/*! 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 6 Two gravitating masses \(m_{1}\)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 6

Two gravitating masses \(m_{1}\) and \(m_{2}\left(m_{1}+m_{2}=M\right)\) are separated by a distance \(r_{0}\) and released from rest. Show that when the separation is \(r\left( < r_{0}\right),\) the speeds are $$v_{1}=m_{2} \sqrt{\frac{2 G}{M}\left(\frac{1}{r}-\frac{1}{r_{0}}\right)}, \quad v_{2}=m_{1} \sqrt{\frac{2 G}{M}\left(\frac{1}{r}-\frac{1}{r_{0}}\right)}$$

Short Answer

Expert verified
Using the conservation of mechanical energy, we found that when two masses $m_1$ and $m_2$ are released from an initial separation $r_0$ and reach a new separation $r$, their speeds are given by the formulas: \[v_1 = m_2\sqrt{\frac{2G}{M}\left(\frac{1}{r} - \frac{1}{r_0}\right)}\] and \[v_2 = m_1\sqrt{\frac{2G}{M}\left(\frac{1}{r} - \frac{1}{r_0}\right)}\] where $G$ is the gravitational constant and $M = m_1 + m_2$ is the total mass of the system.

Step by step solution

01

Write down the mechanical energy conservation equation

(for the given initial and final states): The total mechanical energy (sum of kinetic and potential energies) of the system at the initial state (\(v_1 = 0\), \(v_2 = 0\), \(r = r_0\)) should be equal to the total mechanical energy of the system at the final state, when the masses have new velocities \(v_1\) and \(v_2\) and their separation is \(r\). So, we can write the equation as: Initial mechanical energy = Final mechanical energy
02

Express the initial and final energies in terms of the given variables

: Initially, the masses are at rest, so their kinetic energies are zero and the mechanical energy is only due to their gravitational potential energy: \[E_{initial} = -\frac{Gm_1m_2}{r_0}\] At the final state, when the separation is \(r\), both masses will have kinetic and potential energies: \[E_{final} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 - \frac{Gm_1m_2}{r}\]
03

Substitute the conservation of mechanical energy equation and solve for \(v_1\) and \(v_2\)

: According to the energy conservation equation, \[E_{initial} = E_{final}\] Substitute the expressions for initial and final energies, \[-\frac{Gm_1m_2}{r_0} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 - \frac{Gm_1m_2}{r}\] Now, rearrange the terms and solve for \(v_1\) and \(v_2\), separately: 1) For \(v_1\): \[v_1^2 = \frac{2Gm_1m_2}{m_1}\left(\frac{1}{r} - \frac{1}{r_0}\right)\] Now, divide by \(m_1\) on both sides: \[v_1^2 = \frac{2Gm_2}{M}\left(\frac{1}{r} - \frac{1}{r_0}\right)\] Take the square root of both sides: \[v_1 = m_2\sqrt{\frac{2G}{M}\left(\frac{1}{r} - \frac{1}{r_0}\right)}\] 2) Similarly, for \(v_2\): \[v_2^2 = \frac{2Gm_1m_2}{m_2}\left(\frac{1}{r} - \frac{1}{r_0}\right)\] Divide by \(m_2\) and take square root: \[v_2 = m_1\sqrt{\frac{2G}{M}\left(\frac{1}{r} - \frac{1}{r_0}\right)}\] Thus, we have proved that the given formulas for \(v_1\) and \(v_2\) are correct.

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 91影视!

Key Concepts

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

Mechanical Energy Conservation
Understanding the conservation of mechanical energy is essential for analyzing the behavior of objects within a gravitational system. Simply put, the principle of mechanical energy conservation states that in the absence of non-conservative forces, such as friction, the total mechanical energy of a system remains constant.

Mechanical energy itself is the sum of kinetic energy, which is energy due to motion, and potential energy, which is stored energy based on the object's position or state. In the context of gravitational systems, the relevant form of potential energy is gravitational potential energy.

Consider two objects with masses \(m_1\) and \(m_2\) separated by a distance \(r\). When they are released from rest, as in the exercise given, we expect them to move toward each other due to the force of gravity. Because of conservation of mechanical energy, the decrease in gravitational potential energy as they move closer is transformed into an increase in kinetic energy.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field. For two objects with masses \(m_1\) and \(m_2\) distanced at \(r\), their GPE is given by \(-\frac{Gm_1m_2}{r}\), where \(G\) represents the gravitational constant. This negative sign indicates that gravitational force is attractive.

In our particular problem, as the distance between the masses decreases from \(r_0\) to \(r\), the GPE decreases, signifying that work is done by the gravitational force, leading to a conversion of potential energy into kinetic energy. It is this transfer of energy that causes the masses to accelerate towards each other. The precise calculation of this potential energy is key to solving the problem of determining the speeds of the masses at a new separation \(r\).
Kinetic Energy
Kinetic energy (KE) is the energy of an object due to its motion. It is calculated using the formula \(\frac{1}{2}mv^2\), where \(m\) is the mass of the object and \(v\) its velocity. Importantly, kinetic energy is a scalar quantity; it has magnitude but no direction.

In the context of our problem, the masses \(m_1\) and \(m_2\) start with zero kinetic energy because they are initially at rest. As they gravitationally attract each other and move closer, they accelerate, thereby gaining kinetic energy. This change in kinetic energy is directly linked to the change in gravitational potential energy; as one increases, the other decreases in a closed system according to the law of conservation of energy.

The solution method for this exercise makes use of the fact that the two types of energy, potential and kinetic, transform into one another as the masses fall towards each other. This concept is a cornerstone in classical mechanics and illustrates how forces and motion are interconnected through energy conservation.

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

Consider the problem of the particle moving on the surface of a cone, as discussed in Examples 7.4 and 8.7 . Show that the effective potential is $$V(r)=\frac{l^{2}}{2 m r^{2}}+m g r \cot \alpha$$ (Note that here \(r\) is the radial distance in cylindrical coordinates, not spherical \(c 0\) ordinates; see Figure \(7-2 .\) ) Show that the turning points of the motion can be found from the solution of a cubic equation in \(r\). Show further that only two of the roots are physically meaningful, so that the motion is confined to lie within two horizontal planes that cut the cone.

Perform the integration of Equation 8.38 to obtain Equation 8.39

Use Kepler's results (i.e., his first and second laws) to show that the gravitational force must be central and that the radial dependence must be \(1 / r^{2}\). Thus, perform an inductive derivation of the gravitational force law.

An almost circular orbit (i.e., \(\varepsilon \ll 1\) ) can be considered to be a circular orbit to which a small perturbation has been applied. Then, the frequency of the radial motion is given by Equation \(8.89 .\) Consider a case in which the force law is \(F(r)=-k / r^{n}\) (where \(n\) is an integer), and show that the apsidal angle is \(\pi / \sqrt{3-n}\) Thus, show that a closed orbit generally results only for the harmonic oscillator force and the inverse-square-law force (if values of \(n\) equal to or smaller than -6 are excluded

Discuss the motion of a particle in a central inverse-square-law force field for a superimposed force whose magnitude is inversely proportional to the cube of the distance from the particle to the force center; that is, $$F(r)=-\frac{k}{r^{2}}-\frac{\lambda}{r^{3}} \quad k, \lambda > 0$$ Show that the motion is described by a precessing ellipse. Consider the cases \(\lambda < l^{2} / \mu, \lambda=l^{2} / \mu,\) and \(\lambda > l^{2} / \mu\)
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Thermodynamics

Read Explanation

Force

Read Explanation

Oscillations

Read Explanation

Electricity

Read Explanation

Modern Physics

Read Explanation
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.