Explanation:
We have,
Semimajor axis is [tex]4\times 10^{12}\ m[/tex]
It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :
[tex]T^2=\dfrac{4\pi ^2}{GM}a^3[/tex]
G is universal gravitational constant
M is solar mass
Plugging all the values,
[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s[/tex]
Since,
[tex]1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}[/tex]
So, the orbital period of a dwarf planet is 138.52 years.
