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Chapter 7: Problem 34

Gravity in One Dimension. Two point masses, \(m_{1}\) and \(m_{2}\) , lie on the \(x\) -axis, with \(m_{1}\) held in place at the origin and \(m_{2}\) at position \(x\) and free to move. The gravitational potential energy of these masses is found to be \(U(x)=-G m_{1} m_{2} / x,\) where \(G\) is a constant (called the gravitational constant). You'll learn more about gravitation in Chapter 12 . Find the \(x\) -component of the force acting on \(m_{2}\) due to \(m_{1} .\) Is this force attractive or repulsive? How do you know?

Short Answer

Expert verified
The force is \( F(x) = -\frac{G m_{1} m_{2}}{x^2} \), and it is attractive.

Step by step solution

01

Understand the Problem

We need to find the force acting on mass \(m_2\), which is placed at position \(x\) on the \(x\)-axis. The force is due to the gravitational attraction from mass \(m_1\), which is located at the origin.
02

Recall the Relationship between Potential Energy and Force

The force acting on a particle is related to the potential energy by the formula \( F(x) = -\frac{dU}{dx} \). This means that the force is the negative gradient of the potential energy function.
03

Differentiate the Potential Energy Function

Given the potential energy function \(U(x) = -\frac{G m_{1} m_{2}}{x}\), we differentiate it with respect to \(x\):\[\frac{dU}{dx} = \frac{d}{dx} \left(-\frac{G m_{1} m_{2}}{x}\right) = \frac{G m_{1} m_{2}}{x^2}\].
04

Apply the Negative Sign to Find the Force

Using the relationship \( F(x) = -\frac{dU}{dx} \), substitute the derivative we computed:\[F(x) = -\frac{G m_{1} m_{2}}{x^2}\]This indicates that the force is attractive, as the negative sign shows that it is directed towards \(m_1\).
05

Determine if the Force is Attractive or Repulsive

The force \( F(x) = -\frac{G m_{1} m_{2}}{x^2} \) is negative, implying it acts in the direction to decrease \(x\): towards \(m_1\). This signifies the force is attractive.

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Key Concepts

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

Potential Energy
Potential energy is a form of energy that an object possesses due to its position in a force field or its condition. It is energy that is stored and has the potential to do work. There are various types of potential energy, depending on the forces involved:
  • Gravitational Potential Energy: energy due to position relative to a gravitational source.
  • Elastic Potential Energy: energy stored in objects that can be stretched or compressed.
  • Chemical Potential Energy: energy stored within chemical bonds.
Potential energy is a fundamental concept because it emphasizes that energy can be stored and released at a later point. This stored energy allows systems to perform work once conditions change and energy is transformed into kinetic energy. A classic example of potential energy is a ball held at a height. Due to its elevated position, it has gravitational potential energy, which can change into kinetic energy as it falls.
Gravitational Potential Energy
Gravitational potential energy specifically refers to the energy stored by an object as a result of its position in a gravitational field, typically due to its height above the ground.
  • Falling water in a dam, for example, possesses gravitational potential energy due to the height it has relative to the turbines at the bottom.
  • Any object elevated above the ground, like a bird in flight, holds gravitational potential energy.
Mathematically, the gravitational potential energy (GPE) of an object near the earth鈥檚 surface is given by the formula: \[ U = mgh \]where:
  • \( U \) is the gravitational potential energy,
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity, approximately 9.81 m/s虏 on Earth's surface, and
  • \( h \) is the height above the reference point.
When considering point masses in space, the formula for gravitational potential energy changes slightly to account for the distances between masses and the universal law of gravitation:\[ U(x) = -\frac{G m_1 m_2}{x} \]This formula illustrates that as the distance \( x \) between two masses \( m_1 \) and \( m_2 \) decreases, the magnitude of the gravitational potential energy becomes more negative, indicating stronger attraction.
Newton's Law of Gravitation
Newton's Law of Gravitation plays a crucial role in understanding gravitational forces between masses. This law posits that every point mass attracts every other point mass in the universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. It is mathematically expressed as:\[ F = G \frac{m_1 m_2}{r^2} \]where:
  • \( F \) specifies the gravitational force between two objects,
  • \( G \) represents the gravitational constant, approximately \(6.674 \times 10^{-11} \) N m虏/kg虏,
  • \( m_1 \) and \( m_2 \) are the masses of the two objects, and
  • \( r \) is the distance between the centers of the two masses.
This law elucidates why all objects exert gravitational forces, from planets pulling moons into orbit to objects like a ball being pulled to Earth's surface. One key element of Newton's Law is its universality鈥攊t applies regardless of the distance, provided it's significant compared to the size of the objects involved.

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Most popular questions from this chapter

A crate of mass \(M\) starts from rest at the top of a frictionless ramp inclined at an angle \(\alpha\) above the horizontal. Find its speed at the bottom of the ramp, a distance \(d\) from where it started. Do this in two ways: (a) Take the level at which the potential energy is zero to be at the bottom of the ramp with \(y\) positive upward. (b) Take the zero level for potential energy to be at the top of the ramp with y positive upward. (c) Why did the normal force not enter into your solution?

The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=a / r^{12}-b / r^{6},\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) one atom as a function of \(r\) . Make two graphs, one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r .\) (b) Find the equilibrium distance between the two atoms. Is this cquilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{J}\) per molecule. Find the values of the constants \(a\) and \(b\) .

A force of 800 \(\mathrm{N}\) stretches a certain spring a distance of 0.200 \(\mathrm{m} .\) (a) What is the potential energy of the spring when it is stretched 0.200 \(\mathrm{m} ?(\mathrm{b})\) What is its potential energy when it is compressed 5.00 \(\mathrm{cm} ?\)

You are testing a new amusement park roller coaster with an empty car with mass 120 \(\mathrm{kg}\) . One part of the track is a vertical loop with radius 12.0 \(\mathrm{m}\) . At the bottom of the loop (point \(A\) ) the car has speed \(25.0 \mathrm{m} / \mathrm{s},\) and at the top of the loop (point \(B )\) it has speed 8.0 \(\mathrm{m} / \mathrm{s}\) . As the car rolls from point \(A\) to point \(B,\) how much work is done by friction?

A slingshot will shoot a \(10-8\) pebble 220 \(\mathrm{m}\) straight up. (a) How much potential energy is stored in the slingshot's rubber band? (b) With the same potential energy stored in the rubber band, how high can the slingshot shoot a \(25-g\) pebble? (c) What physical effects did you ignore in solving this problem?
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