Answer:
Time period, T = 403.78 years
Explanation:
It is given that,
Orbital distance, [tex]a=1\ AU=1.496\times 10^{11}\ m[/tex]
Mass of the Earth, [tex]m_e=5.972\times 10^{24}\ kg[/tex]
Mass of the planet, [tex]m_p=2m_e=11.944\times 10^{24}\ kg[/tex]
Let T is the orbital period of this planet. The Kepler's third law of motion gives the relation between the orbital period and the orbital distance.
[tex]T^2=\dfrac{4\pi^2}{Gm_p}a^3[/tex]
[tex]T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 11.944\times 10^{24}}\times (1.496\times 10^{11})^3[/tex]
[tex]T=1.28\times 10^{10}\ s[/tex]
or
T = 403.78 years
So, the orbital period of this planet is 404 years. Hence, this is the required solution.
