Chapter 5, Problem 92P

(a)

To determine

The angular displacement in radians in an elapsed time of 1 day.

Answer to Problem 92P

The angular displacement in radians in an elapsed time of 1 day is $\underset{\_}{0.0172\text{rad}}$.

Explanation of Solution

Write the expression for the angular displacement.

    $\Delta \theta =\omega \Delta t$        (I)

Here, $\Delta \theta$ is the angular displacement, $\omega$ is the angular velocity, $t$ is the time.

Write the expression for the angular speed.

    $\omega =\frac{2\pi }{T}$        (II)

Conclusion:

Substitute $365.25\text{days}$ for

   $\begin{array}{c}\omega =\frac{2\pi }{365.25\text{days}\times \left(\frac{24\text{h}}{1\text{day}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)}\\ =\frac{2\pi }{31557600\text{s}}\\ =1.99\times {10}^{-7}\text{s}\end{array}$

Substitute $1\text{day}$ for $\Delta t$ and S for $\omega$ in equation (I) to find $\Delta \theta$.

    $\begin{array}{c}\Delta \theta =1.99\times {10}^{-7}\text{s}\left(1\text{day}\times \frac{24\text{h}}{1\text{day}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\\ =0.0172\text{rad}\end{array}$

Therefore, angular displacement in radians in an elapsed time of 1 day is $\underset{\_}{0.0172\text{rad}}$.

(b)

To determine

The change in Earths velocity.

Answer to Problem 92P

The change in Earths velocity is $\underset{\_}{513.5\text{m}/\text{s}}$ toward the center of Sun.

Explanation of Solution

Write the expression for the velocity.

    $v=r\omega$        (III)

Here, $\omega$ is the angular velocity, $r$ is the radius.

The magnitude of change in velocity in 1 day is,

    $\left|\Delta \overrightarrow{v}\right|=v\Delta \theta$        (IV)

Conclusion:

Substitute $1.5\times {10}^{11}\text{m}$ for $r$, $1.99\times {10}^{-7}\text{rad}/\text{s}$ for $\omega$ in equation (III) to find $v$.

    $\begin{array}{c}v=1.5\times {10}^{11}\text{m}\times \left(1.99\times {10}^{-7}\text{rad}/\text{s}\right)\\ =2.98\times {10}^4\text{m}/\text{s}\end{array}$

Substitute $2.98\times {10}^4\text{m}/\text{s}$ for $v$ and $0.0172\text{rad}$ for $\Delta \theta$ in equation (IV) to find $\left|\Delta \overrightarrow{v}\right|$.

    $\begin{array}{c}\left|\Delta \overrightarrow{v}\right|=\left(2.98\times {10}^4\text{m}/\text{s}\right)\left(0.0172\text{rad}\right)\\ =513.5\text{m}/\text{s}\end{array}$

Therefore, the change in Earths velocity is $\underset{\_}{513.5\text{m}/\text{s}}$ toward the center of Sun.

(c)

To determine

The average acceleration in one day.

Answer to Problem 92P

The average acceleration in one day is $\underset{\_}{0.00594\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the average acceleration during 1 day,

    $a=\frac{v}{t}$        (V)

Conclusion:

Substitute $513.5\text{m}/\text{s}$ for $v$ and $86400\text{s}$ for $t$ in equation (V) to find $a$.

    $\begin{array}{c}a=\frac{513.5\text{m}/\text{s}}{86400\text{s}}\\ =0.00594\text{m}/{\text{s}}^2\end{array}$

Therefore, The average acceleration in one day is $\underset{\_}{0.00594\text{m}/{\text{s}}^2}$.

(d)

To determine

The radial acceleration of earth.

Answer to Problem 92P

The radial acceleration of earth is $\underset{\_}{0.00594\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the radial acceleration of the Earth.

    $a_R=\frac{v^2}{r}$        (VI)

Conclusion:

Substitute $513.5\text{m}/\text{s}$ for $v$ and $1.5\times {10}^{11}\text{m}$ for $r$ in equation (V) to find $a_R$.

    $\begin{array}{c}a=\frac{{\left(513.5\text{m}/\text{s}\right)}^2}{1.5\times {10}^{11}\text{m}}\\ =0.00594\text{m}/{\text{s}}^2\end{array}$

Therefore, The average acceleration in one day is $\underset{\_}{0.00594\text{m}/{\text{s}}^2}$.