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The magnitude of the gravitational force between a particle of mass \(m_{1}\) and one of mass \(m_{2}\) is given by $$ F(x)=G \frac{m_{1} m_{2}}{x^{2}} $$ where \(G\) is a constant and \(x\) is the distance between the particles. (a) What is the corresponding potential energy function \(U(x)\) ? Assume that \(U(x) \rightarrow 0\) as \(x \rightarrow \infty\) and that \(x\) is positive. (b) How much work is required to increase the separation of the particles from \(x=x_{1}\) to \(x=x_{1}+d ?\)

Short Answer

Expert verified
(a) The potential energy is \( U(x) = -G\frac{m_{1} m_{2}}{x} \). (b) The work required is \( G m_{1} m_{2} \left( \frac{1}{x_1} - \frac{1}{x_1 + d} \right) \).

Step by step solution

01

Understanding the Relation between Force and Potential Energy

Potential energy is related to force by the formula: \[ F(x) = - \frac{dU}{dx} \]which means the gravitational force is the negative derivative of the potential energy with respect to distance \(x\). Our goal is to determine the potential energy function \(U(x)\) given the force function \(F(x)\).
02

Integrate the Force to Find the Potential Energy Function

To find \(U(x)\), integrate the force function with respect to \(x\):\[ U(x) = - \int F(x) \, dx = - \int \left( G \frac{m_{1} m_{2}}{x^2} \right) dx \]Evaluating the integral, we obtain:\[ U(x) = - G m_{1} m_{2} \int \frac{1}{x^2} \, dx = G m_{1} m_{2} \frac{1}{x} + C \]where \(C\) is the integration constant to be determined.
03

Determine the Integration Constant

Given the condition \(U(x) \rightarrow 0\) as \(x \rightarrow \infty\), the potential energy at an infinite distance is zero. Thus,\[ \lim_{x \to \infty} U(x) = \lim_{x \to \infty} \left( G m_{1} m_{2} \frac{1}{x} + C \right) = 0 \]This implies \(C = 0\). Hence, the potential energy function is:\[ U(x) = - G \frac{m_{1} m_{2}}{x} \].
04

Calculate the Work Required to Increase the Separation

The work done \(W\) is the difference in potential energy when separation changes from \(x_1\) to \(x_1 + d\):\[ W = U(x_1 + d) - U(x_1) \]Substitute the potential energy function:\[ W = \left(- G \frac{m_{1} m_{2}}{x_1 + d}\right) - \left(- G \frac{m_{1} m_{2}}{x_1}\right) \]Simplify:\[ W = G m_{1} m_{2} \left( \frac{1}{x_1} - \frac{1}{x_1 + d} \right) \]

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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 interactions in nature. It acts between two bodies of mass: the larger the masses, the stronger the force. The gravitational force is described by Isaac Newton's law of universal gravitation. The formula \[ F(x) = G \frac{m_{1} m_{2}}{x^2} \] indicates that the force depends inversely on the square of the distance between the masses (\(x\)) and directly on the product of the two masses, \(m_1\) and \(m_2\). The gravitational constant \(G\) quantifies the strength of gravity.
- **Inverse Square Law:** As the distance between the two particles increases, the gravitational force decreases rapidly.- **Mass Dependence:** The more massive each object is, the stronger the gravitational attraction between them.
This formula also points to an essential property of gravitational forces: they are attractive and long-ranged. Understanding this helps in visualizing how celestial bodies interact and why orbits of planets are the way they are.
Work-Energy Theorem
The work-energy theorem connects the concepts of work and energy, particularly useful in analysing physical systems. Work is defined as the energy transferred when a force moves an object over a distance. For gravitational force, work done can alter an object's mechanical energy.
When a force acts over a distance \(dx\), it contributes to the work done on the system, which in turn can translate to changes in the potential energy. Consider the equation for work:\[ W = \Delta U = U(x_2) - U(x_1) \]This indicates that work done by or against a force causes the system's potential energy to change. In gravitational terms, increasing the separation between two masses requires work against the gravitational pull, thus increasing potential energy.
  • **Conservation of Energy:** Total mechanical energy (kinetic + potential) remains constant in an isolated system.
  • **Direct Relation:** Work done results in an exact change in energy—increasing separation implies additional work and thus more potential energy.
Utilizing this theorem allows us to calculate the effort needed to change positions in a gravitational field by considering potential energy transformations.
Integration in Physics
Integration serves as a crucial tool in physics for deriving various quantities from rates of change or relationships, directly used for finding potential energy from force functions. In this context, potential energy (\(U(x)\)) is derived from the force (\(F(x)\)) by integrating.
The relationship can be shown as:\[ U(x) = - \int F(x) \, dx \]Here, the negative sign reflects the conservative nature of gravitational forces, which we discuss later. The integral of the gravitational force function \[ G \frac{m_{1} m_{2}}{x^2} \]is performed to derive the potential energy function given by:\[ U(x) = -G \frac{m_{1} m_{2}}{x} \]
  • **Role of Integration:** It accumulates the small force elements over distance to compute overall effect.
  • **Finding Constants:** By setting conditions such as \(U(x) \rightarrow 0\) as \(x \rightarrow \infty\), we find any constants of integration.
This use of integration helps visualize how potential energy behaves as objects move in a gravitational field.
Conservative Forces
Conservative forces, such as gravity, have a unique feature: the work done by these forces is path-independent. It depends only on the initial and final positions and not on the trajectory taken. This implies that potential energy functions can be defined for such forces.
- In a gravitational field, the change in potential energy as two masses move is determined only by their starting and ending distances, not the path.- This is why, in finding potential energy functions, we can integrate force functions considering only these positions without worrying about the exact path taken.
The force function for gravity is conservative, thus:\[ -\frac{dU}{dx} = F(x) \]and indicates potential energy can be recovered when needed—an implication of energy conservation in closed systems.
  • **Energy Recovery:** Any stored potential energy can be converted back fully to kinetic energy.
  • **Defined Potentials:** Potential energy values can be ascertained which are specific to a state or position in a field.
Understanding conservative forces simplifies analysis of mechanical systems, allowing for potential energy calculations and energy conservation applications.

