Question

The planet Neptune travels around the sun with a period of 165 yr. What is the radius of the orbit approximately, if the orbit is considered circular?

Answer 1

We can recall from Kepler’s third law that the square of the period of revolution of the planets around the sun is directly proportional to the cube of their mean distance to the sun. Use the knowledge of the earth's period to get a relation.
Formula used: In this solution we will be using the following formulae;
${T^2} = k{R^3}$ where $T$ is the period of the revolution of a planet around the sun, $R$ is the mean radius of the planet to the sun.

Complete Step-by-Step solution:
To solve this question, we use the Kepler’s third law. This states that the square of the period of the revolution of all the planets about the sun is directly proportional to the cube of the mean distance between the planets and the sun. This can be mathematically given as
${T^2} \propto {R^3}$
$\Rightarrow {T^2} = k{R^3}$ where $T$ is the period of the revolution of a planet around the sun, $R$ is the mean radius of the planet to the sun.
Hence, we can write by comparison between two planets, that
$\dfrac{{{T_1}^2}}{{T_2^2}} = \dfrac{{R_1^3}}{{R_2^3}}$
We can use any planet with a known distance and period as the second planet. We choose earth.
$\dfrac{{{T_N}^2}}{{T_E^2}} = \dfrac{{R_N^3}}{{R_E^3}}$ where the subscript N and E stands for Neptune and Earth respectively.
For earth, the period is 1 year. Hence, write that
$\dfrac{{{{165}^2}}}{{{1^2}}} = \dfrac{{R_N^3}}{{R_E^3}}$
$\Rightarrow R_N^3 = {165^2}R_E^3$
Hence, by finding the cube root of both sides, we have
${R_N} = \sqrt[3]{{{{165}^2}}}{R_E}$
$\Rightarrow {R_N} = 30{R_E}$
Radius of the earth is about $1.50 \times {10^{11}}m$. Hence,
${R_N} = 30\left( {1.50 \times {{10}^{11}}} \right) = 4.5 \times {10^{12}}m$

Note: For understanding, note that although Kepler's law was stated with respect to the period and distance of the planet to the sun, it works for other kinds of orbital system. Any small orbiting body around a massive body will always obey the Kepler’s law including the moon and artificial satellite around the earth.