Problem 81 - University Physics with Modern Physics

Question

Gravity in Three Dimensions. A point mass \(m_{1}\) is held in place at the origin, and another point mass \(m_{2}\) is free to move a distance \(r\) away at a point \(P\) having coordinates \(x, y,\) and \(z\) . The gravitational potential energy of these masses is found to be \(U(r)=-G m_{1} m_{2} / r,\) where \(G\) is the gravitational constant (see Exercises 7.34 and 7.35 . (a) Show that the components of the force on \(m_{2}\) due to \(m_{1}\) are $$ F_{x}=-\frac{G m_{1} m_{2} x}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \quad F_{y}=-\frac{G m_{1} m_{2} y}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} $$ $$ F_{z}=-\frac{G m_{1} m_{2} z}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} $$ (Hint: First write \(r\) in terms of \(x, y,\) and \(z\) ) (b) Show that the magnitude of the force on \(m_{2}\) is \(F=G m_{1} m_{2} / r^{2} .\left(\text { c) Does } m_{1} \text { attract or }\right.\) repel \(m_{2} ?\) How do you know?

Step-by-Step Solution

Answer

(a) Force components: \( F_x, F_y, F_z \) are given. (b) Magnitude: \( F = \frac{G m_1 m_2}{r^2} \). (c) \( m_1 \) attracts \( m_2 \).

Step 1: Express r in terms of x, y, z

The distance between the masses, denoted as \(r\), can be given by the Euclidean distance formula in three dimensions: \(r = \sqrt{x^2 + y^2 + z^2}\).

Step 2: Find the Force Components Derivatives

The force acting on \(m_2\) due to \(m_1\) is the negative gradient of the potential energy \(U(r)\). Thus, \(F = -abla U\). Calculate each component as a partial derivative:- \( F_x = -\frac{\partial U}{\partial x} \)- \( F_y = -\frac{\partial U}{\partial y} \)- \( F_z = -\frac{\partial U}{\partial z} \)

Step 3: Compute Partial Derivative with respect to x

Using \(U(r) = -\frac{G m_1 m_2}{r}\) and \(r = \sqrt{x^2 + y^2 + z^2}\), differentiate to find \(\frac{\partial U}{\partial x}\): \[ \frac{\partial U}{\partial x} = -(-1) G m_1 m_2 \frac{1}{(x^2 + y^2 + z^2)^{3/2}} \cdot x = -\frac{G m_1 m_2 x}{r^3} \] Thus, \( F_x = -\frac{G m_1 m_2 x}{(x^2 + y^2 + z^2)^{3/2}} \).

Step 4: Compute Partial Derivative with respect to y and z

By symmetry, the force components for \(y\) and \(z\) can be obtained similarly. Therefore: \( F_y = -\frac{G m_1 m_2 y}{(x^2 + y^2 + z^2)^{3/2}} \) \( F_z = -\frac{G m_1 m_2 z}{(x^2 + y^2 + z^2)^{3/2}} \)

Step 5: Calculate the Magnitude of the Force

The magnitude of the force \(F\) is given by the vector sum:\[ F = \sqrt{F_x^2 + F_y^2 + F_z^2} \] Substituting the formulas for the force components:\[ F = \sqrt{\left(-\frac{G m_1 m_2 x}{r^3}\right)^2 + \left(-\frac{G m_1 m_2 y}{r^3}\right)^2 + \left(-\frac{G m_1 m_2 z}{r^3}\right)^2} \]\[ F = \frac{G m_1 m_2}{r^2} \] as \[ \sqrt{x^2 + y^2 + z^2} = r. \]

Step 6: Determine Direction of Force

The negative sign in the force components indicates that the force is attractive. Since \( F = -abla U \) and the potential \( U(r) \) decreases as \( r \) increases, \( m_1 \) attracts \( m_2 \).

Key Concepts

Three-Dimensional Motion

Understanding three-dimensional motion is essential in physics, especially when dealing with forces that act in various directions simultaneously. In our scenario, two point masses interact through gravitational force. Each mass has a position assigned by coordinates

• \(x\)
• \(y\)
• \(z\)

which describe their location in a three-dimensional space. The concept of three-dimensional motion allows us to examine how an object moves or interacts, for instance, under the influence of gravity, in all possible directions.
Gravity in this context is not limited to a single direction. Instead, it manifests through combined effects along three axes, necessitating vector representation of location and force. By breaking down forces into components, we can predict how each part of an object's motion is affected independently, helping solve complex movement problems in physics.

Potential Energy

Potential energy in a gravitational field describes the energy stored due to an object's position relative to another object. In gravitational interactions, potential energy \( U(r) \) is calculated using the equation \( U = - \frac{Gm_1m_2}{r} \).Here,

• \( G \) is the gravitational constant
• \( m_1 \) and \( m_2 \) are the masses involved
• \( r \) is the distance between them

The negative sign in the formula indicates that work is done by the gravitational force to move the masses apart. Consequently, potential energy decreases as energy is transferred from gravitational potential to other forms, such as kinetic energy, as the masses move closer together. Understanding potential energy is crucial for analyzing systems where forces act over distances, letting us calculate how such forces affect the motion and position of objects.

Force Components

Force components are essential for resolving how complex forces act on an object. When a force acts in three-dimensional space, it can be decomposed into separate effects along each

• \(x\)
• \(y\)
• \(z\)

axis.
For gravitational forces between two masses, the components are deduced from the negative gradient of the gravitational potential energy. This process leads to the components:

• \( F_x = - \frac{G m_1 m_2 x}{(x^2 + y^2 + z^2)^{3/2}} \)
• \( F_y = - \frac{G m_1 m_2 y}{(x^2 + y^2 + z^2)^{3/2}} \)
• \( F_z = - \frac{G m_1 m_2 z}{(x^2 + y^2 + z^2)^{3/2}} \)

Each component describes how the force applies separately along each axis and enables us to calculate the net effect of the force on an object by combining these individual actions. This division simplifies solving equations of motion in three-dimensional physics, enabling us to understand complex interactions between objects.

Euclidean Distance Formula

The Euclidean distance formula is used to determine the straight-line distance between two points in space. In three-dimensional space, the distance \( r \) between two points with coordinates

• \( (x_1, y_1, z_1) \)
• \( (x_2, y_2, z_2) \)

can be calculated as \( r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \).
In our exercise, one mass is at the origin, simplifying the formula to \( r = \sqrt{x^2 + y^2 + z^2} \).This classic formula is a cornerstone in physics because it translates spatial relationships into quantitative data that can be used to analyze and predict behavior, particularly in problems involving distances and their effect on force, velocity, and potential energy. By applying the Euclidean distance formula, one can determine how changes in spatial positioning influence other physical parameters.