- #1

brotherbobby

- 699

- 163

- 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.

**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##?
Problem Statement :**

**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?