Calculate the orbital distance (semi-major axis) for an extra-solar planet in a 3.312 Earth-day orbit around a 1.3 solar-mass star.?

Select one: a. 0.068 AU b. 0.058 AU c. 0.038 AU d. 0.048 AU e. 0.078 AU

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May 27, 2017

The semi-major axis is 0.048AU, answer d.

Explanation:

The orbital period $p$ in years and the semi-major axis $a$ in AU of an orbiting body are related by Newton's form of Kepler's third law.

$M {p}^{2} = {a}^{3}$

Where $M$ is the mass of the star in solar masses.

Substituting the values and converting days to years gives:

${a}^{3} = 1.3 \cdot {3.312}^{2} / {365}^{2} = 0.000107$

Taking the cube root gives $a = 0.04748$.

So the nearest answer is d. 0.048AU.

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