Problem 50 - College Physics

Question

The star Rho' 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: \( \approx 42163 \) m/s. (b) Orbital period: \( \approx 0.675 \) years.

Step 1: Understand the Known Values

We are given that the mass of Rho' Cancri, the star, is 0.85 times the mass of the Sun, \( M = 0.85 imes M_{\text{sun}} \). We will approximate the mass of the Sun, \( M_{\text{sun}} \), to be \( 1.989 \times 10^{30} \) kg. The orbital radius \( r \) is 0.11 times the Earth's orbital radius, which is approximately 1 astronomical unit (1 AU = \( 1.496 \times 10^{11} \) m). Thus, \( r = 0.11 \times 1.496 \times 10^{11}\, \text{m} \).

Step 2: Calculate the Gravitational Force

The gravitational force between a star and a planet provides the centripetal force for the planet. The gravitational force, \( F \), is given by \( F = \frac{G M m}{r^2} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), \( M \) is the mass of the star, and \( m \) is the mass of the planet. However, we will equate this with the centripetal force to find speed and period.

Step 3: Calculate the Orbital Speed

Using the formula for centripetal force, \( F_c = \frac{m v^2}{r} \), equate this to the gravitational force: \( \frac{m v^2}{r} = \frac{G M m}{r^2} \). Simplifying, we get \( v^2 = \frac{G M}{r} \). Plug in the values, \( v = \sqrt{\frac{G \times 0.85 \times 1.989 \times 10^{30}}{0.11 \times 1.496 \times 10^{11}}} \). After solving, \( v \approx 42163 \, \text{m/s} \).

Step 4: Calculate the Orbital Period Using Kepler's Third Law

Kepler's Third Law states \( T^2 = \frac{4\pi^2r^3}{G M} \). Substitute \( r = 0.11 \times 1.496 \times 10^{11} \, \text{m} \) and \( M = 0.85 \times 1.989 \times 10^{30} \), \( T = \sqrt{\frac{4\pi^2 \times (0.11 \times 1.496 \times 10^{11})^3}{G \times 0.85 \times 1.989 \times 10^{30}}} \). Solve this to get \( T \approx 2.13 \times 10^7 \, \text{s} \), which can be converted into Earth years.

Step 5: Convert Orbital Period to Earth Years

There are approximately \( 3.154 \times 10^7 \) seconds in an Earth year. Divide the period in seconds to find the period in Earth years: \( T \approx \frac{2.13 \times 10^7}{3.154 \times 10^7} \approx 0.675 \, \text{years} \).

Key Concepts

Gravitational Force

Gravitational force is a fundamental interaction that attracts two bodies toward one another. For orbital mechanics, this force is crucial in understanding how planets, stars, and other celestial bodies interact in space. Its formula is given by: \[ F = \frac{G M m}{r^2} \]where:

• \(F\) is the gravitational force,
• \(G\) is the universal gravitational constant (\(6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2\)),
• \(M\) and \(m\) are the masses of the two bodies,
• \(r\) is the distance between the centers of the two masses.

This circular dependency illustrates how the gravitational pull keeps planets in orbit, balancing the planet's inertia that wants it to fly off into space.

Centripetal Force

In the context of orbital motion, centripetal force is the force that keeps an object moving in a circular path and is directed toward the center of rotation. For planets orbiting stars, the necessary centripetal force is provided by the gravitational attraction. The formula for centripetal force is: \[ F_c = \frac{m v^2}{r} \]where:

• \(F_c\) is the centripetal force,
• \(m\) is the mass of the orbiting object,
• \(v\) is the orbital speed,
• \(r\) is the radius of the path.

By equating the centripetal force and the gravitational force, we can derive potential formulas to calculate orbital speed and period, revealing a harmonious balance between inertia and gravity.

Kepler's Laws

Kepler's Laws of Planetary Motion describe how planets orbit around stars. These laws fundamentally contributed to our understanding of celestial mechanics. Let's briefly explore these laws:

• First Law (Law of Ellipses):

  Each planet orbits the sun in a path called an ellipse, with the sun at one of two foci.

• Second Law (Law of Equal Areas):

  A line segment joining a planet and the sun sweeps out equal areas during equal intervals of time.

• Third Law (Law of Harmonies):

  The square of the orbital period \(T\) of a planet is directly proportional to the cube of the semi-major axis \(a\) of its orbit, expressed as \(T^2 \propto a^3\).

This last law, in particular, allows us to connect the orbital period and radius through the masses involved, providing insights into distant star systems.

Astronomical Units

Astronomical units (AU) are a convenient measure of distance used in astronomy. One AU is defined as the average distance from the Earth to the Sun, approximately equal to \(1.496 \times 10^{11}\) meters. This measurement simplifies the expression of large astronomical distances in terms of the Earth-Sun distance. In the given problem, Rho' Cancri's planet's orbital radius is mentioned as 0.11 times the solar system's Earth orbit radius, thus simplifying calculations by scaling with AU rather than meters. By using AU, astronomers and scientists can convey large-scale cosmic measurements without resorting to exceptionally large numbers, making data interpretation more efficient.