Chapter 3, Problem 67P

(a)

To determine

The speed and magnitude of the acceleration of a person standing on the equator.

Answer to Problem 67P

A person standing on the equator has a speed $463.8\text{ m/s}$ and a centripetal acceleration of magnitude $3.37\times {10}^{-2}{\text{m/s}}^2$ which is $0.344\%\text{ of }g$ .The total acceleration of the person is $99.65\%\text{of }g$ .

Explanation of Solution

Given:

Time taken by the Earth to rotate once on its axis, $T=24\text{ h}$

Radius of the Earth, $R_E=6.378\times {10}^6\text{m}$

Formula used:

The speed $v_E$ of the person on the equator is determined by the equation

  $v_E=\frac{2\pi R_E}{T}$.........(1)

The person on the equator experiences a centripetal acceleration $a_{rE}$ given by

  $a_{rE}=\frac{v_E^2}{R_E}$.........(2)

The total acceleration at the equator is given by,

  $a_E=g-a_{rE}$.........(3)

Calculation:

A person standing on the equator, revolves in a circular path of radius $R_E$ , the radius of the Earth as the Earth rotates about its axis. In a time $T$ of 24 h, the earth completes one rotation and in the same time the person completes its circular motion in an orbit of radius $R_E$ .

Calculate speed $v_E$ of the person on the equator using equation (1).

  $\begin{array}{c}{v}_E=\frac{2\pi R_E}{T}\\ =\frac{2\pi \left(6.378\times {10}^6\text{m}\right)}{\left(24\text{ h}\right)\left(3600\text{ s/h}\right)}\\ =463.8\text{ m/s}\end{array}$

All particles fixed on the Earth experience a centripetal force directed towards the center of its circular path. Hence it would experience a centripetal acceleration towards the center of its circular path.

Calculate the centripetal acceleration $a_{rE}$ of the person using equation (2).

  $\begin{array}{c}{a}_{rE}=\frac{v_E^2}{R_E}\\ =\frac{{\left(463.8\text{ m/s}\right)}^2}{\left(6.378\times {10}^6\text{m}\right)}\\ =3.37\times {10}^{-2}{\text{m/s}}^2\end{array}$

Express the acceleration as a percentage of $g$ , the acceleration of free fall.

  $\begin{array}{c}\frac{a_{rE}}{g}\times 100=\frac{3.37\times {10}^{-2}{\text{m/s}}^2}{9.81{\text{ m/s}}^2}\times 100\\ =0.344\%\end{array}$

The total acceleration experienced by the person is given by equation (3).

  $\begin{array}{c}{a}_E=g-a_{rE}\\ =\left(100-0.344\right)\%\text{ of }g\\ =99.6\%\text{ of }g\end{array}$

Conclusion:

Thus, a person standing on the equator has a speed $463.8\text{ m/s}$ and a centripetal acceleration of magnitude $3.37\times {10}^{-2}{\text{m/s}}^2$ which is $0.344\%\text{ of }g$ .The total acceleration of the person is $99.65\%\text{of }g$ .

(b)

To determine

The direction of the acceleration vector of the person on the equator.

Answer to Problem 67P

The acceleration vector of the person is directed towards the center of the Earth.

Explanation of Solution

Introduction:

A person on the surface of the Earth experiences two forces-(i) Gravitational force $W$ directed towards the center of the Earth and (ii) Normal contact force $N$ perpendicular to the surface of the Earth outwards.

An object in order to move in a circular path requires centripetal force. The centripetal force is radial in nature and is directed towards the center of the earth. As the Earth rotates, the difference between $W$ and $N$ should provide the centripetal force $F_C$ . This is shown in Figure1.

  $W-N=F_C$

  

Figure 1

However, the weight, as felt by the person on the surface of the earth, is equal to the Normal force he experiences.

  $N=W-F_C$

Hence, the acceleration $a_E$ ,he feels on the surface of Earth on Equator is less than the acceleration of free fall $g$ . But since both $g$ and $a_{rE}$ are directed towards the center of the earth, the total acceleration is also directed towards the center of the earth.

Conclusion:

Thus, the acceleration vector of the person on the Equator is directed towards the center of the Earth.

(c)

To determine

The speed and magnitude of the acceleration of a person standing at $35°$ latitude.

Answer to Problem 67P

The speed of the person at $35°$ latitude is $379.9\text{ m/s}$ and his acceleration is $9.83{\text{ m/s}}^2$ .

Explanation of Solution

Given:

The latitude where the person was standing, $\theta =35°$

Radius of the earth at equator, $R_E=6.378\times {10}^6\text{m}$

The speed of the person at equator, $v_E=463.8\text{ m/s}$

The centripetal acceleration at the equator, $a_{rE}=3.37\times {10}^{-2}{\text{m/s}}^2$

Formula used:

The person at $35°$ latitude moves in a circle of radius $r$ as the Earth spins around its axis with an angular velocity $\omega$ .

