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Chapter 13: Problem 25

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?

Short Answer

Expert verified
Orbital speed is approximately 47,465 m/s, and the orbital period is about 25.2 days.

Step by step solution

01

Understand the given data

We are given that the star Rho \(^{1}\) Cancri is 57 light-years from Earth, with a mass of \(0.85 \times M_{\text{Sun}}\). A planet orbits the star with a radius of \(0.11 \times a_{\text{Earth}}\). We're tasked to find the orbital speed and period.
02

Use the gravitational force and centripetal force relation

According to Newton's law of gravitation and circular motion, the gravitational force provides the necessary centripetal force for orbital motion. The formula is \( F_{\text{gravity}} = \frac{G M m}{r^2} = \frac{m v^2}{r} \), where \( G \) is the gravitational constant, \( M \) is the mass of the star, \( m \) is the mass of the planet, \( r \) is the radius of the orbit, and \( v \) is the orbital speed.
03

Solve for orbital speed \( v \)

By simplifying the formula \( \frac{G M m}{r^2} = \frac{m v^2}{r} \), we can solve for \( v \):\[ v = \sqrt{\frac{G M}{r}}. \] Substituting \( M = 0.85 \times M_{\text{Sun}} \) and \( r = 0.11 \times a_{\text{Earth}} \), calculate \( v \). This requires knowing \( G = 6.674 \times 10^{-11} \ \text{Nm}^2/\text{kg}^2 \), \( M_{\text{Sun}} = 1.989 \times 10^{30} \text{ kg} \), and \( a_{\text{Earth}} = 1.496 \times 10^{11} \ \text{m} \).
04

Calculate the value of \( v \)

First calculate \( r = 0.11 \times 1.496 \times 10^{11} = 1.64656 \times 10^{10} \ \text{m} \). Then, substitute these into the orbital speed equation: \[ v = \sqrt{\frac{6.674 \times 10^{-11} \times 0.85 \times 1.989 \times 10^{30}}{1.64656 \times 10^{10}}} \approx 47465 \ \text{m/s}. \] Hence, the orbital speed is approximately 47,465 m/s.
05

Use the orbital period formula

The orbital period \( T \) can be found using \( T = \frac{2\pi r}{v} \). We already have \( r = 1.64656 \times 10^{10} \ \text{m} \) and \( v = 47465 \ \text{m/s} \).
06

Calculate the orbital period \( T \)

Substitute the values into the period formula: \[ T = \frac{2 \pi \times 1.64656 \times 10^{10}}{47465} \approx 2176664 \ \text{seconds}. \] Convert this to days: \( \frac{2176664}{86400} \approx 25.2 \ \text{days}. \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Orbital Speed
Orbital speed is a critical concept in astrophysics, describing how fast an object travels along its orbit due to gravitational forces. The speed depends on the mass of the celestial body it orbits and the radius of the orbit itself. In our exercise involving a planet orbiting the star Rho \(^{1}\) Cancri, we determine the orbital speed using the formula: \[v = \sqrt{\frac{G M}{r}},\]where
  • \( v \) is the orbital speed,
  • \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \),
  • \( M \) is the mass of the star the planet orbits,
  • and \( r \) is the radius of the orbit.

In simple terms, the orbital speed increases with a more massive star and a tighter orbit, while it decreases if the path is larger or the star less massive.
Orbital Period
The orbital period is the time a planet takes to complete one full orbit around its star. For the planet orbiting Rho \(^{1}\) Cancri, we used the formula:\[T = \frac{2\pi r}{v},\]where
  • \( T \) is the orbital period,
  • \( 2\pi \) represents the complete circular path around the orbit,
  • \( r \) is the orbital radius,
  • and \( v \) is the orbital speed.

This method helps us understand how quickly a planet travels around its star. The smaller the orbit or the slower the speed, the longer the period. In our calculation, the planet's period is approximately 25.2 days, showing it's relatively fast compared to Earth's familiar year-long orbit.
Gravitational Force
Gravitational force is the attractive force that two masses exert on each other. In the context of our problem, it's the force between the planet and Rho \(^{1}\) Cancri. Newton's law of universal gravitation defines this force as:\[F_{\text{gravity}} = \frac{G M m}{r^2},\]where
  • \( G \) is the gravitational constant,
  • \( M \) is the mass of the star,
  • \( m \) is the mass of the planet,
  • \( r \) is the distance between their centers, or the orbital radius.

This force is crucial because it keeps planets in their orbits, providing the necessary centripetal force. Without it, planets would simply drift away into space. The gravitational pull between two bodies increases with larger masses and diminishes as they move further apart.
Newton's Law of Gravitation
Newton's law of gravitation elegantly explains the force acting between masses. It's responsible for the observable phenomenon of gravity, which though weak on small scales, governs the motion of celestial bodies. The law is expressed as:\[F_{\text{gravity}} = \frac{G M_1 M_2}{d^2},\]where
  • \( F_{\text{gravity}} \) is the gravitational force between two objects,
  • \( M_1 \) and \( M_2 \) are the masses of the objects,
  • \( d \) is the distance between their centers.

This law helped us find the gravitational interaction between the planet of Rho \(^{1}\) Cancri and its star. The concept is pivotal for understanding not only planet-star interactions but also for predicting phenomena such as orbit paths, satellite trajectories, and even tidal effects on Earth. Newton's insight laid the groundwork for later understanding and exploration of the universe.

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Most popular questions from this chapter

The acceleration due to gravity at the north pole of Neptune is approximately 10.7 \(\mathrm{m} / \mathrm{s}^{2} .\) Neptune has mass \(1.0 \times 10^{26} \mathrm{kg}\) and radius \(2.5 \times 10^{4} \mathrm{km}\) and rotates once around its axis in about 16 \(\mathrm{h}\) . (a) What is the gravitational force on a 5.0 -kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be \(15.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the center and \(2.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the surface. What is the acceleration due to gravity at the surface of this planet?

Rhea, one of Saturn's moons, has a radius of 765 \(\mathrm{km}\) and an acceleration due to gravity of 0.278 \(\mathrm{m} / \mathrm{s}^{2}\) at its surface. Calculate its mass and average density.

At the Galaxy's Core. Astronomers have observed a small, massive object at the center of our Milky Way galaxy (see Section 13.8 ). A ring of material orbits this massive object; the ring has a diameter of about 15 light-years and an orbital speed of about 200 \(\mathrm{km} / \mathrm{s} .\) (a) Determine the mass of the object at the center of the Milky Way galaxy. Give your answer both in kilograms and in solar masses (one solar mass is the mass of the sun). (b) Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? (c) Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzchild radius of this black hole be? Would a black hole of this size fit inside the earth's orbit around the sun?

Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: \(A 2.50-\) -kg stone thrown upward from the ground at 12.0 \(\mathrm{m} / \mathrm{s}\) returns to the ground in 6.00 s; the circumference of Mongo at the equator is \(2.00 \times 10^{5} \mathrm{km} ;\) and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information: (a) What is the mass of Mongo? (b) If the Aimless Wanderer goes into a circular orbit \(30,000 \mathrm{km}\) above the surface of Mongo, how many hours will it take the ship to complete one orbit?
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