<p>The commonly used first approximation for the latitude of the Sun's GP is <span class="math-container">$23.5  sin(\frac{2d \pi}{365.25})$</span>,where d is the number of days since the vernal equinox. If your sine lookup is for degrees use 360 instead of <span class="math-container">$2 \pi$</span> in the formula. South latitudes are negative.</p>

CC BY-SA 4.0
2021-09-25T16:11:03.333Z
2021-09-25T16:11:03.333Z
stretch
<p>Here we calculate an approximation of the latitude which experiences zero shadow on (at the start of) a given day of a year.</p>
<p>Assumptions:</p>
<ol>
<li>The Earth is spherical</li>
<li>Its orbit is circular and the eccentricity is zero (the actual value of eccentricity  is <span class="math-container">$0.0167$</span>).</li>
<li>Revolution of the Earth around the Sun occurs with a period of exactly 1 year which is <span class="math-container">$365.25$</span> days and we disregard any the changes in axial tilt.</li>
<li>Our aim will be to calculate the latitude which experiences zero shadow at the start of any given day of a year. To this end we may temporarily disregard the Earth's rotational motion to simplify calculation.</li>
</ol>
<p>The Earth's axis of rotation is inclined (with respect to its orbit) by about <span class="math-container">$23.5°$</span>. As the Earth revolves around the sun, the latitude which experiences zero shadow ranges between <span class="math-container">$23.5°$</span>N and <span class="math-container">$23.5°$</span>S.</p>
<p>If we assume absence of rotation, it implies that there will be one spot on the Earth at any given time of the year which experiences zero shadow.</p>
<p>For initial conditions: assume that the starting time is the day of Spring equinox (which is roughly on March <span class="math-container">$20$</span>) and the point experiencing zero shadow at the start of the zeroth day is P. Therefore P lies on the equator.</p>
<p>As the Earth revolves, the zero shadow spot traces out a circle on the Earth's surface. The circle has exactly one point on the Tropic of Cancer and exactly one point on the Tropic of Capricorn, and it passes through the point P.
Please refer to the figure for further calculations.</p>
<p><a href="https://i.sstatic.net/g4yZw.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/g4yZw.jpg" alt="enter image description here" /></a></p>
<p>The zero shadow spot moves along its circular path on the Earth (shown in red color in the figure) in uniform circular motion, with period equal to 1 year. It's projection along the diameter MR in the figure is a SHM with frequency <span class="math-container">$\frac {1}{365.25}$</span>. per day.</p>
<p>The projection of this SHM onto the segment LM (which is also an SHM) has the same frequency <span class="math-container">$\frac {1}{365.25}$</span>.</p>
<p>The value of x can be used to calculate the angle with the equator, <span class="math-container">$θ$</span>. This is the required latitude.</p>
<p>Thus we have:</p>
<p><span class="math-container">$\tan θ = \frac {x}{a}$</span></p>
<p><span class="math-container">$x=\frac {b}{r} r\sin(\frac {360t}{365.25})° km$</span></p>
<p><span class="math-container">$=b\sin(\frac{360t}{365.25})°km$</span></p>
<p><span class="math-container">$\therefore \tan θ=\frac {b}{a}\sin(\frac{360 t}{365.25})°$</span>
and hence,
<span class="math-container">$θ=(\tan^{-1}(\frac {b}{a}\sin(\frac {360t}{365.25})°))°$</span></p>
<p>where <span class="math-container">$a=5848.72 km,
b=2543.27 km, 
t$</span> is in days and <span class="math-container">$12$</span> p.m. on March <span class="math-container">$20$</span> is the zeroth day.</p>

<equinox><earths-orbit>
Are all the steps below leading this approximation for a zero-shadow latitude correct?
Mausami Mulay
# Are all the steps below leading this approximation for a zero-shadow latitude correct?

(Asked by: username_0 on 2021-09-25)

Here we calculate an approximation of the latitude which experiences zero shadow on (at the start of) a given day of a year.

Assumptions:

1. The Earth is spherical
2. Its orbit is circular and the eccentricity is zero (the actual value of eccentricity is $0.0167$).
3. Revolution of the Earth around the Sun occurs with a period of exactly 1 year which is $365.25$ days and we disregard any the changes in axial tilt.
4. Our aim will be to calculate the latitude which experiences zero shadow at the start of any given day of a year. To this end we may temporarily disregard the Earth's rotational motion to simplify calculation.

The Earth's axis of rotation is inclined (with respect to its orbit) by about $23.5°$. As the Earth revolves around the sun, the latitude which experiences zero shadow ranges between $23.5°$N and $23.5°$S.

If we assume absence of rotation, it implies that there will be one spot on the Earth at any given time of the year which experiences zero shadow.

For initial conditions: assume that the starting time is the day of Spring equinox (which is roughly on March $20$) and the point experiencing zero shadow at the start of the zeroth day is P. Therefore P lies on the equator.

As the Earth revolves, the zero shadow spot traces out a circle on the Earth's surface. The circle has exactly one point on the Tropic of Cancer and exactly one point on the Tropic of Capricorn, and it passes through the point P.
Please refer to the figure for further calculations.

[![enter image description here](https://i.sstatic.net/g4yZw.jpg)](https://i.sstatic.net/g4yZw.jpg)

The zero shadow spot moves along its circular path on the Earth (shown in red color in the figure) in uniform circular motion, with period equal to 1 year. It's projection along the diameter MR in the figure is a SHM with frequency $\frac {1}{365.25}$. per day.

The projection of this SHM onto the segment LM (which is also an SHM) has the same frequency $\frac {1}{365.25}$.

The value of x can be used to calculate the angle with the equator, $θ$. This is the required latitude.

Thus we have:

$\tan θ = \frac {x}{a}$

$x=\frac {b}{r} r\sin(\frac {360t}{365.25})° km$

$=b\sin(\frac{360t}{365.25})°km$

$\therefore \tan θ=\frac {b}{a}\sin(\frac{360 t}{365.25})°$
and hence,
$θ=(\tan^{-1}(\frac {b}{a}\sin(\frac {360t}{365.25})°))°$

where $a=5848.72 km,
b=2543.27 km,
t$ is in days and $12$ p.m. on March $20$ is the zeroth day.

**Tags:** equinox, earths-orbit


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## 1 answers

### Answer by username_1 on 2021-09-25. Score: 1

The commonly used first approximation for the latitude of the Sun's GP is $23.5 sin(\frac{2d \pi}{365.25})$,where d is the number of days since the vernal equinox. If your sine lookup is for degrees use 360 instead of $2 \pi$ in the formula. South latitudes are negative.


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