/*! 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 10 The point masses \(m\) and \(2 m... [FREE SOLUTION] | 91Ó°ÊÓ

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The point masses \(m\) and \(2 m\) lie along the \(x\) -axis, with \(m\) at the origin and \(2 m\) at \(x=L\). A third point mass \(M\) is moved along the \(x\) -axis. (a) At what point is the net gravitational force on \(M\) due to the other two masses equal to zero? (b) Sketch the \(x\) -component of the net force on \(M\) due to \(m\) and \(2 m,\) taking quantities to the right as positive. Include the regions \(x<0,0L\). Be especially careful to show the behavior of the graph on either side of \(x=0\) and \(x=L\)

Short Answer

Expert verified
The point where mass \(M\) experiences no net gravitational force is the point \(x=L\), which is the same position as mass \(2m\). The x-component of the net gravitational force experienced by \(M\) is negative for \(x < 0\) and \(0 < x < L\), and positive for \(x > L\), with points of infinite force at \(X=0\) and \(X=L\).

Step by step solution

01

Set Up Equations

To solve for the position of \(M\), the key is to find when the total gravitational force on \(M\) from masses \(m\) and \(2m\) is zero. The gravitational force between two bodies is given by \(F = G.m_1.m_2 / r^2\) where G is the gravitational constant, \(m_1\) and \(m_2\) are the masses and \(r\) is the distance between the objects. The forces due to \(m\) and \(2m\) are in opposite directions. Thus we can calculate force due to each mass on \(M\) and equate it to zero. Let \(x\) be the distance from mass \(m\) to \(M\). Therefore, the distance from \(2m\) to \(M\) is \(L-x\). The equation should then be: \[F_{m\rightarrow M} + F_{2m\rightarrow M} = 0\] which expands to, \[-(G.M.m / x^2) + (G.M.2m / (L-x)^2) = 0.\]
02

Solve the Equation

From the equation in step 1, solve for \(x\). This equation simplifies to: \[-m / x^2 = 2m / (L-x)^2\] This equation can further simplified to a quadratic equation, \(x^2 - 2Lx + L^2 = 0\]. Solving this quadratic equation yields the solution \(x=L\) . Note that the physical solution for \(x\) should ensure \(0 < x < L\). Thus, the position where \(M\) experiences no net gravitational force is at \(x=L\), which is at the same position as \(2m\).
03

Sketch the Graph

The sketch should display the net gravitational force experienced by \(M\) as a function of \(x\). The forces on either side of \(0\) and \(L\) should be drawn. It should be negative for \(x < 0\) and \(0 < x < L\), as the force is towards the left. At \(x = 0\), the force is infinite because there is a mass \(m\) at this point. As \(x\) approaches \(L\) from left, the force smoothly approaches zero because the attractive force due to \(2m\) begins to outweigh the force due to \(m\). Right after \(x = L\), the net force changes sign because \(M\) is closer to \(2m\) than \(m\). It remains positive (ie. towards right) for \(x > L\) and again becomes infinity when \(x \rightarrow ∞\) because the two masses are perpendicular to each other. Thus force due to \(m\) becomes insignificant as compared to that due to \(2m\).

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Key Concepts

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

Gravitational Force
When we talk about the forces that govern the movements of celestial bodies, satellites, and even the apple that famously fell on Newton's head, we're talking about gravitational force. This force is an invisible pull that all objects with mass exert on one another. The more massive an object is, the stronger its gravitational pull.

For students, understanding gravitational force is critical when studying physics or astronomy. The exercises involving gravitational force often require us to predict the motion of objects, calculate the force between two masses, or determine situations of equilibrium, such as when the gravitational force is zero.

Even though we can't see gravitational forces, we can calculate their effects using mathematical equations. Through these calculations, we can gain a deeper understanding of the fundamental principles that keep planets in orbit and prevent us from drifting into space. As such, mastering the concept of gravitational force is essential for any aspiring physicist or curious mind wanting to unravel the mysteries of the universe.
Newton's Law of Universal Gravitation
Sir Isaac Newton, in his law of universal gravitation, stated that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This law can be expressed as: \[ F = G \frac{m_1 m_2}{r^2} \]

Here, \(F\) is the magnitude of the gravitational force between the two masses, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses of the objects, and \(r\) is the distance between the centers of the masses.

Understanding this law is crucial for students as it allows them to calculate the gravitational force between two masses. Knowledge of this principle is the foundation for exploring the dynamics of our solar system, understanding the principles of orbits, and for applying the concept to problems such as determining the zero point of gravitational force.
Quadratic Equations
Quadratic equations are fundamental in algebra and appear frequently in various scientific calculations. An equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, is known as a quadratic equation.

