Problem 52 - College 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 I 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 "Cancri?

Step-by-Step Solution

Answer

The orbital speed is 55,927 m/s and the orbital period is 27.69 Earth days.

Step 1: Understand the Problem

We are given a star, the mass of which is 0.85 times the mass of our sun, and a planet orbiting it in a circular orbit. The orbital radius of this planet is 0.11 times the Earth's orbital radius around the sun. We need to calculate the orbital speed and period of the planet.

Step 2: Calculate the Orbital Speed

The formula for calculating the orbital speed of a planet is \( v = \sqrt{\frac{GM}{r}} \), where \( G \) is the gravitational constant, \( M \) is the mass of the star, and \( r \) is the orbital radius. First, substitute the values: \( M = 0.85 \times M_{\text{sun}} \) and \( r = 0.11 \times r_{\text{earth}} \). Using \( G = 6.67 \times 10^{-11} \, \text{m}^3\, \text{kg}^{-1}\, \text{s}^{-2} \), \( M_{\text{sun}} = 1.989 \times 10^{30} \, \text{kg} \), and \( r_{\text{earth}} = 1.496 \times 10^{11} \, \text{m} \), calculate \( v \).

Step 3: Calculate the Orbital Period

Use Kepler's Third Law for circular orbits, which can be expressed as \( T = 2\pi \sqrt{\frac{r^3}{GM}} \). Using the previously calculated and substituted values for \( r \) and \( M \) based on the given conditions, solve for \( T \).

Step 4: Compute the Orbital Speed and Period Numerically

Substitute the numerical values into the equations from Steps 2 and 3: - \( v = \sqrt{\frac{(6.67 \times 10^{-11}) \times (0.85 \times 1.989 \times 10^{30})}{0.11 \times 1.496 \times 10^{11}}} \) - \( T = 2\pi \sqrt{\frac{(0.11 \times 1.496 \times 10^{11})^3}{(6.67 \times 10^{-11}) \times (0.85 \times 1.989 \times 10^{30})}} \)Calculate these values to find the orbital speed and period.

Step 5: Analyze Results

After performing the calculations, we find the orbital speed and period of the planet. - The orbital speed is approximately 55,927 m/s. - The orbital period translates to about 27.69 Earth days.

Key Concepts

Orbital Speed

Orbital speed refers to the velocity at which a planet moves in its orbit around a star. It's crucial for understanding how planets interact with their stars and ensuring they remain in stable orbits.
To calculate the orbital speed, we use the formula:

• \( v = \sqrt{\frac{GM}{r}} \)

In this equation:

• \( v \) is the orbital speed,
• \( G \) is the gravitational constant \( (6.67 \times 10^{-11} \, \text{m}^3\, \text{kg}^{-1}\, \text{s}^{-2}) \),
• \( M \) is the mass of the star (e.g., Rho \(^{1}\) Cancri), and
• \( r \) is the orbital radius of the planet.

By substituting the values into this formula, we can determine the exact speed at which the planet orbits its star.

Orbital Period

The orbital period is the time it takes for a planet to complete one full orbit around a star. Understanding it gives us a sense of the rhythm and timing of planetary systems.
The orbital period can be calculated using Kepler's Third Law, which is particularly useful for circular orbits:

• \( T = 2\pi \sqrt{\frac{r^3}{GM}} \)

Here:

• \( T \) is the orbital period,
• \( r \) is the orbital radius,
• \( G \) is the gravitational constant, and
• \( M \) is the mass of the star.

By solving this equation, we find how long it takes the planet to orbit Rho \(^{1}\) Cancri once.

Kepler's Third Law

Kepler's Third Law provides a beautifully simple relationship between the orbital period and the distance of a planet from its star. The law states that the square of a planet's orbital period is proportional to the cube of the average distance from the star.
Mathematically, it is expressed as:

• \( T^2 \propto r^3 \)

This means:

• The further a planet is from its star, the longer it takes to orbit.
• This law helps predict planetary motions without needing complex models.

By utilizing Kepler's Third Law, we gain insight into the dynamic structures of solar systems and the balance of gravitational forces.

Gravitational Constant

The gravitational constant, denoted as \( G \), is a fundamental constant in physics. It quantifies the strength of the gravitational force between two masses.
The value of \( G \) is:

• \( 6.67 \times 10^{-11} \, \text{m}^3\, \text{kg}^{-1}\, \text{s}^{-2} \)

Gravitational constant is crucial for calculating forces between celestial objects, such as stars and planets.
In the context of planetary motion:

• It allows us to compute the gravitational attraction exerted by a star on its planets.
• Along with the mass of the star and the distance of the planet, \( G \) helps determine key orbital properties like speed and period.

Understanding \( G \) gives us the ability to predict and explain the motion of planets in different solar systems.