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Chapter 5: Problem 75

Two equal-mass stars maintain a constant distance apart of 8.0 \(\times\) 10\(^{11}\) m and revolve about a point midway between them at a rate of one revolution every 12.6 yr. (\(a\)) Why don't the two stars crash into one another due to the gravitational force between them? (\(b\)) What must be the mass of each star?

Short Answer

Expert verified
Centripetal force balances gravity, preventing crash; star mass is approximately 3.1 x 10^30 kg.

Step by step solution

01

Understand the problem context

We are given two equal-mass stars maintaining a constant distance apart, revolving around a midway point, hence they have a circular orbit due to gravitational forces. Our task is to understand why they don't crash into one another and determine the mass required for such a system.
02

Relate gravitational force and centripetal force

In a stable orbit, the gravitational force acting on each star equals the required centripetal force to keep the stars in that orbit. This is given by: \[ F_{gravity} = F_{centripetal} \].
03

Express gravitational force

The gravitational force between the two stars is given by Newton's law of universal gravitation: \[ F_{gravity} = \frac{G m^2}{r^2} \], where \(m\) is the mass of each star and \(r\) is the distance between them.
04

Express centripetal force

The centripetal force required for each star in circular motion is given by: \[ F_{centripetal} = \frac{mv^2}{r/2} \], where the radius of each star's orbit is half the distance between them \(\frac{r}{2}\), and \(v\) is the orbital speed of the star.
05

Relate speed to period

The orbital speed \(v\) is related to the orbital period \(T\) by: \[ v = \frac{2\pi(r/2)}{T} \]. Here, \(T\) is the period of revolution (12.6 years) converted into seconds.
06

Equate and solve for mass

By equating the expressions for gravitational and centripetal force, \[ \frac{G m^2}{r^2} = \frac{m \left(\frac{2\pi(r/2)}{T}\right)^2}{r/2} \], and solving for \(m\), we find the mass of each star.
07

Calculate values

Substituting values: \(G = 6.674 \times 10^{-11} \, \text{m}^3\, \text{kg}^{-1}\, \text{s}^{-2}\), \(r = 8.0 \times 10^{11} \, \text{m}\), and \(T = 12.6 \times 365 \times 24 \times 3600 \, \text{s}\), we solve for \(m\).

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Key Concepts

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

Centripetal Force
Centripetal force is the force required to keep an object moving in a circular path. It always points towards the center of the circle and is crucial for maintaining circular motion. In the case of two stars orbiting each other, the gravitational force between them provides the necessary centripetal force for their circular orbits.
  • The formula for centripetal force is given by \( F_{centripetal} = \frac{mv^2}{r} \), where \( m \) is the mass of the object, \( v \) is the object's speed, and \( r \) is the radius of the circle.
  • For the stars in our scenario, since they revolve around a midway point, each star's orbit radius is half the distance between them, which is \( \frac{r}{2} \).
The fact that these stars are held in orbit by this force, instead of crashing into each other, shows how accurately gravitational and centripetal forces must balance. This delicate balance prevents the stars from spiraling inward and colliding.
Orbital Mechanics
Orbital mechanics, or celestial mechanics, is the study of the motion of celestial objects under gravity. It's a key part of understanding how objects like stars, planets, and moons move in space. In our problem, we analyze the orbit formed by two stars that keep a constant distance while revolving around a shared center of mass.
  • Orbits can be circular or elliptical. The stars in the exercise follow a circular orbit, where they maintain a constant radius as they revolve.
  • The orbital speed necessary to keep an object in orbit can be calculated from its orbital period \( T \) using the formula \( v = \frac{2\pi(r/2)}{T} \). Here \( T \) is in seconds, as it converts the 12.6-year period from the exercise into a usable unit.
This analysis is crucial for predicting the future positions of stars, satellites, and planets, helping scientists understand cosmic events better.
Newton's Law of Universal Gravitation
Newton's law of universal gravitation is a fundamental principle that describes the gravitational attraction between two objects. It states that every point mass attracts every other point mass with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
  • The mathematical expression for this force is \( F_{gravity} = \frac{G m_1 m_2}{r^2} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{m}^3\, \text{kg}^{-1}\, \text{s}^{-2} \), \( m_1 \) and \( m_2 \) are the masses of the two stars, and \( r \) is the distance between their centers.
  • In our case, since the masses of the stars are equal, the equation simplifies to \( F_{gravity} = \frac{G m^2}{r^2} \).
This law is vitally important for understanding how stars, such as those in the exercise, remain in stable orbits without colliding, illustrating the harmonious balance between their mutual attractions and the distances separating them.

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