Chapter 7, Problem 45A

To determine

The orbital period of a planet around the Sun in years, if it is located $8.0$ times as far from the Sun as Earth is.

Answer to Problem 45A

  $22.62\text{years}$

Explanation of Solution

Given:

The small planet D is located $8.0$ times as far from the Sun as Earth is.

Formula used:

Period of a Planet orbiting the Sun is given by

  $T=2\pi \sqrt{\frac{r^3}{G{m}_S}}$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and $m_S$ is the mass of the Sun.

Calculation:

The value of gravitational constant is,

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

The orbital period of Earth around the Sun is,

  $T_E=1\text{year = }2\pi \sqrt{\frac{r_E{}^3}{G{m}_S}}$

Now, the orbital radius of the considered planet is,

  $r_p=8r_E$

Then, the orbital period of the planet around the Sun will be,

  $\begin{array}{c}{T}_P=2\pi \sqrt{\frac{r_P{}^3}{G{m}_S}}\\ =2\pi \sqrt{\frac{{\left(8r_E\right)}^3}{G{m}_S}}\\ =22.62T_E\\ =22.62\text{years}\end{array}$

Conclusion:

Thus, the orbital period of the considered planet is $22.62\text{years}$ .