Different sunset times due to elevation ##h## at a point on the Earth

[brotherbobby]

Homework Statement

A man and his friend stand at the bottom and the top of a tower of height 800 m respectively. If the sun sets at 6 pm for the man, at what time will it set for his friend?

Relevant Equations

1. The cosine of an angle $\theta$ in a right angled triangle is defined as : $\cos\theta=\dfrac{\text{Adjacent}}{\text{Hypotenuse}}.$

2. For small $x$, we can write $(1+x)^{-1}\approx 1-x$.

3. As the earth takes 24 hours to rotate once on its axis, we can pretend as if its the sun rotating around the earth completing an angle of $2\pi$ radians in a time of 24 hours.

Problem Statement : I draw a picture of the given problem alongside. P is the location of the man and Q that of his friend at a height $h$ above. If the sun is at a position $\text{S}_1$ at 6 pm, at what time is the sun at position $\text{S}_2$?
Attempt : If $\text{S}_2Q$ is inclined to $\text{S}_1P$ by an angle $\theta$, then so are their perpendiculars. Hence $\measuredangle\text{MOP}=\theta$.
In $\triangle OMQ$, $\cos\theta=\dfrac{R}{R+h}=\dfrac{1}{1+\dfrac{h}{R}}=\left(1+\dfrac{h}{R}\right)^{-1}\approx 1-\dfrac{h}{R}=1-\dfrac{800\;m}{6.4\times10^6\;m}\Rightarrow\theta=0.016^c$, in radians.

The sun "rotates" around the earth in a time of 24 hours. Hence $2\pi$ radians $\rightarrow 24\times 60$ minutes. An angle of 0.016 radians would take about $\Delta t = \dfrac{24\times 60\times 0.016}{2\pi} = 3.7\;\text{mins}\approx 4\;\text{mins}$.

Hence the sun would set for his friend at $\color{green}{\boxed{\color{green}{\text{6.04 pm}}}}$.

Request : Am I correct in my conclusion?

 

[anuttarasammyak]

Have you used
\cos \theta \approx 1- \frac{\theta^2}{2}?

 

[brotherbobby]

No.

 

[anuttarasammyak]

Then how do you get the value of $\theta$ ?

 

[brotherbobby]

Finding cos $\theta$ first and then inverting.

 

[anuttarasammyak]

I see. Your teacher allows you to use calculator which gives you arccos. Then you don't have to make the approximation on R/(R+h). In the approximation I proposed
\theta \approx \sqrt{\frac{2h}{R}} \approx \sqrt{2.468} *10^{-2}\approx 1.5* 10^{-2}
with $1.5^2=2.25$.
Your result seems all right.

 

[brotherbobby]

Thank you.
If I had followed your method instead where $\cos\theta \approx 1-\dfrac{\theta^2}{2}$, I'd still be forced to use a calculator in order to find the square root of a very small number. Here $\theta^2<\theta$.

 

[anuttarasammyak]

You can put $10^{-4}$ out of sqrt which is to be $10^{-2}$ there so that the number in sqrt is 2 to 3. You can roughly estimate the value of $\sqrt{2.468}$ as I did above by recalling
\sqrt{2}=1.4142...
\sqrt{3}=1.7320...

 

[Orodruin]

The problem is underspecified. We are told nothing about where on Earth the tower is or what time of year it is. The way you have solved it here essentially assumes the tower is on the equator during an equinox.

 

[Hill]

If I had followed your method instead where cos⁡θ≈1−θ22, I'd still be forced to use a calculator in order to find the square root of a very small number.

Not necessarily. Using $\theta^2$ you could directly calculate $\Delta t^2$ and get about $13.14 \text { mins}^2$, from which it is easy to estimate $\Delta t \approx 3.6 \text { mins}$.

 

[brotherbobby]

The problem is underspecified. We are told nothing about where on Earth the tower is or what time of year it is. The way you have solved it here essentially assumes the tower is on the equator during an equinox.

Yes good point. I hadn't realised.
Let's say the country is at a latitude of $30^{\circ}\;\text N$ and the time of the year is the equinox. How do I go about solving the problem?
If I "move" the earth so that the country moves to the North Pole, will it not still be true that, during sunset, the sun rays will still be tangent to the surface of the earth at that point?

 

[Orodruin]

Yes good point. I hadn't realised.
Let's say the country is at a latitude of $30^{\circ}\;\text N$ and the time of the year is the equinox. How do I go about solving the problem?

You will have to do a bit more geometry. While the Earth surface is tangential to the Sunlight at sunset, its normal is not orthogonal to the axis of rotation. At the equinoxes, at least the axis of rotation is in the plane perpendicular to the Sunlight. Away from the equinoxes this introduces additional complication.

If I "move" the earth so that the country moves to the North Pole, will it not still be true that, during sunset, the sun rays will still be tangent to the surface of the earth at that point?

Yes, but at the North pole at the equinox, the Sun doesn’t set at top of the tower.

 

[brotherbobby]

You will have to do a bit more geometry. While the Earth surface is tangential to the Sunlight at sunset, its normal is not orthogonal to the axis of rotation. At the equinoxes, at least the axis of rotation is in the plane perpendicular to the Sunlight. Away from the equinoxes this introduces additional complication.

I will have to think about it. I have understood both parts of what you were trying to say.

Yes, but at the North pole at the equinox, the Sun doesn’t set at top of the tower.

I am afraid you have got me wrong there. I am using the symmetry of the earth to "move" the place to the north pole, nothing more than that. To facilitate the visualising. Of course the sun doesn't set at places on the north pole during spring equinox. But that is another matter.

 

[Orodruin]

I am afraid you have got me wrong there. I am using the symmetry of the earth to "move" the place to the north pole, nothing more than that. To facilitate the visualising

If you are saying that you want to switch coordinates to spherical coordinates based on the particular point in question I think that is a mistake because it severely complicates the necessary rotation. Keeping the coordinates based on the axis of rotation makes the rotation significantly easier.