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

• brotherbobby
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?

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

No.

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

Finding cos ##\theta## first and then inverting.

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##.

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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##.

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
$$\sqrt{2}=1.4142...$$
$$\sqrt{3}=1.7320...$$

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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.

PeroK
brotherbobby said:
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}##.

Orodruin said:
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?

brotherbobby said:
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.

brotherbobby said:
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.

Orodruin said:
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.

Orodruin said:
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.

brotherbobby said:
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.

1. How does elevation affect sunset times?

Elevation affects sunset times because the higher you go, the farther you can see along the Earth's curvature. As a result, the sun appears to set later at higher elevations compared to sea level.

2. By how much time can sunset be delayed with an increase in elevation?

The delay in sunset time due to elevation can vary, but a general rule of thumb is that for every 1,000 meters (3,280 feet) increase in elevation, sunset can be delayed by approximately 1 to 2 minutes.

3. Why does the curvature of the Earth influence sunset times at different elevations?

The curvature of the Earth influences sunset times because it determines the horizon line. At higher elevations, you have a lower horizon line, allowing you to see the sun for a longer period before it dips below the horizon.

4. Are there mathematical formulas to calculate the exact change in sunset time due to elevation?

Yes, there are mathematical formulas that take into account the observer's elevation, the Earth's curvature, and atmospheric refraction to calculate the exact change in sunset time. One common approximation is using the formula: ΔT = (h / 7.92) minutes, where h is the elevation in meters.

5. Does atmospheric refraction play a role in sunset times at different elevations?

Yes, atmospheric refraction bends light rays, causing the sun to appear slightly higher in the sky than it actually is. This effect can slightly alter the perceived sunset time, especially at higher elevations where the atmosphere is thinner and refraction effects are reduced.

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