Question

Let \(\mathbf{X}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\) be the position vector of an object with mass \(m,\) expressed in terms of a rectangular coordinate system with origin at Earth's center (Figure 10.1 .3 ). Derive a system of differential equations for \(x, y,\) and \(z,\) assuming that the object moves under Earth's gravitational force (given by Newton's law of gravitation, as in Example 10.1 .3 ) and a resistive force proportional to the speed of the object. Let \(\alpha\) be the constant of proportionality.

Short answer

Question: Write down the system of differential equations describing the motion of an object under Earth's gravitational force and a resistive force proportional to the object's speed. Answer: The system of differential equations is as follows: \begin{align} \dfrac{d^2x}{dt^2} &= -\dfrac{GM_E x}{(x^2+y^2+z^2)^{\frac{3}{2}}} -\alpha \dfrac{dx}{dt}, \\ \dfrac{d^2y}{dt^2} &= -\dfrac{GM_E y}{(x^2+y^2+z^2)^{\frac{3}{2}}} -\alpha \dfrac{dy}{dt}, \\ \dfrac{d^2z}{dt^2} &= -\dfrac{GM_E z}{(x^2+y^2+z^2)^{\frac{3}{2}}} -\alpha \dfrac{dz}{dt} \end{align}

Step-by-step solution

Gravitational force

Newton's law of gravitation states that the gravitational force between two objects with masses m1 and m2 is given by: \begin{equation} F = G\dfrac {m_1 m_2}{r^2} \end{equation} where G is the gravitational constant, and r is the distance between the two objects. In our case, Earth's mass is \(M_E\) and the object's mass is \(m\). Also, \(\mathbf{X}\) represents the position vector of the object in the rectangular coordinate system with origin at Earth's center, so the distance r is simply the magnitude of \(\mathbf{X}\), which can be calculated as \begin{equation} r = ||\mathbf{X}|| = \sqrt{x^2 + y^2 + z^2} \end{equation} Now, we can find the gravitational force vector \(\mathbf{F}_{grav}\) as: \begin{equation} \mathbf{F}_{grav} = -GM_E m \dfrac{\mathbf{X}}{{||\mathbf{X}||}^3} = -GM_E m\dfrac{x \mathbf{i}+y\mathbf{j}+z\mathbf{k}}{{(x^2+y^2+z^2)}^{\frac {3}{2}}} \end{equation}

Resistive force

The problem states that the resistive force is proportional to the speed of the object. Let the object's velocity vector be \(\mathbf{V} = V_x \mathbf{i}+V_y \mathbf{j}+V_z \mathbf{k}\), where \(V_x, V_y,\) and \(V_z\) are the components of the velocity vector. Then the resistive force can be expressed as: \begin{equation} \mathbf{F}_{res} = -\alpha m \mathbf{V} = -\alpha m(V_x \mathbf{i}+V_y \mathbf{j}+V_z \mathbf{k}) \end{equation}

Newton's second law

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration vector: \(\mathbf{F}_{net} = m\mathbf{A}\). For the given problem, the net force is the sum of the gravitational force and the resistive force: \begin{equation} \mathbf{F}_{net} = \mathbf{F}_{grav} + \mathbf{F}_{res} = m\mathbf{A} \end{equation} Substitute the expressions for the gravitational and resistive forces and the acceleration vector \(\mathbf{A} = A_x \mathbf{i}+A_y \mathbf{j}+A_z \mathbf{k}\): \begin{equation} {-GM_E m\dfrac{x \mathbf{i}+y\mathbf{j}+z\mathbf{k}}{{(x^2+y^2+z^2)}^{\frac {3}{2}}} -\alpha m(V_x \mathbf{i}+V_y\mathbf{j}+V_z\mathbf{k}) } = m(A_x\mathbf{i}+A_y\mathbf{j}+A_z\mathbf{k}) \end{equation}

