91Ó°ÊÓ

Kepler’s law is the square of the time period of any plane is directly related to the cube of the semi-major axis.

Kepler’s law tells that each planet is elliptical in nature, and Sun is the focus of the ellipse.

The average distance of Neptune is\(r = 4.5 \times {10^9}\;{\rm{km}}\).

The distance of the Earth from the Sun is \(a = 1.50 \times {10^8}\;{\rm{km}}\).

The expression for the Kepler’s law is given as,

\({T^2} \propto {a^3}\)

The expression for the Neptune is,

\(T_N^2 \propto {a^3}\) … (i)

The expression for the Earth is,

\(T_E^2 \propto {r^3}\) … (ii)

Dividing equation (ii) by equation (i),

\(\frac{{T_E^2}}{{T_N^2}} = \frac{{{a^3}}}{{{r^3}}}\)

Substitute the values in the above equation,

\(\begin{aligned}\frac{{{{\left( {1\;{\rm{year}}} \right)}^2}}}{{T_N^2}} &= \frac{{{{\left( {\left( {1.50 \times {{10}^8}\;{\rm{km}}} \right) \times \frac{{1\;AU}}{{1.50 \times {{10}^8}\;{\rm{km}}}}} \right)}^3}}}{{{{\left( {\left( {4.50 \times {{10}^9}\;{\rm{km}}} \right) \times \frac{{1\;AU}}{{1.50 \times {{10}^8}\;{\rm{km}}}}} \right)}^3}}}\\{T_N} &= 164.3\;{\rm{years}}\end{aligned}\)

Thus, the length of the Neptunium year is \(164.3\;{\rm{years}}\)