Chapter 7.1, Problem 9SSC

To determine

Period of Neptune’s orbit.

Answer to Problem 9SSC

  $T=164.43\text{years}$

Explanation of Solution

Given:

Sun’s mass, $M=1.99\times {10}^{30}\text{kg}$

Average distance of Neptune from Sun, $r=4.496\times {10}^{12}\text{m}$

Formula Used:

From Kepler’s third law of planetary motion:

  $T^2=\frac{4{\pi }^2}{Gm}{r}^3$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and m is the mass of the star and T is the orbital period.

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

Calculation:

  $T=\sqrt{\left(\frac{4{\pi }^2}{GM}\right)r^3}$

Substitute the values and solve:

  $\begin{array}{c}T=\sqrt{\left(\frac{4{\pi }^2}{6.67\times {10}^{-11}\times 1.99\times {10}^{30}}\right){\left(4.496\times {10}^{12}\right)}^3}\\ =\sqrt{270\times {10}^{17}}\\ =\sqrt{27\times {10}^{18}}\\ =5.196\times {10}^9\text{s}\end{array}$

Changing the orbit period to years:

In one year, there are $3.16\times {10}^7\text{s}$

  $\begin{array}{c}T=\frac{5.196\times {10}^9}{3.16\times {10}^7}\\ =164.43\text{years}\end{array}$

Conclusion:

Thus, orbital period of Neptune is 164.43 years.