Please determine the noon sun angle at the given latitudes on April 11:
40 degrees north
0 degrees
57
To determine the noon sun angle at a specific latitude on a given date, you need to know the declination of the sun on that date. The declination is the angle between the sun and the Earth's equator.
April 11th is approximately two weeks after the vernal equinox, which is when the sun is directly over the equator. From then on, the sun's declination increases towards the northern hemisphere until the summer solstice.
For 40 degrees north latitude, we can determine the approximate declination by subtracting the latitude from 90 degrees. So the declination on April 11th at 40 degrees north would be:
90 - 40 = 50 degrees.
Similarly, for 0 degrees latitude (equator), the declination is 0 degrees because the sun is directly overhead at the equator on the equinoxes.
Now, to determine the noon sun angle, we need to subtract the declination from the zenith angle. The zenith angle is 90 degrees minus the latitude.
For 40 degrees north latitude:
Zenith angle = 90 - 40 = 50 degrees
Noon sun angle = Zenith angle - Declination
Noon sun angle = 50 - 50 = 0 degrees
For 0 degrees latitude (equator):
Zenith angle = 90 - 0 = 90 degrees
Noon sun angle = Zenith angle - Declination
Noon sun angle = 90 - 0 = 90 degrees
Therefore, on April 11th, the noon sun angle at 40 degrees north latitude is 0 degrees, while at 0 degrees latitude (equator), the noon sun angle is 90 degrees.
To determine the noon sun angle at a given latitude on a specific date, we need to consider the angle between the sun and the observer's horizontal line. This angle is known as the solar zenith angle.
To calculate the solar zenith angle, we can use the following equation:
sin(θ) = sin(φ) * sin(δ) + cos(φ) * cos(δ) * cos(H)
Where:
θ is the solar zenith angle,
φ is the observer's latitude,
δ is the solar declination, and
H is the hour angle.
First, let's calculate the solar declination (δ) for April 11th. The solar declination varies throughout the year and is determined based on the Earth's tilt. For April 11th, we can use the approximate value of -1.63 degrees.
Next, let's calculate the solar zenith angle at the given latitudes:
1. For 40 degrees north latitude:
φ = 40 degrees
δ = -1.63 degrees
To find the solar zenith angle, we need the hour angle (H) at noon. At noon, the Sun reaches its highest point in the sky. Therefore, H is 0 degrees.
Now substituting the values into the equation:
sin(θ) = sin(40) * sin(-1.63) + cos(40) * cos(-1.63) * cos(0)
After evaluating the equation, you will find that sin(θ) is approximately 0.678.
To find the solar zenith angle (θ), calculate the inverse sine (sin^-1) of 0.678:
θ = sin^-1(0.678)
Thus, the noon sun angle at 40 degrees north latitude on April 11th is approximately 43.6 degrees.
2. For 0 degrees latitude (equator):
φ = 0 degrees
δ = -1.63 degrees
H = 0 degrees
Using the same formula as above:
sin(θ) = sin(0) * sin(-1.63) + cos(0) * cos(-1.63) * cos(0)
After evaluating the equation, sin(θ) is approximately 0.
Therefore, the noon sun angle at 0 degrees latitude (equator) on April 11th is 0 degrees. This means that the sun will be directly overhead at noon.
Remember, these calculations are approximations and may vary slightly depending on the actual solar declination for the given date.