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Chapter 13: Problem 71

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) \( F = \frac{G M^2}{4R^2} \), (b) \( v = \sqrt{\frac{G M}{2R}} \), \( T = 2\pi \sqrt{\frac{2R^3}{G M}} \), (c) \( E = \frac{G M^2}{2R} \).

Step by step solution

01

Understanding the Force Between Stars

The gravitational force between two stars can be calculated using Newton's law of universal gravitation. The formula is: \[ F = \frac{G M^2}{(2R)^2} = \frac{G M^2}{4R^2} \] where \(G\) is the gravitational constant, and \(2R\) is the distance between the centers of the two stars.
02

Calculating Orbital Speed

To find the orbital speed of each star, we use the centripetal force equation where the gravitational force provides the necessary centripetal force: \[ \frac{G M^2}{4R^2} = \frac{M v^2}{R} \] Solving for \(v\), the orbital speed, we get: \[ v = \sqrt{\frac{G M}{2R}} \].
03

Determining the Period of Orbit

The period \(T\) can be found from the relationship between the circumference of the orbit and the orbital speed:\[ T = \frac{2\pi R}{v} \]Substituting \(v = \sqrt{\frac{G M}{2R}} \), we find: \[ T = \frac{2\pi R}{\sqrt{\frac{G M}{2R}}} = 2\pi \sqrt{\frac{2R^3}{G M}} \].
04

Calculating Energy Required to Separate Stars

The energy required is equal to the gravitational potential energy. For two point masses, this is given by: \[ U = -\frac{G M^2}{2R} \]Thus, the energy needed to separate them is \(-U\), which becomes: \[ E = \frac{G M^2}{2R} \].

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Key Concepts

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

Gravitational Force
In a binary star system, two stars exert a gravitational force on each other. This force is determined by Newton's law of universal gravitation. The gravitational force is the attractive force that pulls the two stars towards each other. It is governed by the equation:
  • \( F = \frac{G M^2}{(2R)^2} = \frac{G M^2}{4R^2} \)
Here, \( G \) represents the gravitational constant, \( M \) is the mass of each star, and \( 2R \) is the distance separating the stars. The force arises from their mutual pull due to gravity.
This force helps maintain their stable orbit around the common center of mass. It ensures that despite their motion, they remain at a constant distance apart. Understanding this force is essential in studying celestial mechanics and dynamics of any two-body system.
Orbital Speed
The orbital speed of a star in a binary system is crucial for understanding its trajectory. The speed with which each star orbits the center of mass ensures that it remains in a stable circular orbit. This requires balancing the gravitational force and centripetal force.
The equation for orbital speed \( v \) in this system is derived from equating the gravitational force and the centripetal force:
  • \( \frac{G M^2}{4R^2} = \frac{M v^2}{R} \)
  • Solving for \( v \) gives: \( v = \sqrt{\frac{G M}{2R}} \)
The formula shows that the speed depends on the star's mass \( M \) and the radius \( R \) of the orbit. Understanding this concept helps in determining the motion of stars and planets in various gravitational systems.
Orbital Period
The orbital period is the time taken by a star to complete one full orbit around the center of mass. In our binary star system, this period provides insight into the dynamics and timing of the stars' movements.
To calculate the orbital period \( T \), we must consider the circumference of the orbit and the orbital speed:
  • \( T = \frac{2\pi R}{v} \)
  • Substituting \( v = \sqrt{\frac{G M}{2R}} \) results in: \( T = 2\pi \sqrt{\frac{2R^3}{G M}} \)
The period depends on the radius \( R \) and the mass \( M \), reflecting the time for gravitational interaction to occur over the pathway. Measuring the orbital period is fundamental in astronomy for predicting star positions and system behavior over time.
Gravitational Potential Energy
Gravitational potential energy in a binary star system relates to the work needed to separate the stars to infinity. Initially, the stars are bound by their gravitational attraction, with potential energy given by the formula:
  • \( U = -\frac{G M^2}{2R} \)
This equation quantifies the energy due to their positions in the gravitational field. To separate the stars completely, an amount of energy equal to the magnitude of this potential energy needs to be provided.
The energy required to achieve this separation is:
  • \( E = \frac{G M^2}{2R} \)
This highlights the strength of the gravitational bond between the two masses. Understanding gravitational potential energy helps explain phenomena like star system stability and dynamics when external forces are applied.

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Most popular questions from this chapter

Binary Star-Different Masses. Two stars, with masses \(M_{1}\) and \(M_{2},\) are in circular orbits around their center of mass. The star with mass \(M_{1}\) has an orbit of radius \(R_{1} ;\) the star with mass \(M_{2}\) has an orbit of radius \(R_{2} .\) (a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their masses - that is, \(R_{1} / R_{2}=M_{2} / M_{1}\) . (b) Explain why the two stars have the same orbital period, and show that the period \(T\) is given by \(T=2 \pi\left(R_{1}+R_{2}\right)^{3 / 2} / \sqrt{G\left(M_{1}+M_{2}\right)}\) . (c) The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 \(\mathrm{km} / \mathrm{s}\) . The second star, Beta, has an orbital speed of 12.0 \(\mathrm{km} / \mathrm{s}\) . The orbital period is 137 \(\mathrm{d}\) . What are the masses of each of the two stars? (d) One of the best candidates for a black hole is found in the binary system called A \(0620-0090 .\) The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole (see Fig. 13.27\() .\) The orbital period of \(\mathrm{A} 0620-0090\) is 7.75 hours, the mass of V616 Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of each object's orbit and the orbital speed of each object. Compare these answers to the orbital radius and orbital speed of the earth in its orbit around the sun.

Cavendish Experiment. In the Cavendish balance apparatus shown in Fig. \(13.4,\) suppose that \(m_{1}=1.10 \mathrm{kg}, m_{2}=\) \(25.0 \mathrm{kg},\) and the rod connecting the \(m_{1}\) pairs is 30.0 \(\mathrm{cm}\) long. If, in each pair, \(m_{1}\) and \(m_{2}\) are 12.0 \(\mathrm{cm}\) apart center to center, find (a) the net force and (b) the net torque (about the rotation axis) on the rotating part of the apparatus. (c) Does it seem that the torque in part (b) would be enough to easily rotate the rod? Suggest some ways to improve the sensitivity of this experiment.

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.

A landing craft with mass \(12,500 \mathrm{kg}\) is in a circular orbit \(5.75 \times 10^{3} \mathrm{m}\) above the surface of a planet. The period of the orbit is 5800 s. The astronauts in the lander measure the diameter of the planet to be \(9.60 \times 10^{6} \mathrm{m} .\) The lander sets down at the north pole of the planet. What is the weight of an 85.6 -kg astronaut as he steps out onto the planet's surface?

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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