Question

The period of revolution of the planet Neptune around the sun is _________.
A) \[165\text{ hours}\]
B) \[\text{165 days}\]
C) \[\text{165 years}\]
D) None of these

Answer 1

Hint: Before discussing the period of revolution of Neptune, let’s get acquainted with it. It is the eighth and farthest-known solar planet from the Sun. In the Solar System, it is the fourth-largest planet by diameter, the third-most-massive planet, and the densest giant planet. It is seventeen times the mass of Earth, slightly more massive than its near-twin Uranus. Kepler’s law gives us a way to find the period of revolution of a planet from its semi-major axis or the radius of its orbit.

Complete step by step solution:
As discussed in the hint section, Kepler’s law states that
\[{{T}^{2}}=k\times {{a}^{3}}\] where \[T\] is the period of revolution of a planet, \[a\] is the semi-major axis of the planet and \[k\] is a constant of proportionality
Upon researching, we found that
The semi-major axis of the earth \[\left( {{a}_{earth}} \right)=149.6\times {{10}^{6}}km\]. Similarly, the semi-major axis of Neptune \[\left( {{a}_{neptune}} \right)=4495.06\times {{10}^{6}}km\]

The period of revolution of the earth around the sun is one year. Substituting these values, we get the expression of Kepler’s Law for both planets as follows
\[\begin{align}
  & {{\left( {{T}_{earth}} \right)}^{2}}=k\times {{\left( {{a}_{earth}} \right)}^{3}}---equation(1) \\
 & {{\left( {{T}_{neptune}} \right)}^{2}}=k\times {{\left( {{a}_{neptune}} \right)}^{3}}---equation(2) \\
\end{align}\]
Dividing the two equations, we get
\[{{\left( \dfrac{{{T}_{neptune}}}{{{T}_{earth}}} \right)}^{2}}={{\left( \dfrac{{{a}_{neptune}}}{{{a}_{earth}}} \right)}^{3}}\]
Substituting the values of the semi-major axis of the planets and the period of revolution of the earth, we get
\[\begin{align}
  & {{\left( \dfrac{{{T}_{neptune}}}{1year} \right)}^{2}}={{\left( \dfrac{4495.06\times {{10}^{6}}km}{149.6\times {{10}^{6}}km} \right)}^{3}} \\
 & \Rightarrow {{\left( \dfrac{{{T}_{neptune}}}{1} \right)}^{2}}={{(30.05)}^{3}} \\
 & \Rightarrow {{T}_{neptune}}={{(30.05)}^{\dfrac{3}{2}}}=164.72years \\
\end{align}\]

Hence the period of revolution of the planet Neptune around the sun is approximately and the correct option is (C).

Additional Information: Increasing concentrations of methane, ammonia and water are found in the lower regions of the atmosphere. The core of Neptune is likely composed of iron, nickel and silicates. Absorption of red light by the methane in its atmosphere gives Neptune a blue appearance. Neptune’s thermosphere is at an unusually high temperature of \[\text{750 K}\] . The reason for this anomalous increased temperature is still unknown. Neptune’s weather is characterised by extremely dynamic storm systems.

Note: Upon research into Neptune, we found that the average distance between Neptune and the Sun is \[\text{4}\text{.5}\] billion years and it completes an orbit around the sun on average every \[\text{164}\text{.79 years}\] , with a variance of \[\text{0}\text{.1 years}\]. This data closely resembles our calculations and hence our solution is correct.