Problem 27 - University Physics with Modern Physics

Question

The star Rho\(^1\) Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho\(^1\) Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho\(^1\) Cancri?

Step-by-Step Solution

Answer

(a) Orbital speed: 4.465 × 10^4 m/s. (b) Orbital period: 1.25 years.

Step 1: Identify Constants and Given Values

We start with identifying constants and given values:- Gravitational constant, \( G = 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \).- The mass of the Sun, \( M_\odot = 1.989 \times 10^{30} \, \text{kg} \).- Orbital radius of Earth, \( R_E = 1.496 \times 10^{11} \, \text{m} \).Given:- Mass of Rho\(^1\) Cancri is 0.85 times the mass of the Sun: \( M_{\text{star}} = 0.85 \times M_\odot \).- Orbital radius of the planet is 0.11 times the Earth's orbital radius: \( r = 0.11 \times R_E \).

Step 2: Calculate the Mass of Rho^1 Cancri

Using the mass of the sun, we can calculate the mass of Rho\(^1\) Cancri:\[ M_{\text{star}} = 0.85 \times M_\odot = 0.85 \times 1.989 \times 10^{30} \, \text{kg} \approx 1.690 \times 10^{30} \, \text{kg} \].

Step 3: Calculate the Orbital Radius

Calculate the orbital radius of the planet using Earth's orbital radius:\[ r = 0.11 \times R_E = 0.11 \times 1.496 \times 10^{11} \, \text{m} \approx 1.6456 \times 10^{10} \, \text{m} \].

Step 4: Derive the Orbital Speed of the Planet

Use the formula for orbital speed:\[ v = \sqrt{\frac{G \cdot M_{\text{star}}}{r}} \].Substitute the known values:\[ v = \sqrt{\frac{6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \times 1.690 \times 10^{30} \, \text{kg}}{1.6456 \times 10^{10} \, \text{m}}} \approx 4.465 \times 10^4 \, \text{m/s} \].

Step 5: Derive the Orbital Period of the Planet

To find the orbital period, use Kepler's third law:\[ T = 2\pi \sqrt{\frac{r^3}{G \cdot M_{\text{star}}}} \].Substitute the known values:\[ T = 2\pi \sqrt{\frac{(1.6456 \times 10^{10} \, \text{m})^3}{6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \cdot 1.690 \times 10^{30} \, \text{kg}}} \approx 3.943 \times 10^7 \, \text{s} \].Convert to Earth years:\[ T \approx \frac{3.943 \times 10^7 \, \text{s}}{3.154 \times 10^7 \, \text{s/year}} \approx 1.25 \, \text{years}. \]

Step 6: Conclusion

The calculations show that:(a) The orbital speed of the planet is approximately \( 4.465 \times 10^4 \, \text{m/s} \).(b) The orbital period of the planet is approximately 1.25 Earth years.

Key Concepts

Orbital Speed

Orbital speed refers to the velocity needed for an object to continue orbiting another object without falling into it or flying away. It is a crucial concept when studying celestial bodies in motion. For any planet or satellite revolving around a star, its orbital speed ensures it maintains a stable orbit. This speed is determined by gravitational forces and the distance between the two bodies.

Orbital speed can be calculated using the formula:

• \( v = \sqrt{\frac{G \cdot M_{\text{star}}}{r}} \)

Here, \( G \) stands for the universal gravitational constant, \( M_{\text{star}} \) is the mass of the star, and \( r \) denotes the orbital radius of the planet.

In the example of the planet orbiting Rho\( ^1 \) Cancri, this speed is determined by taking into account the mass of Rho\( ^1 \) Cancri and its distance from the planet. The resulting orbital speed of approximately \( 4.465 \times 10^4 \) meters per second exemplifies how speed varies with these parameters.

Orbital Period

The orbital period is the time a celestial body takes to complete one full orbit around another body. It's related to both the distance of the orbiting body from its center of orbit and the mass of the central body.

Kepler's Third Law provides the foundation for calculating the orbital period, indicating a relationship between the time a planet takes to orbit a body and its orbital radius:

• \( T = 2\pi \sqrt{\frac{r^3}{G \cdot M_{\text{star}}}} \)

In this formula, \( T \) represents the orbital period, \( r \) the orbital radius, \( G \) the gravitational constant, and \( M_{\text{star}} \) the mass of the star.

For instance, the planet around Rho\( ^1 \) Cancri has an orbital period of about 1.25 Earth years, meaning it takes this planet roughly 1.25 years to circle once around its star. Such calculations highlight how the mass of the star and distance to the planet impact the planet's orbital period.

Gravitational Constant

The gravitational constant, often symbolized as \( G \), is a fundamental concept in physics and astronomy. It quantifies the strength of gravity, describing how massive objects attract each other. The value is approximately \( 6.674 \times 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \).

This constant plays a vital role in calculating gravitational forces in the universe. It appears in Newton's law of universal gravitation, which states that every point mass attracts every other point mass by a force acting along the line intersecting both points:

• \( F = G \frac{m_1 m_2}{r^2} \)

Here, \( F \) is the force between the masses, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between their centers.

In the studied example involving Rho\(^1\) Cancri, \( G \) helps calculate the orbital speed and period of its planet, illustrating how gravity influences celestial mechanics. This constant remains a cornerstone of astrophysical calculations, enabling us to understand movements and interactions in the cosmos.