Chapter 5, Problem 91P

(a)

To determine

The radial acceleration of an object at the equator.

Answer to Problem 91P

The radial acceleration of an object at the equator is $\underset{\_}{0.0337\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the acceleration.

    $a_r=r{\omega }^2$        (I)

Here, $a_r$ is the radial acceleration, $r$ is the radius of earth, $\omega$ is the angular velocity.

Write the expression for $\omega$.

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

Here, $T$ is the time period.

Use equation (II) in (I) to solve for $a$

    $a_r=r{\left(\frac{2\pi }{T}\right)}^2$        (III)

Conclusion:

Substitute $6.378\times {10}^8\text{m}$ for $r$, $24\text{h}$ for $T$ in equation (III) to find $a_r$.

    $\begin{array}{c}{a}_r=\left(6.378\times {10}^8\text{m}\right){\left(\frac{2\pi \text{rad}}{24\text{h}\times \frac{3600\text{s}}{1\text{h}}}\right)}^2\\ =0.0337\text{m}/{\text{s}}^2\end{array}$

Therefore, the radial acceleration of an object at the equator is $\underset{\_}{0.0337\text{m}/{\text{s}}^2}$.

(b)

To determine

Whether the object’s apparent weight greater or lesser than its weight.

Answer to Problem 91P

The apparent weight of the object is less than the weight of the object.

Explanation of Solution

Write the expression for the apparent weight of the object.

    $W^{\prime }=m{a}_r$        (IV)

Here, $W^{\prime }$ is the apparent weight.

Write the expression for the actual weight of the object.

    $W=mg$        (V)

Here, $W$ is the actual mass.

Since the obtained value for the radial acceleration is less than the gravitational field strength $\left(a_r<g\right)$. As a result, the apparent weight of the object is less than the weight of the object.

Conclusion:

Therefore, the apparent weight of the object is less than the weight of the object.

(c)

To determine

The percentage of apparent weight differ from the weight at the equator.

Answer to Problem 91P

The percentage of apparent weight differ from the weight at the equator is $\underset{\_}{0.344\%}$.

Explanation of Solution

Write the expression for the percentage of apparent weight differ from the weight at the equator.

    $\text{percent between apparent weight and actual weight}=\frac{W^{\prime }}{W}\times 100$        (VI)

Use equation (IV) and (V) in (VI) to solve for percentage of apparent weight differ from the weight at the equator.

    $\begin{array}{c}\text{percent between apparent weight and actual weight}=\frac{m{a}_r}{mg}\times 100\\ =\frac{a_r}{g}\times 100\end{array}$        (VII)

Conclusion:

Substitute $0.0337\text{m}/{\text{s}}^2$ for $a_r$, $9.8\text{m}/{\text{s}}^2$ for $g$ in equation (VII) to find the percentage of apparent weight differ from the weight at the equator.

    $\begin{array}{c}\text{percent between apparent weight and actual weight}=\frac{0.0337\text{m}/{\text{s}}^2}{9.8\text{m}/{\text{s}}^2}\times 100\\ =0.344\%\end{array}$

Therefore, the percentage of apparent weight differ from the weight at the equator is $\underset{\_}{0.344\%}$.

(d)

To determine

Whether the object’s apparent weight and the actual weight readings are equal on any place on earth.

Answer to Problem 91P

The object’s apparent weight and the actual weight readings are equal on the poles of the earth.

Explanation of Solution

The position on the earth where the object’s apparent weight and the actual weight are equal, the ratio of radial acceleration to the force of gravity is unity.

    $\begin{array}{c}\frac{a_r}{g}=1\\ a_r=g\end{array}$        (VIII)

The poles of the earth, north pole and the south pole have the same radial acceleration and the force of gravity. As a result, the object’s apparent weight and the actual weight readings are equal on the poles of the earth.

Conclusion:

Therefore, the object’s apparent weight and the actual weight readings are equal on the poles of the earth.