Chapter 25, Problem 14P

What is the orbital period of a comet nucleus at the inner edge of the Oort Cloud? What is its orbital velocity? Assume a circular orbit. (Hint: Use Kepler’s third law, Eq. 4-1) (Note: The circumference of a circle is 2πr.)

To determine

The orbital period and orbital velocity of comet nucleus in the Oort Cloud.

Answer to Problem 14P

The orbital period of the comet nucleus in the Oort Cloud is from $1\text{million}\text{year}$ to $30\text{million}\text{years}$ and the orbital velocity of the comet nucleus in the Oort Cloud is from $300\text{m}/\text{s}$ to $100\text{m}/\text{s}$.

Explanation of Solution

The expression for the orbital period of comet nucleus in the Oort Cloud using Kepler’s law is,

    $\begin{array}{c}{P}_c^2=a^3\\ P_c=a^{\frac{3}{2}}\end{array}$        (I)

Here, $P_c$ is the orbital period of comet nucleus in the Oort Cloud and $a$ is the range of distance from the Sun of the comet nucleus in the Oort Cloud from $10000\text{AU}$ to $100000\text{AU}$

The expression for the orbital velocity of comet nucleus in the Oort Cloud is,

    $v=\frac{2\pi a}{P_c}$        (II)

Here, $v$ is the orbital velocity of comet nucleus in the Oort Cloud

Conclusion:

For $a=10000\text{AU}$,

Substitute $10000\text{AU}$ for $a$ in equation (I) to find $P_c$

    $\begin{array}{c}{P}_c={\left(10000\text{AU}\right)}^{\frac{3}{2}}\\ =1\times {10}^6\text{y}\left(\frac{1\text{million}\text{year}}{{10}^6\text{y}}\right)\\ =1\text{million}\text{years}\end{array}$

Substitute $10000\text{AU}$ for $a$ and $1\times {10}^6\text{y}$ for $P_c$ in equation (II) to find $v$

    $\begin{array}{c}v=\frac{2\pi \left(10000\text{AU}\right)}{1\times {10}^6\text{y}}\\ =\frac{2\pi \left(10000\text{AU}\right)\left(\frac{1.5\times {10}^{11}\text{m}}{1\text{AU}}\right)}{1\times {10}^6\text{y}\left(\frac{365.25\text{days}}{1\text{y}}\right)\left(\frac{86400\text{s}}{1\text{day}}\right)}\\ =298.5\text{m}/\text{s}\\ \approx 300\text{m}/\text{s}\end{array}$

For $a=100000\text{AU}$,

Substitute $100000\text{AU}$ for $a$ in equation (I) to find $P_c$

    $\begin{array}{c}{P}_c={\left(100000\text{AU}\right)}^{\frac{3}{2}}\\ =30\times {10}^6\text{y}\left(\frac{1\text{million}\text{year}}{{10}^6\text{y}}\right)\\ =30\text{million}\text{years}\end{array}$

Substitute $10000\text{AU}$ for $a$ and $30\times {10}^6\text{y}$ for $P_c$ in equation (II) to find $v$

    $\begin{array}{c}v=\frac{2\pi \left(100000\text{AU}\right)}{30\times {10}^6\text{y}}\\ =\frac{2\pi \left(100000\text{AU}\right)\left(\frac{1.5\times {10}^{11}\text{m}}{1\text{AU}}\right)}{30\times {10}^6\text{y}\left(\frac{365.25\text{days}}{1\text{y}}\right)\left(\frac{86400\text{s}}{1\text{day}}\right)}\\ =99.5\text{m}/\text{s}\\ \approx 100\text{m}/\text{s}\end{array}$

Therefore. the orbital period of the comet nucleus in the Oort Cloud is from $1\text{million}\text{year}$ to $30\text{million}\text{years}$ and the orbital velocity of the comet nucleus in the Oort Cloud is from $300\text{m}/\text{s}$ to $100\text{m}/\text{s}$.