Chapter 5, Problem 92P

(a)

To determine

The angular displacement of Earth in one day.

Answer to Problem 92P

The angular displacement of Earth in one day is $1.72\times {10}^{-2}\text{rad}$.

Explanation of Solution

Write an expression to calculate the angular displacement of Earth in one day.

$\Delta \theta =\frac{2\pi }{T}\Delta t$ (I)

Here, the angular displacement of Earth in one day is $\Delta \theta$, the time taken for complete revolution is $T$ and the time taken is $\Delta t$.

Conclusion:

Substitute $365.25\text{d}$ for $T$, and $1\text{d}$ in equation (I) to find $\Delta \theta$.

$\begin{array}{c}\Delta \theta =\frac{2\pi }{\left(365.25\text{d}\right)}\left(1\text{d}\right)\\ =2\pi \left(2.74\times {10}^{-3}\right)\\ =1.72\times {10}^{-2}\text{rad}\end{array}$

Thus, the angular displacement of Earth in one day is $1.72\times {10}^{-2}\text{rad}$.

(b)

To determine

The change in Earth’s velocity.

Answer to Problem 92P

The change in Earth’s velocity is  $514\text{m/s towards the Sun}$.

Explanation of Solution

Write an expression to calculate the change in Earth’s velocity.

$\left|\Delta \overrightarrow{v}\right|=\frac{2\pi r}{T}\Delta \theta$ (II)

Here, the change in Earth’s velocity is $\left|\Delta \overrightarrow{v}\right|$ and the distance between Sun and Earth is $r$.

Conclusion:

Substitute $1.50\times {10}^{11}\text{m}$ for $r$, $365.25\text{d}$ for $T$, and $1.72\times {10}^{-2}\text{rad}$ for $\Delta \theta$ in equation (II) to find $\left|\Delta \overrightarrow{v}\right|$ .

$\begin{array}{c}\left|\Delta \overrightarrow{v}\right|=\frac{2\pi \left(1.50\times {10}^{11}\text{m}\right)}{\left(\left(365.25\text{d}\right)\left(\frac{24\text{h}}{1\text{d}}\right)\left(\frac{60\text{min}}{1\text{h}}\right)\left(\frac{60\text{s}}{1\text{min}}\right)\right)}\left(1.72\times {10}^{-2}\text{rad}\right)\\ =\frac{2\pi \left(1.50\times {10}^{11}\text{m}\right)\left(1.72\times {10}^{-2}\text{rad}\right)}{3.156\times {10}^7\text{s}}\\ =514\text{m/s}\end{array}$

Thus, the change in Earth’s velocity is  $514\text{m/s towards the Sun}$.

(c)

To determine

The average acceleration during one day.

Answer to Problem 92P

The average acceleration during one day is $0.00595{\text{m/s}}^2\text{perpendicular to the average velocity}$.

Explanation of Solution

Write an expression to calculate the average acceleration during one day.

$\left|{\overrightarrow{a}}_{\text{av}}\right|=\frac{2\pi r\Delta \theta }{T\Delta t}$ (III)

Here, the average acceleration during one day is $\left|{\overrightarrow{a}}_{\text{av}}\right|$.

Conclusion:

Substitute $1.50\times {10}^{11}\text{m}$ for $r$, $1.72\times {10}^{-2}\text{rad}$ for $\Delta \theta$, $365.25\text{d}$ for $T$, and $1\text{d}$ for $\Delta t$ equation (III) to find $\left|{\overrightarrow{a}}_{\text{av}}\right|$.

$\begin{array}{c}\left|{\overrightarrow{a}}_{\text{av}}\right|=\frac{2\pi \left(1.50\times {10}^{11}\text{m}\right)\left(1.72\times {10}^{-2}\text{rad}\right)}{\left(\left(365.25\text{d}\right)\left(\frac{24\text{h}}{1\text{d}}\right)\left(\frac{60\text{min}}{1\text{h}}\right)\left(\frac{60\text{s}}{1\text{min}}\right)\right)\left(\left(1\text{d}\right)\left(\frac{24\text{h}}{1\text{d}}\right)\left(\frac{60\text{min}}{1\text{h}}\right)\left(\frac{60\text{s}}{1\text{min}}\right)\right)}\\ =\frac{2\pi \left(1.50\times {10}^{11}\text{m}\right)\left(1.72\times {10}^{-2}\text{rad}\right)}{\left(365.25\text{d}\right)\left(1\text{d}\right){\left(\left(\frac{24\text{h}}{1\text{d}}\right)\left(\frac{60\text{min}}{1\text{h}}\right)\left(\frac{60\text{s}}{1\text{min}}\right)\right)}^2}\\ =0.00595{\text{m/s}}^2\end{array}$

Thus, the average acceleration during one day is $0.00595{\text{m/s}}^2\text{perpendicular to the average velocity}$.

(d)

To determine

Compare the average acceleration with Earth’s instantaneous radial acceleration.

Answer to Problem 92P

The average acceleration with Earth’s is equal to the instantaneous radial acceleration.

Explanation of Solution

Write an expression to calculate the radial acceleration.

$a_r=\frac{4{\pi }^{2}r}{T^2}$ (IV)

Here, the radial acceleration is $a_r$.

Conclusion:

Substitute $1.50\times {10}^{11}\text{m}$ for $r$, and $365.25\text{d}$ for $T$ equation (IV) to find $a_r$.

$\begin{array}{c}{a}_r=\frac{4{\pi }^2\left(1.50\times {10}^{11}\text{m}\right)}{{\left(\left(365.25\text{d}\right)\left(\frac{24\text{h}}{1\text{d}}\right)\left(\frac{60\text{min}}{1\text{h}}\right)\left(\frac{60\text{s}}{1\text{min}}\right)\right)}^2}\\ =\frac{4{\pi }^2\left(1.50\times {10}^{11}\text{m}\right)}{{\left(\left(365.25\text{d}\right)\left(\frac{86400\text{s}}{1\text{d}}\right)\right)}^2}\\ =0.00595{\text{m/s}}^2\end{array}$

Thus, Earth’s motion for one day tends to make small angle at the Sun, thus, the arc length and the chord length along the Earth's orbit are very nearly equal.