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Chapter 12: Problem 11

A particle of mass 3\(m\) is located 1.00 \(\mathrm{m}\) from a particle of mass \(m .\) (a) Where should you put a third mass \(M\) so that the net gravitational force on \(M\) due to the two masses is exactly zero? (b) Is the equilibrium of \(M\) at this point stable or unstable (i) for points along the line connecting \(m\) and \(3 m,\) and (ii) for points along the line passing through \(M\) and perpendicular to the line connecting \(m\) and 3\(m ?\)

Short Answer

Expert verified
Place M at \(\frac{-1 + \sqrt{3}}{2}\) m from 3m. Equilibrium is unstable along the line connecting m and 3m, stable perpendicular to it.

Step by step solution

01

Identify Forces Acting on M

To find where the third mass \(M\) should be placed, we must identify the forces acting on it. Both the mass \(m\) and mass \(3m\) exert gravitational forces on the mass \(M\). According to the law of universal gravitation, the force exerted on \(M\) by another mass \(m_i\) is given by \(F = \frac{Gm_iM}{r^2}\), where \(r\) is the distance between the two masses.
02

Set Forces to Achieve Equilibrium

To achieve net zero gravitational force on \(M\), the force exerted by mass \(m\) should equal the force exerted by mass \(3m\) but in opposite directions. Assume the distance from \(3m\) to \(M\) is \(x\), this makes the distance from \(m\) to \(M\) \((1-x)\). Therefore, set the forces equal: \[\frac{GmM}{(1-x)^2} = \frac{G(3m)M}{x^2}.\] Simplify this to \[(1-x)^2 = 3x^2.\]
03

Solve for x

Solve the equation \((1-x)^2 = 3x^2\) to find \(x\). Expanding and rearranging gives: \[1 - 2x + x^2 = 3x^2.\] Simplifying the quadratic: \[2x^2 + 2x - 1 = 0.\] Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = 2, c = -1\), we find the solutions for \(x\).
04

Evaluate x Using the Quadratic Formula

Apply the quadratic formula: \(a = 2, b = 2, c = -1\). This gives: \[x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-2 \pm \sqrt{4 + 8}}{4}.\] Simplify to \[x = \frac{-2 \pm 2\sqrt{3}}{4},\] which results in \(x = \frac{-1 + \sqrt{3}}{2}\) or \(x = \frac{-1 - \sqrt{3}}{2}\). Since \(x\) must be positive and less than 1, we take \(x = \frac{-1 + \sqrt{3}}{2}\).
05

Examine Stability (Part i)

Analyzing equilibrium along the line connecting \(m\) and \(3m\), we find that such positioning is unstable. Moving \(M\) slightly will create a net force pulling it further from its equilibrium position due to increase in attraction from closer mass.
06

Examine Stability (Part ii)

For points along a line perpendicular to the connecting line, parallel forces maintain equilibrium, making it stable. Small perturbations will not create a resultant force favoring drift away from the original position.

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Key Concepts

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

Gravitational Force
Gravitational force is one of the fundamental forces of nature. It dictates how objects with mass interact with each other at a distance. Simply put, every object in the universe attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The equation that represents this is given by:
  • \( F = \frac{Gm_1m_2}{r^2} \)
  • \( F \) is the gravitational force between the two masses.
  • \( G \) is the universal gravitational constant.
  • \( m_1 \) and \( m_2 \) are the masses involved.
  • \( r \) is the distance between the centers of the two masses.
Gravitational force is always attractive, never repulsive. Understanding this concept is crucial since it helps us predict how objects will move towards each other in space.
Equilibrium
Equilibrium occurs when a particle is in a position where forces acting on it are balanced. This means the net force is zero, and hence, there is no acceleration. In our context, for the third mass \( M \) to be in an equilibrium along the line connecting masses \( m \) and \( 3m \), we need the gravitational forces exerted by these two masses on \( M \) to cancel out each other. Thus:
  • The force by \( m \) should equal the force by \( 3m \), but in opposite directions.
  • The balance is achieved when \( \frac{GmM}{(1-x)^2} = \frac{G(3m)M}{x^2} \).
  • Solving this yields the desired po sition of \( M \).
Without achieving equilibrium, \( M \) would be accelerated towards one of the two masses.
Stability Analysis
Stability analysis determines whether the equilibrium position of an object is stable, unstable, or neutral. Understanding stability is crucial, especially in physics, to predict whether an object will return to its initial position after being slightly disturbed.
  • For equilibrium to be **stable**, a slight displacement should cause a restoring force to return the object to its original position.
  • For it to be **unstable**, a slight displacement causes further displacement away from equilibrium.
  • In the discussed exercise, when mass \( M \) lies along the line connecting \( m \) and \( 3m \), slight shifts lead to further forces pulling \( M \) away, highlighting **unstability**.
  • Conversely, along a line perpendicular to this connection, the interactions stabilize disturbances, marking a **stable** equilibrium.
Quadratic Equation Solution
Often, physics problems lead us to solving quadratic equations to find unknowns. A quadratic equation takes the form \( ax^2 + bx + c = 0 \). Solving it involves using the quadratic formula:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
  • This formula gives the values of \( x \) that satisfy the quadratic equation.
  • The discriminant, \( b^2 - 4ac \), determines the nature of the solutions.
Applying this to the exercise, we solve \( 2x^2 + 2x - 1 = 0 \) to find the position \( x \) where mass \( M \) achieves equilibrium. Understanding quadratic solutions is fundamental in physics and many practical applications. This skill extends beyond theoretical problems and into real-world scenarios.

