Okay, let's tackle this problem step by step. So, we have three exoplanets with circular orbits, and we need to find the factor by which Planet3's orbital period is larger than Planet1's. The given ratios are for equilibrium temperatures, masses, and we need to use the TTV method. Hmm, TTV stands for Transit Timing Variations, but wait, maybe that's a red herring here? The question mentions the ratio of equilibrium temperatures and masses, but since the albedo is the same for all, maybe we can use Kepler's third law here.

First, I remember that the equilibrium temperature of a planet is related to the luminosity of the host star and the planet's distance from it. The formula for equilibrium temperature is T = (L / (16π²σD²))^(1/4), where L is the star's luminosity, D is the distance, and σ is the Stefan-Boltzmann constant. But since the albedo is the same for all planets, the ratio of temperatures would depend on the ratio of the stellar fluxes and the inverse square of their distances. Wait, but the problem gives the ratio of equilibrium temperatures between the planets. Let me think.

The equilibrium temperature ratio between Planet1 and Planet2 is 1.4. So, T1/T2 = 1.4. Similarly, T2/T3 = 2.3, so T1/T3 = 1.4 * 2.3 = 3.22. So the temperature of Planet1 is about 3.22 times that of Planet3. But how does this relate to their orbital periods?

Kepler's third law states that the square of the orbital period is proportional to the cube of the semi-major axis (distance from the star) divided by the mass of the star. Wait, but the mass of the star isn't given here. However, the problem says the albedo is the same, which affects the temperature. But perhaps the key is to relate the distances first using the temperature ratios and then use Kepler's law to find the period ratio.

The equilibrium temperature formula is T = (L * (1 - a))^(1/4) / sqrt(D), where a is the albedo. Since albedo is the same for all planets, the ratio of temperatures would be (L1/L2)^(1/4) * sqrt(D2/D1). Wait, but here L is the same for all planets since they orbit the same star. So L1 = L2 = L3. Therefore, the temperature ratio simplifies to sqrt(D2/D1) * (1 - a)^(1/4). But since albedo a is the same, the (1 - a) term cancels out. Wait, but if all have the same albedo, then the ratio of temperatures would depend only on the distances. So T1/T2 = sqrt(D2/D1). Wait, no, let me check.

Equilibrium temperature formula: T = (L (1 - a) / (16πσ D²))^(1/4). So T ∝ sqrt(1/D) * (L (1 - a))^(1/4). Since L and a are the same for all, T ∝ 1/D^(1/4). Wait, no, because (1/D²)^(1/4) is 1/D^(1/2). So T ∝ 1/sqrt(D). Therefore, T1/T2 = sqrt(D2/D1). So the ratio of temperatures is inverse square root of the ratio of distances. So, T1/T2 = sqrt(D2/D1) = 1.4. Then, D2/D1 = (1.4)^2 = 1.96. So Planet2 is 1.96 times farther from the star than Planet1.

Similarly, T2/T3 = 2.3 = sqrt(D3/D2). So D3/D2 = (2.3)^2 = 5.29. Therefore, D3 = D2 * 5.29. But since D2 = 1.96 D1, then D3 = 1.96 * 5.29 D1. Let's compute that: 1.96 * 5.29 ≈ 1.96 * 5 = 9.8, plus 1.96 * 0.29 ≈ 0.568, so total ≈ 10.368 D1. So the distance of Planet3 is approximately 10.368 times that of Planet1.

Now, Kepler's third law states that the orbital period squared is proportional to the cube of the semi-major axis (distance) divided by the mass of the star. But in this case, the mass of the star isn't given, but we can assume that all three planets orbit the same star, so the mass of the star is constant. Therefore, (P3/P1)^2 = (D3/D1)^3. So P3/P1 = (D3/D1)^(3/2). Since D3/D1 ≈ 10.368, then (10.368)^(3/2). Let's compute that.

First, sqrt(10.368) ≈ 3.22. Then, 3.22^3 ≈ 3.22 * 3.22 = 10.368, then 10.368 * 3.22 ≈ 33.4. So P3/P1 ≈ 33.4. Therefore, the answer should be B. ~33.4.

Wait, but let me double-check the steps. Starting from the temperature ratios:

T1/T2 = 1.4, which gives D2/D1 = (T1/T2)^(-2) = (1.4)^-2 = 1/1.96 ≈ 0.5102. Wait, wait, no. Wait, T ∝ 1/sqrt(D), so T1/T2 = sqrt(D2/D1). Therefore, D2/D1 = (T1/T2)^2 = (1.4)^2 = 1.96. So Planet2 is 1.96 times farther than Planet1. Then, T2/T3 = 2.3 = sqrt(D3/D2) => D3/D2 = (2.3)^2 = 5.29. Therefore, D3 = 5.29 * D2 = 5.29 * 1.96 D1 ≈ 10.368 D1. Then, the period ratio is (D3/D1)^(3/2) = (10.368)^(3/2). Let's compute this more accurately.

First, sqrt(10.368) = approx 3.22. Then, 3.22^3 = 3.22 * 3.22 = 10.368, then 10.368 * 3.22 ≈ 33.4. So yes, that's correct. Therefore, the factor is approximately 33.4, which is option B. But wait, let me check if there's any other factors involved, like the mass of the planet. The problem states that the masses of the planets are in ratios 1.15 between Planet1 and Planet2, and 1.35 between Planet2 and Planet3. But Kepler's third law is P² ∝ (a³)/(M), where M is the mass of the star. Since all planets are around the same star, M is constant. Therefore, the mass of the planets doesn't affect their orbital periods. The mass ratio is given, but since the star's mass is the same, the planet masses don't influence the period. So the mass ratios are irrelevant here. Therefore, the answer is B. ~33.4.
Therefore, the orbital period of Planet3 is larger than that of Planet1 by a factor of approximately $\boxed{B}$.
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