Chapter 25, Problem 7P

To determine

The orbital period, average distance from the Sun and orbital velocity of asteroid.

Answer to Problem 7P

The orbital period of the asteroid is $5.94\text{years}$, the average distance of asteroid from the Sun is $2.07\times {10}^{11}\text{m}$ and the orbital velocity of the asteroid is $25.38\text{km}/\text{s}$.

Explanation of Solution

Write the expression for the orbital period of the asteroid.

    $p=\frac{P}{2}$        (I)

Here, $p$ is the orbital period of the asteroid and $P$ is the orbital period of the Jupiter.

Write the expression for the Kepler’s third law.

    $p^2=\frac{4{\pi }^2}{GM}{d}^3$        (II)

Here, $d$ is the average distance of asteroid from the Sun, $G$ is the gravitational constant, and $M$ is the mass of the Sun.

Write the expression for the orbital velocity of the asteroid.

    $v=\sqrt{\frac{GM}{d}}$        (III)

Here, $v$ is the orbital velocity of the asteroid.

Refer to the Celestial Profile of Jupiter to obtain the orbital period of Jupiter as $11.87\text{years}$.

Conclusion:

Substitute $11.87\text{years}$ for $P$ in equation (I) to find the orbital period of the asteroid.

    $\begin{array}{c}p=\frac{11.87\text{years}}{2}\\ =5.94\text{years}\end{array}$

Substitute $5.94\text{years}$ for $p$, $6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}$ for $G$, and $2\times {10}^{30}\text{kg}$ for $M$ in equation (II) to find the average distance of asteroid from the Sun.

    $\begin{array}{c}{\left(5.94\text{years}\times \frac{3.154\times {10}^7\text{s}}{1\text{year}}\right)}^2=\frac{4{\pi }^2}{\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(2\times {10}^{30}\text{kg}\right)}{d}^3\\ d^3=\frac{351\times {10}^{14}{\text{s}}^{\text{2}}\left(13.34\times {10}^{19}{\text{m}}^3/{\text{s}}^{\text{2}}\right)}{4{\pi }^2}\\ d=\sqrt[3]{8.9\times {10}^{33}{\text{m}}^{\text{3}}}\\ =2.07\times {10}^{11}\text{m}\end{array}$

Substitute $2.07\times {10}^{11}\text{m}$ for $d$, $6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}$ for $G$, and $2\times {10}^{30}\text{kg}$ for $M$ in equation (III) to find the orbital velocity of the asteroid.

    $\begin{array}{c}v=\sqrt{\frac{\left(6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(2\times {10}^{30}\text{kg}\right)}{2.07\times {10}^{11}\text{m}}}\\ =\sqrt{6.44\times {10}^8{\text{m}}^2/{\text{s}}^{\text{2}}}\\ =\left(25377.16\text{m}/\text{s}\times \frac{{10}^{-3}\text{km}}{1\text{m}}\right)\\ =25.38\text{km}/\text{s}\end{array}$

Therefore, the orbital period of the asteroid is $5.94\text{years}$, the average distance of asteroid from the Sun is $2.07\times {10}^{11}\text{m}$ and the orbital velocity of the asteroid is $25.38\text{km}/\text{s}$.