/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Binary Star-Equal Masses. Two id... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Binary Star-Equal Masses. Two identical stars with mass \(M\) orbit around their center of mass. Each orbit is circular and has radius \(R\) , so that the two stars are always on opposite sides of the circle. (a) Find the gravitational force of one star on the other. (b) Find the orbital speed of each star and the period of the orbit. (c) How much energy would be required to separate the two stars to infinity?

Short Answer

Expert verified
(a) Gravitational force: \( \frac{G M^2}{4R^2} \). (b) Orbital speed: \( \sqrt{\frac{G M}{4R}} \), Period: \( \frac{4\pi R^{3/2}}{\sqrt{G M}} \). (c) Energy to separate: \( \frac{G M^2}{2R} \).

Step by step solution

01

Understand the Gravitational Force Between Two Stars

The gravitational force acting on one star due to the other can be determined by Newton's law of universal gravitation. The force is given by the formula: \[ F = \frac{G M^2}{(2R)^2} \] where \( G \) is the gravitational constant, \( M \) is the mass of each star, and \( 2R \) is the distance between the two stars (since they are on opposite sides of the orbit). Simplifying, we have: \[ F = \frac{G M^2}{4R^2} \]
02

Calculate Orbital Speed

To calculate the orbital speed \( v \), we need the centripetal force, which for circular motion is \( F_c = \frac{Mv^2}{R} \). This is provided by the gravitational force, so setting \( F_c = F \) gives us:\[ \frac{Mv^2}{R} = \frac{G M^2}{4R^2} \]Solving for \( v \), we find:\[ v = \sqrt{\frac{G M}{4R}} \]
03

Find the Orbital Period

The period \( T \) is the time it takes for one complete orbit. Since the orbit is circular, the circumference is \( 2\pi R \) and speed is \( v \), so:\[ T = \frac{2\pi R}{v} \]Substituting for \( v \), we get:\[ T = \frac{2\pi R}{\sqrt{\frac{G M}{4R}}} = \frac{2\pi \sqrt{4R^3}}{\sqrt{G M}} = \frac{2\pi \cdot 2R^{3/2}}{\sqrt{G M}} = \frac{4\pi R^{3/2}}{\sqrt{G M}} \]
04

