Chapter 3, Problem 40P

(a)

To determine

The magnitudes of the velocity and acceleration of the Moon relative to the Earth.

Answer to Problem 40P

The velocity of the Moon as it travels in a circular orbit around the Earth is found to be $1.023\times {10}^3\text{ m/s}$ and its acceleration has a magnitude $2.75\times {10}^{-3}{\text{ m/s}}^2$ directed towards the Earth.

Explanation of Solution

Given:

The radius of the Moon’s orbit around the Earth, $r_M=3.8\times {10}^5\text{km}$

The time period of revolution of the moon around the Earth, $T_M=27\text{ days}$

Formula used:

The Moon travels a distance equal to the circumference of its orbit in the time equal to its time period. Its velocity $v_M$ is given by,

  $v_M=\frac{2\pi r_M}{T_M}$.....(1)

As it revolves around the Earth, it experiences a centripetal force towards the center of its orbit. Its centripetal acceleration is given by,

  $a_M=\frac{v_M^2}{r_M}$.....(2)

Calculation:

Express the radius of the Moon’s orbit around the Earth in meters.

  $\begin{array}{c}{r}_M=\left(3.8\times {10}^5\text{km}\right)\left(\frac{{10}^3\text{ m}}{1\text{ km}}\right)\\ =3.8\times {10}^8\text{m}\end{array}$

Express the time period of the Moon’s revolution around the Earth.

  $\begin{array}{c}{T}_M=\left(27\text{ days}\right)\left(\frac{24\text{ h}}{1\text{ d}}\right)\left(\frac{3600\text{ s}}{1\text{ h}}\right)\\ =2.33\times {10}^6\text{s}\end{array}$

Substitute the values of $r_M$ and $T_M$ in equation (1) and calculate the magnitude of the Moon’s velocity.

  $\begin{array}{c}{v}_M=\frac{2\pi r_M}{T_M}\\ =\frac{2\pi \left(3.8\times {10}^8\text{m}\right)}{\left(2.33\times {10}^6\text{s}\right)}\\ =1.023\times {10}^3\text{m/s}\end{array}$

Substitute the values of $v_M$ and $T_M$ in equation (2) and calculate the magnitude of the Moon’s acceleration.

  $\begin{array}{c}{a}_M=\frac{v_M^2}{r_M}\\ =\frac{{\left(1.023\times {10}^3\text{m/s}\right)}^2}{\left(3.8\times {10}^8\text{m}\right)}\\ =2.75\times {10}^{-3}{\text{m/s}}^2\end{array}$

Conclusion:

Thus, the velocity of the Moon as it travels in a circular orbit around the Earth is found to be $1.023\times {10}^3\text{ m/s}$ and its acceleration has a magnitude $2.75\times {10}^{-3}{\text{ m/s}}^2$ directed towards the Earth.

(b)

To determine

The magnitudes of the velocity and acceleration of the Earth relative to the Sun

Answer to Problem 40P

The velocity of the Earth as it travels in a circular orbit around the sun is found to be $3.0\times {10}^4\text{ m/s}$ and its acceleration has a magnitude $6.0\times {10}^{-3}{\text{m/s}}^2$ directed towards the Sun.

Explanation of Solution

Given:

The radius of the Earth’s orbit around the Sun, $r_E=1.5\times {10}^8\text{km}$

The time period of revolution of the Earth around the Sun, $T_E=365\text{ days}$

Formula used:

The Earth travels a distance equal to the circumference of its orbit in the time equal to its time period. Its velocity $v_E$ is given by,

  $v_E=\frac{2\pi r_E}{T_E}$.....(3)

As it revolves around the Sun, it experiences a centripetal force towards the center of its orbit. Its centripetal acceleration is given by,

  $a_E=\frac{v_E^2}{r_E}$.....(4)

Calculation:

Express the radius of the Earth’s orbit around the sun in meters.

  $\begin{array}{c}{r}_E=\left(1.5\times {10}^8\text{km}\right)\left(\frac{{10}^3\text{ m}}{1\text{ km}}\right)\\ =1.5\times {10}^{11}\text{m}\end{array}$

Express the time period of the Earth’s revolution around the Sun.

