Question

Imagine another solar system, with a star of the same mass as the Sun. Suppose a planet with a mass twice that of Earth (2 Earth) orbits at a distance of 1 AU from the star. What is the orbital period of this planet?

Answer 1

Answer:

Time period, T = 403.78 years

Explanation:

It is given that,

Orbital distance, $a=1\ AU=1.496\times 10^{11}\ m$

Mass of the Earth, $m_e=5.972\times 10^{24}\ kg$

Mass of the planet, $m_p=2m_e=11.944\times 10^{24}\ kg$

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.

$T^2=\dfrac{4\pi^2}{Gm_p}a^3$

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 11.944\times 10^{24}}\times (1.496\times 10^{11})^3$

$T=1.28\times 10^{10}\ s$

or

T = 403.78 years

So, the orbital period of this planet is 404 years. Hence, this is the required solution.