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

Time period, T = 403.78 years

Step-by-step explanation:

It is given that,

Orbital distance,
`a=1\ AU=1.496* 10^(11)\ m`

Mass of the Earth,
`m_e=5.972* 10^(24)\ kg`

Mass of the planet,
`m_p=2m_e=11.944* 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=(4\pi^2)/(Gm_p)a^3`

`T^2=(4\pi^2)/(6.67* 10^(-11)* 11.944* 10^(24))* (1.496* 10^(11))^3`

`T=1.28* 10^(10)\ s`

or

T = 403.78 years

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