Problem 34
Question
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?
Step-by-Step Solution
Verified Answer
The force is \( F(x) = -\frac{G m_{1} m_{2}}{x^2} \), and it is attractive.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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 \(x\):\[\frac{dU}{dx} = \frac{d}{dx} \left(-\frac{G m_{1} m_{2}}{x}\right) = \frac{G m_{1} m_{2}}{x^2}\].
4Step 4: 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 \(m_1\).
5Step 5: 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.
Key Concepts
Potential EnergyGravitational Potential EnergyNewton's Law of Gravitation
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.
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.
- \( U \) is the gravitational potential energy,
- \( m \) is the mass of the object,
- \( g \) is the acceleration due to gravity, approximately 9.81 m/s² on Earth's surface, and
- \( h \) is the height above the reference point.
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 m²/kg²,
- \( m_1 \) and \( m_2 \) are the masses of the two objects, and
- \( r \) is the distance between the centers of the two masses.
Other exercises in this chapter
Problem 32
The potential energy of a pair of hydrogen atoms separated by a large distance \(x\) is given by \(U(x)=-C_{6} / x^{6},\) where \(C_{6}\) is a positive constant
View solution Problem 33
A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},
View solution Problem 35
Gravity in Two Dimensions. Two point masses, \(m_{1}\) and \(m_{2},\) lie in the \(x y\) -plane, with \(m_{1}\) held in place at the origin and \(m_{2}\) free t
View solution Problem 36
An object moving in the \(x y\) -plane is acted on by a conservative force described by the potential energy function \(U(x, y)=\) \(\alpha\left(1 / x^{2}+1 / y
View solution