0.0069\times 0.0186\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0.0000128\right)\right)/\pi \right]$

$D=24\left[1–\left({89.999}^{\circ }\right)/\pi \right]$

$D=24\left[1–0.4999\right]$

$D=24\left[0.5000\right]$

$D=12.00$ hours.