Question

Gravitational force due to a mass The gravitational force on a point mass \(m\) due to a point mass \(M\) at the origin is a gradient field with potential \(U(r)=\frac{G M m}{r},\) where \(G\) is the gravitational constant and \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) is the distance between the masses. a. Find the components of the gravitational force in the \(x-, y-\), and \(z\) -directions, where \(\mathbf{F}(x, y, z)=-\nabla U(x, y, z)\) b. Show that the gravitational force points in the radial direction (outward from point mass \(M\) ) and the radial component is \(F(r)=\frac{G M m}{r^{2}}\) c. Show that the vector field is orthogonal to the equipotential surfaces at all points in the domain of \(U.\)

Short answer

A: The radial component of the gravitational force at any point (x, y, z) based on the distance r is given by: \(F(r) = \frac{G M m}{r^{2}}\).

Step-by-step solution

(Step 1: Find the components of the gravitational force)

(To find the components of the gravitational force in the \(x, y\), and \(z\) directions, we will first find the gradient of the potential \(U(r)\) and then use the given relation \(\mathbf{F}(x, y, z)=-\nabla U(x, y, z)\). )

(Step 2: Calculate the gradient of the potential U)

(To calculate the gradient of the potential \(U(r)=\frac{G M m}{r}\), we need to find the partial derivatives with respect to \(x, y\), and \(z\). We have \(r=\sqrt{x^{2}+y^{2}+z^{2}}\). Then, \(\frac{\partial U}{\partial x} = -\frac{G M m}{r^3} x,\) \(\frac{\partial U}{\partial y} = -\frac{G M m}{r^3} y,\) \(\frac{\partial U}{\partial z} = -\frac{G M m}{r^3} z.\))

(Step 3: Find the components of the gravitational force)

(Since we have found the partial derivatives, we can find the components of the gravitational force by using the relation \(\mathbf{F}(x, y, z)=-\nabla U(x, y, z)\). Thus, the components of the gravitational force are: \(F_x = -\frac{\partial U}{\partial x} = \frac{G M m x}{(x^2+y^2+z^2)^{\frac{3}{2}}},\) \(F_y = -\frac{\partial U}{\partial y} = \frac{G M m y}{(x^2+y^2+z^2)^{\frac{3}{2}}},\) \(F_z = -\frac{\partial U}{\partial z} = \frac{G M m z}{(x^2+y^2+z^2)^{\frac{3}{2}}}.\))

(Step 4: Show that the gravitational force points in the radial direction)

(By examining the components of the force from the previous step, we can see that the gravitational force points in the radial direction since it's proportional to \(x, y,\) and \(z\) which are the coordinates from point mass \(M\) to point mass \(m\). )

(Step 5: Find the radial component of the gravitational force)

(Since the gravitational force points in the radial direction with the components found in Step 3, we can calculate the radial component of the gravitational force at any point \((x, y, z)\) based on the distance \(r\). We have: \(F(r) = \sqrt{F_x^2 + F_y^2 + F_z^2} = \sqrt{\left(\frac{G M m x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\right)^2+\left(\frac{G M m y}{(x^2+y^2+z^2)^{\frac{3}{2}}}\right)^2+\left(\frac{G M m z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\right)^2}=\frac{G M m}{r^{2}}.\))

(Step 6: Show that the vector field is orthogonal to the equipotential surfaces at all points)

( Equipotential surfaces are the surfaces where the potential function \(U(r)\) has constant values. Consider two points on the equipotential surface, \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\). As they are on the same equipotential surface, we have \(U(x_1, y_1, z_1) = U(x_2, y_2, z_2)\). Taking the gradient of the potential function at both points, we have: \(-\nabla U(x_1, y_1, z_1) = -\nabla U(x_2, y_2, z_2)\). Thus, the vector field \(\mathbf{F}(x, y, z)\) is orthogonal to the equipotential surfaces because the difference of gradients between two points on the same equipotential surface is also orthogonal. In simpler terms, since gravitational potential depends only on the distance \(r\) from the origin, we can say that surfaces with constant \(U(r)\) will be the surfaces with the same distance from the origin (i.e., spheres). The force vector \(\mathbf{F}(x, y, z)\) always points in a radial direction, which is normal (orthogonal) to these spherical surfaces. )

Key concepts

Gradient Field

When learning about gravitational forces and their interaction with masses, a fundamental concept that comes into play is the gradient field. In physics, a gradient field is related to the variation of a particular quantity over space. In the case of gravity, this quantity is the gravitational potential energy.

The gradient of a scalar field like gravitational potential energy gives us a vector field which points in the direction of the greatest rate of increase of that scalar. For the gravitational field, this gradient tells us the direction and magnitude of the force experienced by a mass due to another mass. In essence, the gradient field is a mathematical tool that helps us transition from knowing the potential energy at a point to understanding the force experienced at that point.

Understanding the gradient is crucial because it allows us to calculate how objects will move under the influence of gravity. The negative sign in the gradient indicates that the force is attractive and that the force vector always points toward lower potential energy.

Gravitational Potential Energy

The concept of gravitational potential energy (GPE) is intrinsic to the study of celestial mechanics and any system where gravity is a significant force. GPE is the energy that an object possesses because of its position in a gravitational field. In our case, it's represented by the equation \(U(r) = \frac{G M m}{r}\), where \(G\) is the gravitational constant, \(M\) is the mass of one object, \(m\) is the mass of the second object, and \(r\) is the distance between the two masses.

This potential energy is pivotal because it is what gets converted to kinetic energy as masses move toward each other due to gravity's pull. It also serves as a conservative field which means that the total energy (kinetic plus potential) in a system is conserved, making calculations of motion possible without needing to consider the path taken by the mass.

Equipotential Surfaces

Equipotential surfaces are incredibly insightful when understanding gravitational fields and potential energy. These are imaginary surfaces where every point on the surface has the same potential energy. When dealing with a point mass, these surfaces are spherical. The significance of equipotential surfaces in the context of gravity is that they allow us to visualize the potential energy landscape.

Why is this important? If an object moves along an equipotential surface, it doesn't change its potential energy; therefore, no work is done on or by the object. This is a handy principle because it helps us understand that the gravitational force must act in a direction perpendicular to these surfaces. Through such visualization, we can have a much clearer picture of how forces interact with masses and the resultant movement.

Partial Derivatives

Partial derivatives play a critical role in physics, particularly in the context of fields like a gravitational field. They are used to describe how a function changes as one of its variables changes, while the other variables are held constant. In the gravitational potential energy function \(U(r)\), which depends on \(x, y,\) and \(z\), the partial derivatives represent the rate of change of the potential energy with respect to each of these spatial variables.

Understanding partial derivatives allows us to find the gravitational force components in the \(x, y,\) and \(z\) directions, providing us with the direction and magnitude of the force due to the gradient of the potential energy. The application of partial derivatives exemplifies the use of calculus in physics as an essential tool for analyzing variable interactions and their effects on the physical world.