The speed of person at $35°$ latitude is given by,

  $v_P=r\omega$.........(4)

His speed at equator is given by

  $v_E=R_E\omega$.........(5)

From equations (4) and (5),

  $v_P=\frac{r}{R_E}{v}_E$......(6)

The centripetal acceleration at the equator is given by the expression,

  $a_{rE}=R_E{\omega }^2$.........(7)

The centripetal acceleration at $35°$ latitude is given by the expression,

  $a_{rP}=r{\omega }^2$.........(8)

From equations (4) and (5),

  $a_{rP}=\frac{r}{R_E}{a}_{rE}$.........(9)

The person also experiences acceleration $g$ due to the gravitational force exerted on him by the Earth and this is directed towards the center of Earth.

The magnitude of the resultant acceleration is determined by using parallelogram law of vectors.

  $a_P=\sqrt{a_{rP}^2+g^2+2a_{rP}g\cos \theta }$.....(10)

Calculation:

The motion of the person at $35°$latitude is represented in the diagram below:

  

Figure 2

From Figure 2, it can be seen that

  $\frac{r}{R_E}=\cos \theta$.........(11)

Use equation (11) in equation (6).

  $v_P=\frac{r}{R_E}{v}_E=v_E\cos \theta$

Substitute the value of the variables in the above equation.

  $\begin{array}{c}{v}_P=v_E\cos \theta \\ =\left(463.8\text{ m/s}\right)\left(\cos 35°\right)\\ =379.9\text{ m/s}\end{array}$

Use equation (11) in equation (9) and substitute the values of the variables to calculate the value of $a_{rP}$ .

  $\begin{array}{c}{a}_{rP}=\frac{r}{R_E}{a}_{rE}\\ =a_{rE}\cos \theta \\ =\left(3.37\times {10}^{-2}{\text{m/s}}^2\right)\left(\cos 35°\right)\\ =2.76\times {10}^{-2}{\text{m/s}}^2\end{array}$

The vector ${\overrightarrow{a}}_{rP}$ is directed towards the center of its circular path and is at an angle $35°$ to the direction of the acceleration of free fall $\overrightarrow{g}$ .

Substitute the values of variables in equation (10) and calculate the magnitude of the resultant acceleration $a_P$ .

  $\begin{array}{c}{a}_P=\sqrt{a_{rP}^2+g^2+2a_{rP}g\cos \theta }\\ =\sqrt{{\left(2.76\times {10}^{-2}{\text{m/s}}^2\right)}^2+{\left(9.81{\text{m/s}}^2\right)}^2+2\left(2.76\times {10}^{-2}{\text{m/s}}^2\right)\left(9.81{\text{m/s}}^2\right)\left(\cos 35°\right)}\\ =9.83{\text{ m/s}}^2\end{array}$

Conclusion:

Thus, the speed of the person at $35°$ latitude is $379.9\text{ m/s}$ and his acceleration is $9.83{\text{ m/s}}^2$ .

(d)

To determine

The angle between the direction of the acceleration at $35°$ latitude and acceleration at equator at the same longitude.

Answer to Problem 67P

The angle between the direction of the acceleration at $35°$ latitude and acceleration at equator at the same longitude is found to be $34.9°$ .

Explanation of Solution

Given:

The magnitude of centripetal acceleration at $35°$ latitude, $a_{rP}=2.76\times {10}^{-2}{\text{m/s}}^2$

The value of acceleration of free fall, $g=9.81{\text{m/s}}^2$

The latitude where the person was standing, $\theta =35°$

Formula used:

The direction of the acceleration at the equator is along the horizontal direction parallel to the direction of the centripetal acceleration at $35°$ latitude.

  ${\overrightarrow{a}}_E\parallel {\overrightarrow{a}}_{rP}$

Hence the angle $\alpha$ between the directions of ${\overrightarrow{a}}_{rP}$ and ${\overrightarrow{a}}_P$ is equal to the angle between ${\overrightarrow{a}}_P$ and ${\overrightarrow{a}}_E$ .

The vector diagram representing the accelerations is shown below:

  

Figure 3

This is given by

  $\alpha ={\tan }^{-1}\left(\frac{g\sin \theta }{a_{rP}+g\cos \theta }\right)$.........(12)

Calculation:

Substitute the values of the variables in equation (12).

  $\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{g\sin \theta }{a_{rP}+g\cos \theta }\right)\\ ={\tan }^{-1}\left[\frac{\left(9.81{\text{m/s}}^2\right)\left(\sin 35°\right)}{\left(2.76\times {10}^{-2}{\text{m/s}}^2\right)+\left(9.81{\text{m/s}}^2\right)\left(\cos 35°\right)}\right]\\ =34.9°\end{array}$

Conclusion:

Thus, the angle between the direction of the acceleration at $35°$ latitude and acceleration at equator at the same longitude is found to be $34.9°$ .