In the context of gravitational force problems, a quadratic equation can arise when solving for distances that result in a net force of zero. For example, when equating the forces of attraction due to different masses to find a balance point, the resulting equation requiring solution may well be quadratic.

To solve a quadratic equation, students can factorize, complete the square, or use the quadratic formula. The solutions to these equations can reveal important information about physical systems, such as the position where gravitational forces cancel out, which is the central focus of the problem in our textbook example. Thus, becoming proficient with quadratic equations is essential for physics students and enhances their problem-solving toolkit.
Force Diagrams
A force diagram, also known as a free-body diagram, is a graphical illustration used to visualize the forces acting on an object. By representing these forces as vectors, students can analyze the effects of these forces on the motion of the object.

In our textbook problem, creating a force diagram helps visualize how the net gravitational force behaves on the mass \(M\) as it moves along the x-axis. The regions, such as \(x<0\), \(0L\), depict different forces acting on \(M\) from the point masses \(m\) and \(2m\). It's critical for students to understand the signs and magnitudes of these forces to correctly sketch the graph.

Mastering force diagrams allows students to deconstruct complex problems, simplify the analysis of forces, and predict the result of these forces on an object's motion, thereby gaining a more concrete understanding of theoretical physics concepts applied in real-world scenarios.

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Most popular questions from this chapter

A satellite with mass \(848 \mathrm{~kg}\) is in a circular orbit with an orbital speed of \(9640 \mathrm{~m} / \mathrm{s}\) around the earth. What is the new orbital speed after friction from the earth's upper atmosphere has done \(-7.50 \times 10^{9} \mathrm{~J}\) of work on the satellite? Does the speed increase or decrease?

Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was \(2.87 \times 10^{6} \mathrm{~km}\) from the earth and traveling at \(1.20 \times 10^{4} \mathrm{~km} / \mathrm{h}\) relative to the earth. At this time, what were (a) the spacecraft's kinetic energy relative to the earth and (b) the potential energy of the earthspacecraft system?

Aura Mission. On July 15, 2004 NASA launched the Aura spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit \(705 \mathrm{~km}\) above the earth's surface. Assume a circular orbit. (a) How many hours does it take this satellite to make one orbit? (b) How fast (in \(\mathrm{km} / \mathrm{s})\) is the Aura spacecraft moving?

DATA For a spherical planet with mass \(M,\) volume \(V,\) and radius \(R,\) derive an expression for the acceleration due to gravity at the planet's surface, \(g\), in terms of the average density of the planet, \(\rho=M / V,\) and the planet's diameter, \(D=2 R .\) The table gives the values of \(D\) and \(g\) for the eight major planets: $$ \begin{array}{lrc} \text { Planet } & D(\mathrm{~km}) & g\left(\mathrm{~m} / \mathrm{s}^{2}\right) \\ \hline \text { Mercury } & 4879 & 3.7 \\ \text { Venus } & 12,104 & 8.9 \\ \text { Earth } & 12,756 & 9.8 \\ \text { Mars } & 6792 & 3.7 \\ \text { Jupiter } & 142,984 & 23.1 \\ \text { Saturn } & 120,536 & 9.0 \\ \text { Uranus } & 51,118 & 8.7 \\ \text { Neptune } & 49.528 & 11.0 \end{array} $$ (a) Treat the planets as spheres. Your equation for \(g\) as a function of \(\rho\) and \(D\) shows that if the average density of the planets is constant, a graph of \(g\) versus \(D\) will be well represented by a straight line. Graph 8 as a function of \(D\) for the eight major planets. What does the graph tell you about the variation in average density? (b) Calculate the average density for each major planet. List the planets in order of decreasing density, and give the calculated average density of each. (c) The earth is not a uniform sphere and has greater density near its center. It is reasonable to assume this might be true for the other planets. Discuss the effect this has on your analysis. (d) If Saturn had the same average density as the earth, what would be the value of \(g\) at Saturn's surface?

A rocket with mass \(5.00 \times 10^{3} \mathrm{~kg}\) is in a circular orbit of radius \(7.20 \times 10^{6} \mathrm{~m}\) around the earth. The rocket's engines fire for a period of time to increase that radius to \(8.80 \times 10^{6} \mathrm{~m},\) with the orbit again circular. (a) What is the change in the rocket's kinetic energy? Does the kinetic energy increase or decrease? (b) What is the change in the rocket's gravitational potential energy? Does the potential energy increase or decrease? (c) How much work is done by the rocket engines in changing the orbital radius?

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