Chapter 7.1, Problem 74AYU

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.

that is ${66}^{\circ }30\text{'}$ 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 $\left(i={22}^{\circ }48\text{'}\right)$

(d) Thanks to the symmetry of the orbital path of Earth around the Sun, the number of hours of daylight on the winter solstice may be found by computing the number of hours of daylight on the summer solstice and subtracting this result from 24 hours. Compute the number of hours of daylight for this location on the winter solstice. What do you conclude about daylight for a location at ${66}^{\circ }30\text{'}$ north latitude?

To determine

To calculate: The number of hours of daylight $D$ at the location $\theta$ north latitude Approximate the number of hours of daylight at the any location (${66}^{\circ }30’$ north latitude), for the following dates:

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

Answer to Problem 74AYU

Solution:

a. $12.060$ 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 $i={23.5}^{\circ }$.

Formula used:

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

Calculation:

a. Summer solstice $i={23.5}^{\circ };\text{ Θ }={\text{ 66.5}}^{\circ }$ north latitude.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{23.5}^{\circ }\text{tan}{66.5}^{\circ }\right)\right)/\pi \right]$

Convert degree into radians.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0.410}^c\text{tan}{1.6064}^c\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0.0071\times 1.1255\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0.0079\right)\right)/\pi \right]$

$D=24\left[1–\left(89.54\right)/\pi \right]$

$D=24\left[1–0.4974\right]$

$D=24\left[0.5025\right]$

$D=12.060$ 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 at the any location (${66}^{\circ }30’$ north latitude), for the following dates:

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

Answer to Problem 74AYU

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 $i=0^{\circ }$.

Formula used:

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

Calculation:

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

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0}^{\circ }\text{tan}{66.5}^{\circ }\right)\right)/\pi \right]$

Convert degree into radians.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0}^c\text{tan}{1.60640}^c\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0\times 1.1255\right)\right)/\pi \right]$

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

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

$D=24\left[1–0.0055\right]$

$D=24\left[1–0.9944\right]$

$D=23.86$ 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 at the any location (${66}^{\circ }30’$ north latitude), for the following dates:

c. July 4 $i={22}^{\circ }48’$.

Answer to Problem 74AYU

Solution:

c. $12.059$ 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 $i={22}^{\circ }48’$.

Formula used:

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

Calculation:

c. July 4 $i={22}^{\circ }48’\text{ = 22}{\text{.8}}^{\circ };\text{ Θ }=0^{\circ }$.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{22.8}^{\circ }\text{tan}{66.5}^{\circ }\right)\right)/\pi \right]$

Convert degree into radians.

$D=24\left[1–\left({\text{cos}}^{-1}\left(\text{tan}{0.3979}^c\text{tan}{1.6064}^c\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0.0069\times 1.1255\right)\right)/\pi \right]$

$D=24\left[1–\left({\text{cos}}^{-1}\left(0.0077\right)\right)/\pi \right]$

$D=24\left[1–\left({89.555}^{}\right)/\pi \right]$

$D=24\left[1–0.4975\right]$

$D=24\left[0.5024\right]$

$D=12.059$ 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 at the any location (${66}^{\circ }30’$ north latitude), for the following dates:

d. The number of hours of daylight throughout the year for a location at the equator.

Answer to Problem 74AYU

Solution:

d. To compute the number of hours of daylight on the winter solstice $=11.94$ 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.

d. $\text{Θ }={66}^{\circ }30\prime$ north latitude $={66.5}^{\circ }$.

Formula used:

d. The number of hours of daylight on the winter solstice $=24-$ the number of hours of daylight on the summer solstice.

The number of hours of daylight on the winter solstice $=24–12.060=11.94$ hours.

Calculation:

The number of hours of daylight on the summer solstice winter solstice and july 4 is almost half the day whereas in vernal equinox is almost full day at any location $\text{66}{\text{.5}}^{\circ }$ north latitude.