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Most popular questions from this chapter

An object in the shape of a thin ring has radius \(a\) and mass \(M .\) A uniform sphere with mass \(m\) and radius \(R\) is placed with its center at a distance \(x\) to the right of the center of the ring, along a line through the center of the ring, and perpendicular to its plane (Fig. 12.35 ). What is the gravitational force that the sphere exerts on the ring- shaped object? Show that your result reduces to the expected result when \(x\) is much larger than \(a\) .

Deimos, a moon of Mars, is about 12 \(\mathrm{km}\) in diameter with mass \(2.0 \times 10^{15} \mathrm{kg}\) . Suppose you are stranded alone on Deimos and want to play a one-person game of baseball. You would be the pitcher, and you would be the batter! (a) With what speed would you have to throw a baseball so that it would go into a circular orbit just above the surface and return to you so you could hit it? Do you think you could actually throw it at this speed? (b) How long (in hours) after throwing the ball should you be ready to hit it? Would this be an action-packed bascball game?

The planet Uranus has a radius of \(25,560 \mathrm{km}\) and a surface acceleration due to gravity of 11.1 \(\mathrm{m} / \mathrm{s}^{2}\) at its poles. Its moon Miranda (discovered by Kniper in 1948 ) is in a circular orbit about Uranus at an altitude of \(104,000 \mathrm{km}\) above the planet's surface. Miranda has a mass of \(6.6 \times 10^{19} \mathrm{kg}\) and a radius of 235 \(\mathrm{km}\) . (a) Calculate the mass of Uranus from the given data, (b) Calculate the magnitude of Miranda's acceleration due to its orbital motion about Uranus. (c) Calculate the acceleration due to Miranda's gravity at the surface of Miranda. (d) Do the answers to parts (b) and (c) mean that an object released 1 \(\mathrm{m}\) above Miranda's surface on the side toward Uranus will fall up relative to Miranda? Explain.

The weight of Santa Claus at the North Pole, as determined by a spring balance, is 875 \(\mathrm{N}\) . What would this spring balance read for his weight at the equator, assuming that the earth is spherically symmetric?

Kirkwood Gaps. Hundreds of thousands of asteroids orbit the sun within the asteroid belt, which extends from about \(3 \times 10^{8} \mathrm{km}\) to about \(5 \times 10^{8} \mathrm{km}\) from the sun. (a) Find the orbital period (in years) of (i) an asteroid at the inside of the belt and (ii) an asteroid at the outside of the belt. Assume circular orbits. (b) In 1867 the American astronomer Daniel Kirkwood pointed out that several gaps exist in the asteroid belt where relatively few asteroids are found. It is now understood that these Kirkwood gaps are caused by the gravitational attraction of Jupiter, the largest planet, which orbits the sun once every 11.86 years. As an example, if an asteroid has an orbital period half that of Jupiter, or 5.93 years, on every other orbit this asteroid would be at its closest to Jupiter and feel a strong attraction toward the planet. This attraction, acting over and over on successive orbits, could sweep asteroids out of the Kirkwood gap. Use this hypothesis to determine the orbital radius for this Kirkwood gap. (c) One of several other Kirkwood gaps appears at a distance from the sun where the orbital period is 0.400 that of Jupiter. Explain why this happens, and find the orbital radius for this Kirkwood gap.
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