Question

A newly discovered planet orbits a distant star with the same mass as the Sun at an average distance of 101 million kilometers. Its orbital eccentricity is 0.4. Find the planet's orbital period.

Answer 1

Kepler's third law described the relation between semi-major axis (or average distance to the star) and the orbital period (how long it takes to complete one lap) as follows: 
                                           a^3 / p^2 = constant
In the case of our Solar system the constant is 1 
This means that, for this problem:
a^3 / p^2 = 1
p^2 = a^3
p = a^(3/2)
The semi major axis is given as 101 million km. We need to convert this into AU where 1 AU is approximately 150 million Km
101 million Km = (101x1) / 150 = 0.67 AU

Now, we substitute in the equation to get the orbital period as follows:
p = (0.67)^(3/2) = 0.548 earth years