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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

An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

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 partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a y^{2}+2 b y z+c z^{2},(\mathbf{b}) g(x, y, z)=\cos \left(a x y^{2} z^{3}\right),(\mathbf{c}) h(x, y, z)=a r,\) where \(a, b,\) and \(c\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}} .\) Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

(a) Consider a mass \(m\) in a uniform gravitational field \(\mathbf{g},\) so that the force on \(m\) is \(m \mathbf{g},\) where \(\mathbf{g}\) is a constant vector pointing vertically down. If the mass moves by an arbitrary path from point 1 to point \(2,\) show that the work done by gravity is \(W_{\mathrm{grav}}(1 \rightarrow 2)=-m g h\) where \(h\) is the vertical height gained between points 1 and 2. Use this result to prove that the force of gravity is conservative (at least in a region small enough so that \(\mathrm{g}\) can be considered constant). (b) Show that, if we choose axes with \(y\) measured vertically up, the gravitational potential energy is \(U=m g y\) (if we choose \(U=0\) at the origin).

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):(\mathbf{a}) f=\ln (r),\) (b) \(f=r^{n}\), (c) \(f=g(r),\) where \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) and \(g(r)\) is some unspecified function of \(r .\) [Hint: Use the chain rule.]

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