Newton’s version of Kepler’s third law is P2=4π2G(M1+M2)a3. For the Earth orbiting the Sun, the orbital period is 1 year = 12 months. Since the square of the period P varies inversely with the sum of the masses (M1+M2), the period P depends on the inverse square root of the object masses. If a solar system has a star that is 3.1 times as massive as our Sun, and has Earth's exact twin, what is the orbital period of Earth's twin in months? Use two significant figures in your answer.

What is the escape speed in m/s from a tiny moon that has a mass of 5.2e16 kg and radius of 8.3 km? The gravitational constant is G = 6.67e-11 m3 kg-1 s-2. Use two significant figures in your answer.

At rest, carbon emits a spectral line at 388.2 nm. If this line is observed at 415.4 nm from exoplanet host star Helvetios, how fast is Helvetios moving toward or away from us in km/s? The speed of light is c = 3e5 km/s. Use two significant figures in your answer.

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**.

---

### 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).

---

### 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**.

---

### 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.

Newton’s version of Kepler’s third law is P^2 = 4π^2G(M1 + M2)a^3. For the Earth orbiting the Sun, the orbital period is 1 year = 12 months. Since the square of the period P varies inversely with the sum of the masses (M1 + M2), the period P depends on the inverse square root of the object masses. If a solar system has a star that is 3.1 times as massive as our Sun, and has Earth's exact twin, what is the orbital period of Earth's twin in months? Use two significant figures in your answer.

What is the escape speed in m/s from a tiny moon that has a mass of 5.2e16 kg and radius of 8.3 km? The gravitational constant is G = 6.67e-11 m^3 kg^-1 s^-2. Use two significant figures in your answer.

At rest, carbon emits a spectral line at 388.2 nm. If this line is observed at 415.4 nm from exoplanet host star Helvetios, how fast is Helvetios moving toward or away from us in km/s? The speed of light is c = 3e5 km/s. Use two significant figures in your answer.

Solve advanced physics problems involving orbital mechanics, gravitational physics, and the Doppler effect. Learn how to calculate the orbital period of Earth's twin in months, escape speed from a tiny moon, and the radial velocity of exoplanet host star Helvetios. This guide is designed for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation