Let's break down and solve each part of the question systematically.

1. Orbital Period of Earth's Twin

We are using Newton's version of Kepler's third law: $P^2 = \frac{4\pi^2}{G(M_1 + M_2)} a^3$ Where:

• $P$ = orbital period of the planet
• $G$ = gravitational constant
• $M_1$ = mass of the star
• $M_2$ = mass of the planet (Earth's twin)
• $a$ = semi-major axis (distance between the planet and the star)

Since we are comparing the orbital period in two systems where the distance $a$ is the same, we can simplify the equation to: $\frac{P_2}{P_1} = \sqrt{\frac{M_1}{M_2}}$ In this case:

• $M_1$ = 3.1 times the mass of the Sun
• $M_2$ = mass of our Sun (since Earth's twin is orbiting a Sun-like star in the original system)
• $P_1$ = 12 months (orbital period of Earth)

Now, solving for $P_2$: $P_2 = P_1 \cdot \sqrt{\frac{M_2}{M_1}} = 12 \cdot \sqrt{\frac{1}{3.1}} = 12 \cdot \sqrt{0.3226} \approx 12 \cdot 0.568 = 6.8 \text{ months}$ Thus, the orbital period of Earth's twin is 6.8 months.

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2. Escape Speed from a Tiny Moon

The formula for the escape speed $v$ from a celestial body is: $v = \sqrt{\frac{2GM}{R}}$ Where:

• $G = 6.67 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}$
• $M = 5.2 \times 10^{16} \, \text{kg}$
• $R = 8.3 \, \text{km} = 8.3 \times 10^3 \, \text{m}$

Now, plugging in the values: $v = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.2 \times 10^{16}}{8.3 \times 10^3}} = \sqrt{\frac{6.9368 \times 10^6}{8.3 \times 10^3}} = \sqrt{835.63 \, \text{m}^2/\text{s}^2} \approx 28.9 \, \text{m/s}$ Thus, the escape speed is 29 m/s (rounded to two significant figures).

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3. Radial Velocity of Helvetios (Doppler Effect)

The formula for the Doppler shift is: $v = c \cdot \frac{\Delta \lambda}{\lambda_0}$ Where:

• $v$ = velocity of the star
• $c = 3 \times 10^5 \, \text{km/s}$ (speed of light)
• $\Delta \lambda = \lambda_{\text{observed}} - \lambda_{\text{rest}} = 415.4 \, \text{nm} - 388.2 \, \text{nm} = 27.2 \, \text{nm}$
• $\lambda_0 = 388.2 \, \text{nm}$

Now, plugging in the values: $v = 3 \times 10^5 \cdot \frac{27.2}{388.2} = 3 \times 10^5 \cdot 0.07 \approx 21 \, \text{km/s}$ Since the observed wavelength is longer than the rest wavelength, this means Helvetios is moving away from us.

Thus, the star is receding at 21 km/s.

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Summary of Answers:

1. Orbital period of Earth's twin: 6.8 months
2. Escape speed from the tiny moon: 29 m/s
3. Radial velocity of Helvetios: 21 km/s (away from us)

Would you like further details or explanations on any of the calculations? Here are five related questions you might find interesting:

1. How would the orbital period change if the star were twice as massive as the Sun?
2. How does escape velocity vary with mass and radius?
3. What is the relationship between escape velocity and orbital speed at the surface of a planet?
4. What would the Doppler shift be for an object moving toward us?
5. How does the Doppler effect change with different wavelengths of light?

Tip: The escape velocity is directly proportional to the square root of mass and inversely proportional to the square root of radius.