Imagine we've discovered a planet orbiting another star at 1 AU every 6 months. The planet has a moon that orbits the planet at the same distance as our Moon, but it takes 2 months. What can we infer about this planet? (Hint: Use Newton's Version of Kepler's Third Law)

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BlackTom AI · Accepted Answer

We start by parsing the system: a planet orbits its star at a distance of 1 AU with a orbital period of 6 months. Using Newton's form of Kepler's Third Law for a planet-star system, P^2 = a^3 / (M_star + M_planet), where P is in years, a in AU, and masses are in solar masses. Here P = 0.5 years and a = 1 AU, so a^3 / P^2 = 1 / 0.25 = 4. This means the total mass (star plus planet) is about 4 solar masses if we assume the planet’s mass is negligible; if the planet’s mass is not negligible, it would reduce the inferred stellar mass slightly but not by an amount that would contradict a fairly massive star overall. In short, the star appears to be significantly more massive than the Sun, on the order of a few solar masses, with the planet a small contributor to the total mass.

Next, consider the moon of the planet. The moon orbits the planet at a distance comparable to the Earth–Moon distance, but with an orbital period of 2 months (about 60 days). For a moon, the orbital period around the planet is governed by P_m^2 = (4π^2 a_m^3) / (G M_planet). If we take a_m to be similar to the Earth’s Moon distance and compare the observed P_m to the Moon’s actual period (~27.3 days), a longer period implies a smaller host mass. Since P_m scales roughly with 1/√M_planet for a fixed orbital radius, a moon with a ~2.2× longer period suggests the planet’s mass is about (1/2.2^2) ≈ 0.21 times Earth's mass. Put differently, M_planet is likely well below one Earth mass.

Putting these pieces together:
- The star is probably quite massive (several solar masses), inferred from the planet’s 1 AU orbit with a 6-month period.
- The planet itself, given the moon’s ~60-day orbit at a similar distance to the Earth–Moon distance, would have a mass well below Earth’s mass, roughly a few tenths of an Earth mass in a rough estimate.

Therefore, a key inference about the planet is that it is less massive than Earth. The other parts of the system (the star’s mass) do not contradict this conclusion; they mainly affect what kind of star the planet orbits and the dynamical environment, not the conclusion about the planet’s own mass. Any incorrect option would likely misinterpret the moon’s period, the scaling of P with M_planet, or the impact of the star’s mass on the planet’s mass inference, leading to overestimating the planet’s mass or ignoring the qualitative relationship P^2 ∝ a^3 / M_total.

COS SPR 25 class 2 Jan 30 Test1 (USE GOOGLE CHROME)

Imagine we've discovered a planet orbiting another star at 1 AU every 6 months. The planet has a moon that orbits the planet at the same distance as our Moon, but it takes 2 months. What can we infer about this planet? (Hint: Use Newton's Version of Kepler's Third Law)

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