A planet has two moons.
The first moon has an orbital period of 1.262 Earth days and an orbital radius of \(2.346 \times 10^4\) km.
The second moon has an orbital radius of \(9.378 \times 10^3\) km.
What is the orbital period of the second moon?
Answer (2)
Kepler's third law hypothesizes that for all the small bodies in orbit around the same central body, the ratio of (orbital period squared) / (orbital radius cubed) is the same number.
Moon #1: (1.262 days)² / (2.346 x 10^4 km)³
Moon #2: (orbital period)² / (9.378 x 10^3 km)³
If Kepler knew what he was talking about ... and Newton showed that he did ... then these two fractions are equal, and may be written as a proportion.
Cross multiply the proportion:
(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³
Divide each side by (2.346 x 10^4)³:
(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³
= 0.1017 day²
Orbital period = 0.319 Earth day = about 7.6 hours.
Using Kepler's Third Law, we find the orbital period of the second moon, which has an orbital radius of 9.378 x 10^3 km, to be approximately 0.31 Earth days or about 7.44 hours. This is calculated by setting the ratios of the squares of the periods to the cubes of their respective radii equal. After performing the necessary calculations, we arrive at the final value for the second moon's orbital period.
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