Chapter 2, Problem 11Q

To determine

(a)

The sketch showing how Polaris, the Little Dipper and the Big Dipper would appear at $8\text{P}\text{.M}\text{.}$ on August 1.

Explanation of Solution

Introduction:

It is known that the Earth rotates from West to East. One complete rotation around its axis is completed by Earth in $24\text{hrs}$. As seen from Earth it appears as if all the objects are traveling from the East to the West; that is, in the anti-clockwise direction.

Earth rotates by $360°$ in $24\text{hours}$. Therefore, the angle covered by the Earth around its axis in $1\text{hr}$ is calculated as,

$\begin{array}{c}\theta =\frac{360°}{24\text{hrs}}\\ =15°/\text{hr}\end{array}$

The time difference between $11\text{P}\text{.M}\text{.}$ and $8\text{P}\text{.M}\text{.}$ is calculated as,

$\begin{array}{c}t=11\text{P}\text{.M}\text{.}-8\text{P}\text{.M}\text{.}\\ =3\text{hrs}\end{array}$

Therefore, the angle covered by the Earth around its axis within this time period of $3\text{hrs}$ is calculated as,

$\begin{array}{c}\text{Angle}=\left(\theta \right)\left(t\right)\\ =15°/\text{hr}\times 3\text{hrs}\\ =45°\end{array}$

Thus, at $8\text{P}\text{.M}\text{.}$ ,the Earth rotated by an angle of $45°$ towards the West. This implies that in the sky, the objects were towards the East by $45°$ at $8\text{P}\text{.M}\text{.}$ in comparison to the position where it is at $11\text{P}\text{.M}\text{.}$ Therefore, the sky would seem to have rotated by $45°$ about the Polaris in the clockwise direction at $8\text{P}\text{.M}\text{.}$

The following figure shows the sketch showing how Polaris, the Little Dipper and the Big Dipper would appear at $8\text{P}\text{.M}\text{.}$ on August 1.

Figure (1)

Conclusion:

Therefore, Figure (1) shows the sketch of how the Polaris, the Little Dipper and the Big Dipper would appear at $8\text{P}\text{.M}\text{.}$ on August 1.

To determine

(b)

The sketch showing how Polaris, the Little Dipper and the Big Dipper would appear at $2\text{A}\text{.M}\text{.}$ on August 1.

Explanation of Solution

Introduction:

It is known that the Earth rotates from West to East. One complete rotation around its axis is completed by Earth in $24\text{hrs}$. As seen from Earth, it appears as if all the objects are traveling from the East to the West, that is, in the anti-clockwise direction.

Earth rotates by $360°$ in $24\text{hours}$. Therefore, the angle covered by the Earth around its axis in $1\text{hr}$ is calculated as

$\begin{array}{c}\theta =\frac{360°}{24\text{hrs}}\\ =15°/\text{hr}\end{array}$

The time difference between $11\text{P}\text{.M}\text{.}$ and $2\text{A}\text{.M}\text{.}$ is calculated as

$\begin{array}{c}t=2\text{A}\text{.M}\text{.}-11\text{P}\text{.M}\text{.}\\ =-3\text{hrs}\end{array}$

Therefore, the angle covered by the Earth around its axis within this time period of $3\text{hrs}$ is calculated as

$\begin{array}{c}\text{Angle}=\left(\theta \right)\left(t\right)\\ =15°/\text{hr}\times \left(-3\text{hrs}\right)\\ =-45°\end{array}$

Thus, at $2\text{A}\text{.M}\text{.}$ the Earth rotated by an angle of $45°$ towards the East. This implies that in the sky, the objects were towards the West by $45°$ at $2\text{A}\text{.M}\text{.}$ in comparison to the position where it is at $11\text{P}\text{.M}\text{.}$. Therefore, the sky would seem to have rotated by $45°$ about the Polaris in the anti-clockwise direction at $2\text{A}\text{.M}\text{.}$

The following figure shows the sketch showing how Polaris, the Little Dipper and the Big Dipper would appear at $2\text{A}\text{.M}\text{.}$ on August 1.

Figure (2)

Conclusion:

Therefore, Figure (2) shows the sketch of how the Polaris, the Little Dipper and the Big Dipper would appear at $2\text{A}\text{.M}\text{.}$ on August 1.