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Gauss's Law for Gravitation. The gravitational force between two point masses separated by a distance \(r\) is proportional to \(1 / r^{2}\), just like the electric force between two point charges. Because of this similarity between gravitational and electric interactions, there is also a Gauss's law for gravitation. (a) Let \(\vec{g}\) be the acceleration due to gravity caused by a point mass \(m\) at the origin, so that \(\vec{g}=-\left(G m / r^{2}\right) \hat{r} .\) Consider a spherical Gaussian surface with radius \(r\) centered on this point mass, and show that the flux of \(\vec{g}\) through this surface is given by $$ \oint \vec{g} \cdot d \vec{A}=-4 \pi G m $$ (b) By following the same logical steps used in Section \(2.3\) to obtain Gauss's law for the electric field, show that the flux of \(\vec{g}\) through any closed surface is given by $$ \oint \vec{g} \cdot d \overrightarrow{\boldsymbol{A}}=-4 \pi G M_{\text {encl }} $$ where \(M_{\text {encl }}\) is the total mass enclosed within the closed surface.

Step by step solution

01

Understand the Gravitational Field

The gravitational field \( \vec{g} \) due to a point mass \( m \), located at the origin, is given by \( \vec{g} = -\left(\frac{G m}{r^2}\right) \hat{r} \), where \( G \) is the gravitational constant and \( \hat{r} \) is the radial unit vector.

02

Set Up the Gaussian Surface

Consider a spherical surface with radius \( r \) centered on the point mass \( m \). This setup matches the symmetry of the problem and will allow us to apply Gauss's Law for gravitation.

03

Calculate the Gravitational Flux

The gravitational flux through a closed surface is given by the integral \( \oint \vec{g} \cdot d \vec{A} \), where \( d \vec{A} \) is the outward-pointing area element of the sphere. Since the gravitational field is radial and uniform over the spherical surface, \( \vec{g} \cdot d \vec{A} = g \cdot dA \), and \( g = \frac{G m}{r^2} \). The surface area of a sphere is \( 4 \pi r^2 \).

04

Integrate Over the Spherical Surface

The flux is calculated as follows: \(\oint \vec{g} \cdot d \vec{A} = \int g \cdot dA = \int \left( \frac{G m}{r^2} \right) dA = \frac{G m}{r^2} \cdot 4 \pi r^2 = 4 \pi G m. \)Including the direction of \( \vec{g} \), since it points inward, the flux is negative: \( \oint \vec{g} \cdot d \vec{A} = -4 \pi G m. \)

05

Generalize for Any Closed Surface

Similar to electric fields, the gravitational flux through any closed surface, enclosing mass \( M_{\text{encl}} \), can be generalized. The enclosed mass affects the total flux through the surface, providing: \( \oint \vec{g} \cdot d \vec{A} = -4 \pi G M_{\text{encl}} \). This mirrors the pattern found in Gauss's Law for electric fields.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

gravitational field

The gravitational field, denoted as \( \vec{g} \), represents the gravitational force experienced by a unit mass at a point in space. Imagine placing a small test mass in the vicinity of another mass. The pull it feels is determined by the gravitational field. For a point mass \( m \) positioned at the origin, the field is defined mathematically by the formula:

• \( \vec{g} = -\left( \frac{G m}{r^2} \right) \hat{r} \)

Here, \(G\) is the gravitational constant, \(r\) is the distance from the mass, and \(\hat{r}\) is the radial direction. The negative sign signifies that gravity is attractive. This means the field directs inward toward the mass, indicating that the force will pull any object toward the point mass. Visualizing this field is akin to picturing arrows pointing radially inward, shrinking in magnitude further away from the mass.

flux through a surface

Flux through a surface measures how much of a field passes through that surface. Think of it as the number of field lines intersecting the surface. For gravitational fields, the concept of flux helps us understand the influence a mass has beyond its immediate surroundings. Consider a closed surface, like a sphere, centered on a mass. We calculate gravitational flux via an integral:

• \( \oint \vec{g} \cdot d \vec{A} \)

Here, \(d \vec{A}\) is an infinitesimally small piece of the surface area, pointing outward. The dot product \(\vec{g} \cdot d \vec{A}\) simplifies in symmetrical conditions, such as a sphere, as all the field lines either enter or exit perpendicularly. Thus, evaluating this integral involves multiplying the field's strength by the entire surface area, providing insight into how the field interacts with different geometrical spaces.

point mass

A point mass is an idealized object with mass concentrated at a single point in space. This simplifies many gravitational calculations, serving as a foundational concept in physics. Despite existing only in theory, envisioning masses this way helps dissect complex gravitational interactions. When considering gravitational fields or flux, treating objects as point masses offers a straightforward approach to modeling:

• Spheres and uniform objects behave like point masses at their centers for external calculations.
• Point mass assumptions facilitate symmetrical shaping of the gravitational field, easing problem-solving.

Through this lens, the effects of mass become clear, allowing for calculations, such as gravitational flux through spherical surfaces, to be performed more readily.

gravitational flux

Gravitational flux reflects how much of the gravitational field spreads through a surface. It captures the influence a mass exerts over a surrounding area. Calculating this is particularly useful in understanding large systems like planets or stars. Using Gauss's Law for Gravitation, we find:

• For a single point mass, the expression is \( \oint \vec{g} \cdot d \vec{A} = -4 \pi G m \).
• Generalized for multiple masses, \( \oint \vec{g} \cdot d \vec{A} = -4 \pi G M_{\text{encl}} \).

where \(M_{\text{encl}}\) stands for the total mass enclosed by the surface. Gravitational flux, like its electric counterpart, provides a powerful tool for evaluating field effects beyond direct measurement. It reveals that total mass equilibrium within the surface dictates the field's uniformity and intensity.