System of Differential Equations

Now we can equate the i, j, and k components of the net force equation to derive the system of differential equations. For the i component: \begin{equation} -GM_E m\dfrac{x}{{(x^2+y^2+z^2)}^{\frac {3}{2}}} -\alpha mV_x = mA_x \end{equation} For the j component: \begin{equation} -GM_E m\dfrac{y}{{(x^2+y^2+z^2)}^{\frac {3}{2}}} -\alpha mV_y = mA_y \end{equation} For the k component: \begin{equation} -GM_E m\dfrac{z}{{(x^2+y^2+z^2)}^{\frac {3}{2}}} -\alpha mV_z = mA_z \end{equation} Finally, divide each equation by the mass \(m\), and remember that \(A_x = \dfrac{d^2x}{dt^2}, A_y = \dfrac{d^2y}{dt^2},\) and \(A_z = \dfrac{d^2z}{dt^2}\), and \(V_x = \dfrac{dx}{dt}, V_y = \dfrac{dy}{dt},\) and \(V_z = \dfrac{dz}{dt}\): \begin{align} \dfrac{d^2x}{dt^2} &= -\dfrac{GM_E x}{(x^2+y^2+z^2)^{\frac{3}{2}}} -\alpha \dfrac{dx}{dt}, \\ \dfrac{d^2y}{dt^2} &= -\dfrac{GM_E y}{(x^2+y^2+z^2)^{\frac{3}{2}}} -\alpha \dfrac{dy}{dt}, \\ \dfrac{d^2z}{dt^2} &= -\dfrac{GM_E z}{(x^2+y^2+z^2)^{\frac{3}{2}}} -\alpha \dfrac{dz}{dt} \end{align} This system of differential equations describes the motion of the object under Earth's gravitational force and a resistive force proportional to the object's speed.

Key concepts

Newton's law of gravitation

Newton's law of gravitation describes the force of attraction between two masses. This law was formulated by Sir Isaac Newton and states that every particle of matter in the universe attracts every other particle with a force. This gravitational force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. We can express this mathematically as:

• \( F = G\frac {m_1 m_2}{r^2} \)

Here, \(F\) is the gravitational force, \(G\) is the gravitational constant, \(m_1\) and \(m_2\) are the masses, and \(r\) is the distance between the two masses.
In the context of the exercise, we are looking at the gravitational force between Earth and an object in space. Earth's mass is \(M_E\), and the object's mass is \(m\). The position vector \( \mathbf{X} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) gives us the location of the object. The distance from Earth's center to the object is \( r = \sqrt{x^2 + y^2 + z^2} \). This allows us to write the gravitational force experienced by the object as:

• \( \mathbf{F}_{grav} = -GM_E m \frac{\mathbf{X}}{{||\mathbf{X}||}^3} = -GM_E m \frac{x \mathbf{i}+y\mathbf{j}+z\mathbf{k}}{{(x^2+y^2+z^2)}^{\frac {3}{2}}} \)

Understanding how gravity works on objects from a distance and the functional form of its effect helps us predict the movement of celestial bodies and objects under gravitation.

Resistive force

Resistive forces arise when an object moves through a medium, like air or water, and experience forces opposing its motion. In mechanics, the resistive force is often considered to be proportional to velocity. In other words, the faster an object moves, the greater the resistive force it experiences.
The proportional constant \( \alpha \) helps to quantify the resistive force on an object of mass \( m \). Mathematically, if the object's velocity vector is \( \mathbf{V} = V_x \mathbf{i} + V_y \mathbf{j} + V_z \mathbf{k} \), then the resistive force can be expressed as:

• \( \mathbf{F}_{res} = -\alpha m \mathbf{V} = -\alpha m (V_x \mathbf{i} + V_y \mathbf{j} + V_z \mathbf{k}) \)

This means that the resistive force not only opposes the motion of the object but is also directly linked to how quickly it’s moving.
Recognizing the role of resistive forces is important because these forces are always present in real-world scenarios. They limit and alter the motion of objects, making them an important component in understanding dynamics.

Newton's second law of motion

Newton's second law of motion is a fundamental principle in classical mechanics. It connects the forces acting on an object to its motion. This law states that the net force acting on an object is equal to the product of its mass and acceleration, expressed as:

• \( \mathbf{F}_{net} = m \mathbf{A} \)

This equation highlights that a force will change the velocity of an object, causing it to accelerate. In the context of the exercise, the net force is the combination of gravitational force and resistive force:

• \( \mathbf{F}_{net} = \mathbf{F}_{grav} + \mathbf{F}_{res} \)

By substituting the expressions for the gravitational and resistive forces, and knowing the acceleration \( \mathbf{A} = A_x \mathbf{i} + A_y \mathbf{j} + A_z \mathbf{k} \), the motion can be described by the differential equations:

• \( \dfrac{d^2x}{dt^2} = -\dfrac{GM_E x}{(x^2+y^2+z^2)^{\frac{3}{2}}} -\alpha \dfrac{dx}{dt} \)
• \( \dfrac{d^2y}{dt^2} = -\dfrac{GM_E y}{(x^2+y^2+z^2)^{\frac{3}{2}}} -\alpha \dfrac{dy}{dt} \)
• \( \dfrac{d^2z}{dt^2} = -\dfrac{GM_E z}{(x^2+y^2+z^2)^{\frac{3}{2}}} -\alpha \dfrac{dz}{dt} \)

These equations show how an object's motion changes over time due to the combined effects of gravitational pull and resistance from the environment, illustrating Newton's second law in action.