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Verify that the gravitational force \(-G M m \hat{\mathbf{r}} / r^{2}\) on a point mass \(m\) at \(\mathbf{r},\) due to a fixed point mass \(M\) at the origin, is conservative and calculate the corresponding potential energy.

Short Answer

Expert verified
The gravitational force is conservative; potential energy is \( V(r) = -\frac{GMm}{r} \).

Step by step solution

01

Identify the Gravitational Force

The gravitational force on a point mass \( m \) at position \( \mathbf{r} \) due to another fixed point mass \( M \) at the origin is given by: \( \mathbf{F} = -\frac{GMm}{r^2} \hat{\mathbf{r}} \), where \( G \) is the gravitational constant, and \( \hat{\mathbf{r}} \) is the unit vector in the direction of \( \mathbf{r} \).
02

Check for Conservatism of the Force

A force field is conservative if the curl of the force field is zero. For a central force like gravity, \( \mathbf{F} = -abla V \), where \( V \) is the potential energy. We express the curl condition: \( abla \times \mathbf{F} = 0 \). In Cartesian coordinates, the curl of a radial force field like this is always zero, confirming the force is conservative.
03

Determine the Potential Energy

For a conservative force, potential energy \( V \) is found by integrating the force with respect to \( r \). Using the relation: \( \mathbf{F} = -abla V \), integrate \( F \) with respect to \( r \): \( V = - \int \mathbf{F} \cdot d\mathbf{r} = - \int -\frac{GMm}{r^2} dr = -\left[-\frac{GMm}{r}\right] + C \), where \( C \) is a constant of integration.
04

Finalize the Potential Energy Expression

It's common to set the potential energy at infinity to zero, \( V(\infty) = 0 \). Thus, \( C = 0 \), and we have the gravitational potential energy as: \( V(r) = -\frac{GMm}{r} \).

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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 a fundamental interaction that pulls objects with mass towards each other. In the formula for gravitational force between two masses, \[ \mathbf{F} = -\frac{GMm}{r^2} \hat{\mathbf{r}} \]
  • \( G \) is the gravitational constant, a value that represents the strength of gravity universally.
  • \( M \) and \( m \) are the masses of the two objects involved.
  • \( r \) is the distance between the centers of these two masses.
  • \( \hat{\mathbf{r}} \) is a unit vector pointing from the mass \( M \) to the mass \( m \).
This formula expresses how the gravitational force decreases with the square of the distance (\( r^2 \)) between two masses. As distance increases, the force rapidly decreases. Gravitational forces are crucial in celestial mechanics, keeping planets in orbit around stars, for instance.

Understanding gravitational force gives insight into fields like astronomy and physics, explaining phenomena from apple falling to planetary orbits.
Potential Energy
Potential energy represents the stored energy of an object due to its position relative to other objects. In the context of gravitational force, we often calculate it by finding the work done against or by gravitational forces. The potential energy \( V \) due to gravity is given by:\[ V(r) = -\frac{GMm}{r} \]
  • This form arises because the gravitational force is conservative, meaning the work done around any closed path is zero.
  • Potential energy is defined up to an arbitrary constant, which often is chosen such that \( V(\infty) = 0 \), leading to no constant in our final expression.
In practical terms, this potential energy indicates how much energy would be required to bring a mass from infinity to a point a distance \( r \) from a massive body. This energy playing a critical role in physics scenarios such as launching satellites or understanding escape velocities.

Recognizing how potential energy operates allows prediction and mapping of motion under gravity, essential for many real-world applications.
Integral Calculus
Integral calculus is a mathematical branch used to calculate quantities like areas under curves, among other applications. When dealing with forces and energies, integration helps in determining quantities such as potential energy.

Calculating Potential Energy

In our context, we used integration to find the potential energy due to a gravitational force. Specifically, the relationship between force \( \mathbf{F} \) and potential energy \( V \) can be expressed as:

\[ V = - \int \mathbf{F} \cdot d\mathbf{r} \]This integral signifies accumulation over the path from infinity to a distance \( r \), effectively measuring the total effect of the force over that distance.

The Importance of the Integrals

  • The process of integration helps translate local force interactions into global energy levels.
  • Especially for conservative forces, such as gravity, it ensures that energy is conserved and allows for predictions in movement and energy requirements.
Using integral calculus in this way provides a powerful method for understanding how systems evolve, making it fundamental in physics and engineering disciplines.

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Most popular questions from this chapter

(a) Consider an electron (charge \(-e\) and mass \(m\) ) in a circular orbit of radius \(r\) around a fixed proton (charge \(+e\) ). Remembering that the inward Coulomb force \(k e^{2} / r^{2}\) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\). (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius \(r\) around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy \(T_{2} .\) When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius \(r^{\prime} .\) (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy \(E\) long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables \(T_{2}, r,\) and \(r^{\prime}\).

Near to the point where I am standing on the surface of Planet \(X\), the gravitational force on a mass \(m\) is vertically down but has magnitude \(m \gamma y^{2}\) where \(\gamma\) is a constant and \(y\) is the mass's height above the horizontal ground. (a) Find the work done by gravity on a mass \(m\) moving from \(\mathbf{r}_{1}\) to \(\mathbf{r}_{2}\), and use your answer to show that gravity on Planet \(X,\) although most unusual, is still conservative. Find the corresponding potential energy. (b) Still on the same planet, I thread a bead on a curved, frictionless, rigid wire, which extends from ground level to a height \(h\) above the ground. Show clearly in a picture the forces on the bead when it is somewhere on the wire. (Just name the forces so it's clear what they are; don't worry about their magnitude.) Which of the forces are conservative and which are not? (c) If I release the bead from rest at a height \(h\), how fast will it be going when it reaches the ground?

A mass \(m\) moves in a circular orbit (centered ón the origin) in the field of an attractive central force with potential energy \(U=k r^{n} .\) Prove the virial theorem that \(T=n U / 2\).

Find the curl, \(\nabla \times \mathbf{F},\) for the following forces: \((\mathbf{a}) \mathbf{F}=k \mathbf{r} ;(\mathbf{b}) \mathbf{F}=\left(A x, B y^{2}, C z^{3}\right) ;(\mathbf{c}) \mathbf{F}=\) \(\left(A y^{2}, B x, C z\right),\) where \(A, B, C\) and \(k\) are constants.

By writing a \(\cdot \mathbf{b}\) in terms of components prove that the product rule for differentiation applies to the dot product of two vectors; that is, \(\frac{d}{d t}(\mathbf{a} \cdot \mathbf{b})=\frac{d \mathbf{a}}{d t} \cdot \mathbf{b}+\mathbf{a} \cdot \frac{d \mathbf{b}}{d t}\).

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