Problem 38 - Thomas Calculus

Question

Gravitational field a. Find a potential function for the gravitational field $$\mathbf{F}=-G m M \frac{x \mathbf{i}+y \mathbf{j}+z \mathbf{k}}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}}$$ \((G, m,\) and \(M\) are constants \().\) b. Let \(P_{1}\) and \(P_{2}\) be points at distance \(s_{1}\) and \(s_{2}\) from the origin. Show that the work done by the gravitational field in part (a) in moving a particle from \(P_{1}\) to \(P_{2}\) is $$\operatorname{GmM}\left(\frac{1}{s_{2}}-\frac{1}{s_{1}}\right).$$

Step-by-Step Solution

Answer

The potential function is \(-\frac{GmM}{r}\). Work done is \(GmM\left(\frac{1}{s_2} - \frac{1}{s_1}\right)\).

Step 1: Identify the expression for gravitational field

The gravitational field given is \(\mathbf{F} = -G m M \frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{\left(x^{2} + y^{2} + z^{2}\right)^{3 / 2}}\). This field is directed towards the origin and its magnitude depends on the inverse square of the distance from the origin.

Step 2: Define the general potential function

A vector field \(\mathbf{F}\) can be expressed as the gradient of a potential function \(U\), specifically \(\mathbf{F} = -abla U\). For a gravitational field, \(U\) represents the gravitational potential energy.

Step 3: Find the potential function

Assume a potential function \(U = -\frac{G m M}{r}\) where \(r = \sqrt{x^2 + y^2 + z^2}\). Calculate the gradient of \(U\) which, denoting as \(\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}\), can be \(abla U = \frac{G m M \mathbf{r}}{r^3}\). Verify that this is the negative of the given field \(\mathbf{F}\).

Step 4: Verify the potential expression

The field \(\mathbf{F}\) derived from the potential \(U\) matches the initial given force field, confirming \(U = -\frac{G m M}{\sqrt{x^2 + y^2 + z^2}}\) properly describes the potential.

Step 5: Calculate work done using the potential

Use the potential function \(U(r) = -\frac{G m M}{r}\) to find the work done moving from \(P_1\) to \(P_2\). This work done is given by the change in potential energy, \(W = U(s_2) - U(s_1) = -G m M \left(\frac{1}{s_2} - \frac{1}{s_1}\right)\).

Step 6: Confirm work done expression

The expression \(W = G m M \left(\frac{1}{s_1} - \frac{1}{s_2}\right)\) aligns with the provided formula, \(\operatorname{GmM}\left(\frac{1}{s_{2}}-\frac{1}{s_{1}}\right)\), indicating correct calculation.

Key Concepts

Potential Function

A potential function is a scalar function that helps in describing a vector field. In the context of gravitational fields, it is used to represent the gravitational potential energy. The potential function for a gravitational field is typically denoted as \(U\). The relationship between the gravitational field and the potential function is given by \( \mathbf{F} = -abla U \). This means the gravitational field \( \mathbf{F} \) is equal to the negative gradient of the potential function \(U\). For gravitational fields, the potential function takes the form \( U = -\frac{G m M}{r} \), where \( r = \sqrt{x^2 + y^2 + z^2} \). This expression captures how potential energy is inversely proportional to the distance from the origin.

Work Done

The concept of work done in a gravitational field involves finding the change in gravitational potential energy as a particle moves from one point to another. Work done is simply the difference in the potential energy at two given points, denoted as \( W = U(s_2) - U(s_1) \). Calculating work done in moving a particle from \(P_1\) to \(P_2\) in a gravitational field is straightforward. If the potential function is \( U = -\frac{G m M}{r} \), then the work done is \( W = -G m M \left(\frac{1}{s_2} - \frac{1}{s_1}\right) \). The negative sign arises because gravitational force is an attractive force, inherently doing negative work when the particle moves away from the gravitational source.

Inverse Square Law

The inverse square law is pivotal in understanding gravitational fields. This law states that the force exerted by a point source diminishes with the square of the distance from the source. Mathematically, it can be expressed as \( F \propto \frac{1}{r^2} \). In the exercise, the gravitational field \( \mathbf{F} = -G m M \frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{(x^2 + y^2 + z^2)^{3/2}} \) corresponds to this law. Here, the denominator \((x^2 + y^2 + z^2)^{3/2}\) ensures that as the distance \(r\) increases, the gravitational force decreases proportionally as the square of \(r\). This inverse relationship explains why gravitational effects are stronger closer to the source.

Vector Field

A vector field is a collection of vectors that are defined across a region of space. These vectors represent the direction and magnitude of some quantity, such as force. In our gravitational field exercise, \( \mathbf{F} = -G m M \frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{(x^2 + y^2 + z^2)^{3/2}} \) is a vector field representing gravitational force. Each vector points towards the origin, indicating the gravitational pull. This vector field is defined using components \(x \mathbf{i}, y \mathbf{j},\) and \(z \mathbf{k}\), which describe how the field varies in 3D space. Gravitational vector fields are essential in visualizing how forces act on objects within a field and provide a comprehensive understanding of force interactions in space.