Calculate Energy Required to Separate Stars

The energy required to separate the two stars to infinity is equal to the gravitational potential energy of the system. The potential energy of the two-star system is:\[ U = -\frac{G M^2}{2R} \]Since they are in a stable orbit, there is no kinetic energy needed, just enough to overcome this gravitational binding energy. Thus, energy required is:\[ E_{required} = \frac{G M^2}{2R} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Binary Stars
Binary stars are systems comprised of two stars orbiting a common center of mass. Imagine a cosmic dance where each star affects the other's path, creating a unique gravitational relationship. These systems are not only beautiful but also essential for the study of astrophysics, as they offer insights into stellar masses and life cycles.
Understanding binary stars helps us learn not just about the individual stars but also about the effects of gravity on large scales. When the binary stars have identical masses, like in the exercise, each traces a circular path with the same radius, remaining on opposite sides of the orbit. The center of this orbit does not coincide with either star but is located precisely midway between them, acting as a pivot around which the stars revolve.
  • Each star's gravitational pull on the other governs their orbital mechanics.
  • In equal-mass binaries, the gravitational attraction is balanced by the stars' motion, keeping them in a stable orbit.
This dynamic setup allows for complex observations and calculations, further enriching our understanding of gravitational forces and celestial mechanics.
Orbital Mechanics
Orbital mechanics, or celestial mechanics, is the branch of astronomy that deals with the motions of celestial bodies under the influence of gravitational forces. In the context of binary stars that move along circular orbits, the gravitational force between the two stars provides the necessary centripetal force for circular motion.
For binary stars, the orbital mechanics are dictated by Newton's laws of motion and his law of universal gravitation. Each star experiences a gravitational pull towards the other, calculated as: \[ F = \frac{G M^2}{4R^2} \] where each star orbits around the mid-point at a distance, known as the orbital radius. In these circumstances, the orbital speed, denoted by \( v \), can be derived from the balance of gravitational and centripetal forces. Using the relation: \[ \frac{Mv^2}{R} = \frac{G M^2}{4R^2} \], we solve to find the speed: \[ v = \sqrt{\frac{G M}{4R}} \] The period \( T \), or the time it takes for a star to complete one full orbit, is related to both the orbit's circumference \( 2\pi R \) and the speed \( v \) as follows: \[ T = \frac{4\pi R^{3/2}}{\sqrt{G M}} \] Orbital mechanics reveal the natural balance and predictability inherent in celestial movements, offering a window into the effective choreography enforced by gravitational forces.
Gravitational Potential Energy
Gravitational potential energy is the energy stored within a system of masses due to their positions relative to each other under the influence of gravity. In a binary star system, this concept is crucial to understanding the 'binding energy' that holds the stars together in their orbit.
The gravitational potential energy \( U \) for our binary star system is expressed as: \[ U = -\frac{G M^2}{2R} \] The negative sign indicates that work must be done against the gravitational field to move the stars apart. For the task of separating the stars to infinity, this potential energy equates directly to the energy required, marked as \( E_{required} \): \[ E_{required} = \frac{G M^2}{2R} \] Essentially, this is the amount of mechanical energy one would need to provide to overcome the gravitational pull tethering the stars together.
  • It quantifies the energy needed to disrupt the gravitational balance.
  • This concept helps in understanding stellar operations and energy transformations in astrophysical phenomena.
Gravitational potential energy offers vital insights into the energetic demands and dynamics within binary systems, assisting in the broader exploration of the universe's mass-energy interactions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The planet Uranus has a radius of \(25,560 \mathrm{km}\) and a surface acceleration due to gravity of 11.1 \(\mathrm{m} / \mathrm{s}^{2}\) at its poles. Its moon Miranda (discovered by Kniper in 1948 ) is in a circular orbit about Uranus at an altitude of \(104,000 \mathrm{km}\) above the planet's surface. Miranda has a mass of \(6.6 \times 10^{19} \mathrm{kg}\) and a radius of 235 \(\mathrm{km}\) . (a) Calculate the mass of Uranus from the given data, (b) Calculate the magnitude of Miranda's acceleration due to its orbital motion about Uranus. (c) Calculate the acceleration due to Miranda's gravity at the surface of Miranda. (d) Do the answers to parts (b) and (c) mean that an object released 1 \(\mathrm{m}\) above Miranda's surface on the side toward Uranus will fall up relative to Miranda? Explain.

A hammer with mass \(m\) is dropped from rest from a height \(h\) above the earth's surface. This height is not necessarily small compared with the radius \(R_{\mathrm{E}}\) of the earth. If you ignore air resistance, derive an expression for the speed \(v\) of the hammer when it reaches the surface of the earth. Your expression should involve \(h, R_{\mathrm{E}},\) and \(m_{\mathrm{E}},\) the mass of the earth.

One of the brightest comets of the 20 th century was Comet Hyakutake, which passed close to the sun in early 1996 . The orbital period of this comet is estimated to be about \(30,000\) years. Find the semi-major axis of this comet's orbit. Compare it to the average sun-Pluto distance and to the distance to Alpha Centauri, the nearest star to the sun, which is 4.3 light-years distant.

An \(8.00-\mathrm{kg}\) point mass and a \(15.0-\mathrm{kg}\) point mass are held in place 50.0 \(\mathrm{cm}\) apart. A particle of mass \(m\) is released from a point between the two masses 20.0 \(\mathrm{cm}\) from the \(8.00-\mathrm{kg}\) mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

Your starship, the Aimless Wanderer, lands on the mysteri- ous planet Mongo. As chief scientist-engineer, you make the following measurements: A \(2.50-\mathrm{kg}\) stone thrown upward from the ground at 12.0 \(\mathrm{m} / \mathrm{s}\) returns to the ground in \(8.00 \mathrm{s} ;\) 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?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.