/*! 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 63 Two identical stars with mass \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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) The gravitational force is \(\frac{G M^2}{4R^2}\). (b) The orbital speed is \(\sqrt{\frac{G M}{4R}}\) and the period is \(\frac{4\pi \sqrt{R^3}}{\sqrt{G M}}\). (c) The energy required is \(\frac{G M^2}{2R}\).

Step by step solution

01

Gravitational Force Between Two Stars

To find the gravitational force between the two stars, we use Newton's law of universal gravitation. The formula is:\[F = \frac{G M_1 M_2}{r^2}\]Since both stars have the same mass \(M\) and they are separated by a distance \(2R\) (as they are on opposite sides of the circular orbit), we can substitute \(M_1 = M_2 = M\) and \(r = 2R\). Thus, the gravitational force is:\[F = \frac{G M^2}{(2R)^2} = \frac{G M^2}{4R^2}\]
02

Orbital Speed of Each Star

To find the orbital speed of each star, we use the centripetal force equation, which states that the gravitational force provides the necessary centripetal force for circular motion.\[F = \frac{G M^2}{4R^2}\]Also, the centripetal force needed to keep a star in circular motion is given by:\[F = \frac{M v^2}{R}\]Equating the two expressions for the force and solving for \(v\), we get:\[\frac{G M^2}{4R^2} = \frac{M v^2}{R}\]\[\Rightarrow v^2 = \frac{G M}{4R}\]\[\Rightarrow v = \sqrt{\frac{G M}{4R}}\]
03

Period of the Orbit

The period of the orbit \(T\) is the time it takes for a star to complete one full circle. Using the relationship between speed, distance, and time, we get:\[v = \frac{2\pi R}{T}\]Substituting the expression for \(v\) from the previous step, we have:\[\sqrt{\frac{G M}{4R}} = \frac{2\pi R}{T}\]Solving for \(T\),\[\Rightarrow T = \frac{2\pi R}{\sqrt{\frac{G M}{4R}}} = \frac{2\pi R \sqrt{4R}}{\sqrt{G M}} = \frac{2\pi \cdot 2R\sqrt{R}}{\sqrt{G M}}\]\[\Rightarrow T = \frac{4\pi R\sqrt{R}}{\sqrt{G M}} = \frac{4\pi \sqrt{R^3}}{\sqrt{G M}}\]
04

Energy Required to Separate the Stars

The energy required to separate the two stars to infinity is the gravitational potential energy at distance \(2R\), which is given by:\[U = -\frac{G M^2}{2R}\]Since the stars are initially bound by this potential energy, the work done to move them to infinity (where potential energy is zero) is equal to the absolute value of this energy:\[E_\text{required} = -U = \frac{G M^2}{2R}\]

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.

Gravitational Force
Gravitational force is the attractive force between two masses. It is one of the fundamental forces in nature and is described by Newton's law of universal gravitation. For two stars, this force can be calculated using the formula: \[F = \frac{G M_1 M_2}{r^2}\]Here, \(G\) represents the gravitational constant, \(M_1\) and \(M_2\) are the masses of the two stars, and \(r\) is the distance between their centers. In this exercise, since the stars have equal masses \(M\) and are separated by a distance of \(2R\), the gravitational force reduces to: \[F = \frac{G M^2}{4R^2}\]Understanding gravitational force helps us explain how two bodies in space, like stars, attract each other. This attraction keeps them in orbit around their common center of mass, preventing them from drifting apart.
Orbital Speed
Orbital speed is the velocity required for an object to stay in a stable orbit around another object. It is the speed that ensures the gravitational force acting as the centripetal force keeps the star moving in a circular path. For a star in orbit, the centripetal force necessary to maintain its path is given by:\[F = \frac{M v^2}{R}\]Where \(M\) is the mass of the star, \(v\) is the orbital speed, and \(R\) is the radius of the orbit. By equating this to the gravitational force computed earlier, we solve for the orbital speed and find:\[v = \sqrt{\frac{G M}{4R}}\]This crucial concept tells us how fast an object needs to move to remain in orbit, counterbalancing the pull of gravity with its motion's inertia.
Centripetal Force
Centripetal force is the inward force necessary for an object to follow a circular path. In our stars example, the gravitational force between the stars provides this necessary centripetal force. This means:\[F_{\text{centripetal}} = \frac{M v^2}{R}\]It's important to note that although it seems like there's an outward force acting on the orbiting star, there is none. The "outward" sensation or effect is due to inertia — the tendency of the star to move in a straight line is what creates the need for an inward force to keep its path circular.So, without the gravitational force acting as the centripetal force, the star would fly off in a straight line, escaping its orbit.
Gravitational Potential Energy
Gravitational potential energy represents the energy due to the position of an object in a gravitational field. For two stars separated by a distance, it can be expressed as: \[U = -\frac{G M^2}{2R}\]Here, the negative sign indicates that the energy is lower than when the stars are infinitely far apart (zero energy point). To separate these stars to infinity, one must do work equal to the magnitude of this energy:\[E_{\text{required}} = \frac{G M^2}{2R}\]This concept is essential for understanding how much energy is needed to overcome gravitational interactions, effectively de-binding two massive objects held together by their mutual attraction. This is why massive bodies like stars require enormous amounts of energy to be separated.

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

An astronaut, whose mission is to go where no one has gone before, lands on a spherical planet in a distant galaxy. As she stands on the surface of the planet, she releases a small rock from rest and finds that it takes the rock 0.480 s to fall 1.90 m. If the radius of the planet is 8.60 \(\times\) 10\(^7\) m, what is the mass of the planet?

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?

In its orbit each day, the International Space Station makes 15.65 revolutions around the earth. Assuming a circular orbit, how high is this satellite above the surface of the earth?

Planet Vulcan. Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to \(\frac{2}{3}\) of the average orbit radius of Mercury. What would be the orbital period of such a planet? (Such a planet was once postulated, in part to explain the precession of Mercury's orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury's precession has been explained by general relativity.)

A uniform sphere with mass 50.0 kg is held with its center at the origin, and a second uniform sphere with mass 80.0 kg is held with its center at the point \(x =\) 0, \(y =\) 3.00 m. (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.500 kg placed at the point \(x =\) 4.00 m, \(y =\) 0? (b) Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

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.