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Most popular questions from this chapter

A horizontal force of magnitude \(35.0 \mathrm{~N}\) pushes a block of mass \(4.00 \mathrm{~kg}\) across a floor where the coefficient of kinetic friction is \(0.600 .\) (a) How much work is done by that applied force on the block- floor system when the block slides through a displacement of \(3.00 \mathrm{~m}\) across the floor? (b) During that displacement, the thermal energy of the block increases by \(40.0 \mathrm{~J}\). What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

A spring with spring constant \(k=200 \mathrm{~N} / \mathrm{m}\) is suspended vertically with its upper end fixed to the ceiling and its lower end at position \(y=0 .\) A block of weight \(20 \mathrm{~N}\) is attached to the lower end, held still for a moment, and then released. What are (a) the kinetic energy \(K,(\mathrm{~b})\) the change (from the initial value) in the gravitational potential energy \(\Delta U_{g}\), and (c) the change in the elastic potential energy \(\Delta U_{e}\) of the spring-block system when the block is at \(y=-5.0 \mathrm{~cm}\) ? What are (d) \(K\), (e) \(\Delta U_{g}\), and (f) \(\Delta U_{e}\) when \(y=-10 \mathrm{~cm},(\mathrm{~g}) K,(\mathrm{~h}) \Delta U_{g}\), and \((\mathrm{i}) \Delta U_{e}\) when \(y=-15 \mathrm{~cm}\), and \((\mathrm{j}) K,(\mathrm{k}) \Delta U_{g}\), and (1) \(\Delta U_{e}\) when \(y=-20 \mathrm{~cm}\) ?

A uniform cord of length \(25 \mathrm{~cm}\) and mass \(15 \mathrm{~g}\) is initially stuck to a ceiling. Later, it hangs vertically from the ceiling with only one end still stuck. What is the change in the gravitational potential energy of the cord with this change in orientation? (Hint: Consider a differential slice of the cord and then use integral calculus.)

We move a particle along an \(x\) axis, first outward from \(x=1.0 \mathrm{~m}\) to \(x=4.0 \mathrm{~m}\) and then back to \(x=1.0 \mathrm{~m}\), while an external force acts on it. That force is directed along the \(x\) axis, and its \(x\) component can have different values for the outward trip and for the return trip. Here are the values (in newtons) for four situations, where \(x\) is in meters: $$ \begin{array}{ll} \hline \text { Outward } & \text { Inward } \\ \hline \text { (a) }+3.0 & -3.0 \\ \text { (b) }+5.0 & +5.0 \\ \text { (c) }+2.0 x & -2.0 x \\ \text { (d) }+3.0 x^{2} & +3.0 x^{2} \\ \hline \end{array} $$ Find the net work done on the particle by the external force for the round trip for each of the four situations. (e) For which, if any, is the external force conservative?

The luxury liner Queen Elizabeth 2 has a diesel-electric power plant with a maximum power of \(92 \mathrm{MW}\) at a cruising speed of \(32.5\) knots. What forward force is exerted on the ship at this speed? \((1 \mathrm{knot}=1.852 \mathrm{~km} / \mathrm{h} .)\)

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