Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In a binary star system consisting of two stars of equal mass, where is the gravitational potential equal to zero? a) exactly halfway between the stars b) along a line bisecting the line connecting the stars c) infinitely far from the stars d) none of the above

Short Answer

Expert verified
Answer: a) exactly halfway between the stars

Step by step solution

01

Consider a point along the line connecting the two stars

Let's say that each star has a mass M, and the distance between them is d. We consider a point along the line that connects the two stars, and say this point is at a distance x away from the first star and (d-x) away from the second star.
02

Calculate the potential at this point due to each star

According to the formula V = -GM/r, the potential at this point due to the first star is V1 = -GM/x and the potential due to the second star is V2 = -GM/(d-x).
03

Calculate the total potential at this point

The total potential at this point is the sum of the potentials due to the two stars: V_total = V1 + V2 = -GM/x - GM/(d-x).
04

Set the total potential equal to zero

We need to find the location where the total potential is equal to zero, so we set V_total equal to zero and solve for x: 0 = -GM/x - GM/(d-x)
05

Solve for x

To solve for x, we can first multiply each side by x(d-x) to get rid of the denominators: 0 = -GM(d-x) - GMx Now, distribute the terms and combine like terms: 0 = -GMd + GMx - GMx Adding GMd to both sides, we get: GMd = 2GMx
06

Find the location where the potential is zero

Dividing both sides by 2GM, we arrive at the result: x = d/2 This shows that the location where the gravitational potential is equal to zero is exactly halfway between the two stars.
07

Answer

So, the correct option is: a) exactly halfway between the stars

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 Vaia!

Key Concepts

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

Binary Star System
A binary star system is a pair of stars that orbit around a common center of mass, where the gravitational attraction between them keeps them in orbit. It's a ballet of celestial bodies, held together and choreographed by gravity. These systems are the laboratories of the universe, providing insights into the nature of stars, stellar evolution, and gravitational interactions.

Observing binary systems allows astronomers to calculate the mass and other important properties of stars. Since both stars in a binary system orbit their mutual center of mass, the point at which they are perfectly balanced against each other's pull, studying their motion can reveal much about their individual masses and the nature of their gravitational bond. This dynamic interaction is a fundamental part of celestial mechanics, giving rise to various beautiful phenomena in the cosmos.
Gravitational Potential
Gravitational potential is a measure of the potential energy per unit mass at a point in a gravitational field, indicating the work done to bring a mass to that point from infinity. Given in the formula as
\[ V = -\frac{GM}{r} \],
where \( V \) is the gravitational potential, \(G\) is the gravitational constant, \( M \) is the mass of the celestial body creating the field, and \( r \) is the distance from the mass to the point in question. The negative sign signifies that gravitational forces are attractive, and energy is released as objects fall towards each other.

Understanding gravitational potential is crucial in problems involving the motion of bodies under gravity, particularly in systems with more than one mass, such as binary star systems. All stars exert a gravitational pull, and thus have a sphere of influence within which their potential can dominate.
Physics Problem Solving
Physics problem solving is all about applying fundamental principles to analyze, understand, and predict the behavior of physical systems. The step-by-step approach used in the given binary star system problem is a classic example of effective problem solving in physics.

This process generally involves identifying relevant physical equations, breaking down the system into manageable parts, and systematically solving for the unknowns. In our binary star problem, gravity equations and steps that involve algebraic manipulation to find where the gravitational potential is zero demonstrate the logical application of physical laws in a stepwise manner.

The key is to understand concepts at a fundamental level and then apply them to complex scenarios methodically. This approach not only helps in answering textbook questions but also in developing a deep understanding of physical phenomena and their governing laws.
Celestial Mechanics
Celestial mechanics is the branch of astronomy that deals with the motions and gravitational interactions of celestial bodies. This field applies principles of physics, particularly Newton's laws of motion and universal gravitation, to describe and predict the orbits of planets, moons, stars, and even man-made spacecraft.

In the context of a binary star system, celestial mechanics explains how the stars orbit around their center of mass and how their gravitational potentials interact. As seen in our problem, midpoints in binary systems are significant not just for the balance of mass, but also as points where gravitational influence equals out, making potential zero. This understanding of celestial mechanics allows astronomers to track stars' paths and predict their future positions and behaviors within their galaxy. For anyone intrigued by the dance of celestial objects across the night sky, a grasp of celestial mechanics is essential.

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

Newton's Law of Gravity specifies the magnitude of the interaction force between two point masses, \(m_{1}\) and \(m_{2}\), separated by a distance \(r\) as \(F(r)=G m_{1} m_{2} / r^{2} .\) The gravitational constant \(G\) can be determined by directly measuring the interaction force (gravitational attraction) between two sets of spheres by using the apparatus constructed in the late 18th century by the English scientist Henry Cavendish. This apparatus was a torsion balance consisting 6.00-ft wooden rod suspended fr a torsion wire, with a lead sphere having a diameter of 2.00 in and weight of \(1.61 \mathrm{lb}\) attached to each end. Two 12.0 -in, 348 -lb lead ball were located near the smaller bal about 9.00 in away, and held in place with a separate suspension system. Today's accepted value for \(G\) is \(6.674 \cdot 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\) Determine the force of attraction between the larger and smaller balls that had to be measured by this balance. Compare this force to the weight of the small balls.

You have been sent in a small spacecraft to rendezvous with a space station that is in a circular orbit of radius \(2.5000 \cdot 10^{4} \mathrm{~km}\) from the Earth's center. Due to a mishandling of units by a technician, you find yourself in the same orbit as the station but exactly halfway around the orbit from it! You do not apply forward thrust in an attempt to chase the station; that would be fatal folly. Instead, you apply a brief braking force against the direction of your motion, to put you into an elliptical orbit, whose highest point is your present position, and whose period is half that of your present orbit. Thus, you will return to your present position when the space station has come halfway around the circle to meet you. Is the minimum radius from the Earth's center-the low point-of your new elliptical orbit greater than the radius of the Earth \((6370 \mathrm{~km}),\) or have you botched your last physics problem?

A space shuttle is initially in a circular orbit at a radius of \(r=6.60 \cdot 10^{6} \mathrm{~m}\) from the center of the Earth. A retrorocket is fired forward reducing the total energy of the space shuttle by \(10.0 \%\) (that is, increasing the magnitude of the negative total energy by \(10.0 \%\) ), and the space shuttle moves to a new circular orbit with a radius that is smaller than \(r\). Find the speed of the space shuttle (a) before and (b) after the retrorocket is fired.

For two identical satellites in circular motion around the Earth, which statement is true? a) The one in the lower orbit has less total energy. b) The one in the higher orbit has more kinetic energy. c) The one in the lower orbit has more total energy. d) Both have the same total energy.

A spherical asteroid has a mass of \(1.869 \cdot 10^{20} \mathrm{~kg}\) and a radius of \(358.9 \mathrm{~km} .\) What is the escape speed from its surface?

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.

Sign-up for free