  $\begin{array}{c}{T}_E=\left(365\text{ days}\right)\left(\frac{24\text{ h}}{1\text{ d}}\right)\left(\frac{3600\text{ s}}{1\text{ h}}\right)\\ =3.15\times {10}^7\text{s}\end{array}$

Substitute the values of $r_E$ and $T_E$ in equation (3) and calculate the magnitude of the Earth’s velocity.

  $\begin{array}{c}{v}_E=\frac{2\pi r_E}{T_E}\\ =\frac{2\pi \left(1.5\times {10}^{11}\text{m}\right)}{\left(3.15\times {10}^7\text{s}\right)}\\ =3.0\times {10}^4\text{m/s}\end{array}$

Substitute the values of $v_E$ and $T_E$ in equation (4) and calculate the magnitude of the Earth’s acceleration.

  $\begin{array}{c}{a}_E=\frac{v_E^2}{r_E}\\ =\frac{{\left(3.0\times {10}^4\text{m/s}\right)}^2}{\left(1.5\times {10}^{11}\text{m}\right)}\\ =6.0\times {10}^{-3}{\text{m/s}}^2\end{array}$

Conclusion:

Thus, velocity of the Earth as it travels in a circular orbit around the sun is found to be $3.0\times {10}^4\text{ m/s}$ and its acceleration has a magnitude $6.0\times {10}^{-3}{\text{m/s}}^2$ directed towards the Sun.

(c)

To determine

The value of the maximum acceleration of the Moon relative to Sun and the phase of the Moon this occurs.

Answer to Problem 40P

The value of the maximum acceleration of the Moon relative to Sun is found to be $8.75\times {10}^{-3}{\text{m/s}}^2$ and this occurs during Full moon.

Explanation of Solution

Given:

The acceleration of the Moon relative to Earth, $a_M=2.75\times {10}^{-3}{\text{m/s}}^2$

The acceleration of the Earth relative to the Sun, $a_E=6.0\times {10}^{-3}{\text{m/s}}^2$

The radius of the Earth’s orbit around the Sun, $r_E=1.5\times {10}^8\text{km}$

The radius of the Moon’s orbit around the Earth, $r_M=3.8\times {10}^5\text{km}$

Calculation:

The Moon revolves around the Earth in a circular orbit and the Earth revolves around the Sun. The Moon experiences accelerations relative to both Earth and the sun.

This is shown in the diagram below:

  

At New Moon day, the moon is farthest from the Sun and its acceleration towards the Earth and that towards the Sun point in the same direction. Hence, at this time, the accelerations add up and the Moon’s acceleration relative to the Sun is maximum.

Calculate the distance of the moon from the Sun when it is at its farthest position from the Sun.

  $\begin{array}{c}r=r_M+r_E\\ =\left(3.8\times {10}^8\text{m}\right)+\left(1.5\times {10}^{11}\text{m}\right)\\ =1.5038\times {10}^{11}\text{m}\\ \approx 1.5\times {10}^{11}\text{m}\end{array}$

The Moon’s distance from the Sun is nearly equal to the Earth’s distance from the Sun.

The Moon experiences an acceleration directed towards the Earth due its revolution around the Earth and since it moves around the Sun along with the Earth, it experiences an acceleration towards the Sun. During New moon day, the Moon, Earth and the Sun are in a straight line with the Moon at the farthest from the Sun. In this position, the acceleration of the Moon towards the Earth and towards the Sun are directed along the same straight line, towards the Sun. Hence, they add up.

Since the distance of the Moon from the Sun is nearly equal to that of the Earth’s distance from it, the Moon’s acceleration due its revolution around the Sun can be taken to be equal to that of the Earth’s around the Sun.

The Moon’s maximum acceleration in a direction towards the Sun $a$ is given by,

  $a=a_M+a_E$

Substitute the values of the accelerations in the above expression.

  $\begin{array}{c}a=a_M+a_E\\ =\left(2.75\times {10}^{-3}{\text{m/s}}^2\right)+\left(6.0\times {10}^{-3}{\text{m/s}}^2\right)\\ =8.75\times {10}^{-3}{\text{m/s}}^2\end{array}$

Conclusion:

Thus, the value of the maximum acceleration of the Moon relative to Sun is found to be $8.75\times {10}^{-3}{\text{m/s}}^2$ and this occurs during Full moon.