Final answer:
By applying Kepler's Third Law, we can calculate the average distances from the Sun where Kirkwood gaps would occur for asteroids with orbital periods that are three-sevenths and two-sevenths of Jupiter's period. The calculation involves determining the semi-major axis based on the revised orbital periods using Kepler's formula.
Step-by-step explanation:
The question at hand requires the application of Kepler's Third Law of planetary motion, which states that the square of the orbital period (P) of a body in the solar system is directly proportional to the cube of the semi-major axis (a) of its orbit, such that P² ≈ a³. Since we're given the orbital periods as fractions of Jupiter's period, we can use this relationship to find the average distances from the Sun where the corresponding Kirkwood gaps occur. Jupiter's orbital period is approximately 11.86 years, and using Kepler's Third Law, we can calculate the average distances for the given orbital periods.
For an orbital period of three-sevenths of Jupiter's period, that would be 11.86 years × (3/7) = approximately 5.08 years. Now, if P² = a³, then a = √(P²)^(1/3). Using the calculated period, we would find the distance for the first gap.
Similarly, for an orbital period of two-sevenths of Jupiter's, we have 11.86 years × (2/7) = approximately 3.39 years. Again, applying Kepler's Third Law, we calculate the semi-major axis to ascertain the average distance for the second gap. This method gives us the averagedistance at which these asteroidal gaps occur.