Problem 44 - Thomas Calculus

Question

A satellite in circular orbit \(A\) satellite of mass \(m\) is revolving at a constant speed \(v\) around a body of mass \(M\) (Earth, for example \()\) in a circular orbit of radius \(r_{0}\) (measured from the body's center of mass). Determine the satellite's orbital period \(T\) (the time to complete one full orbit), as follows: a. Coordinatize the orbital plane by placing the origin at the body's center of mass, with the satellite on the \(x\) -axis at \(t=0\) and moving counterclockwise, as in the accompanying figure. Let \(\mathbf{r}(t)\) be the satellite's position vector at time \(t .\) Show that \(\theta=y t / r_{0}\) and hence that $$ r(t)=\left(r_{0} \cos \frac{v t}{r_{0}}\right) \mathbf{i}+\left(r_{0} \sin \frac{v t}{r_{0}}\right) \mathbf{j} $$ b. Find the acceleration of the satellite. c. According to Newton's law of gravitation, the gravitational force exerted on the satellite is directed toward \(M\) and is given by $$ \mathbf{F}=\left(-\frac{G m M}{r_{0}^{2}}\right) \frac{\mathbf{r}}{r_{0}} $$ where \(G\) is the universal constant of gravitation. Using Newton's second law, \(\mathbf{F}=m \mathbf{a},\) show that \(v^{2}=G M / r_{0}\) . d. Show that the orbital period \(T\) satisfies \(v T=2 \pi r_{0}\) . e. From parts (c) and (d), deduce that $$ T^{2}=\frac{4 \pi^{2}}{G M} r_{0}^{3} $$ That is, the square of the period of a satellite in circular orbit is proportional to the cube of the radius from the orbital center.

Step-by-Step Solution

Answer

The orbital period \(T\) is given by \(T^2 = \frac{4\pi^2}{GM} r_0^3\). This illustrates that \(T^2\) is proportional to \(r_0^3\).

Step 1: Define the Position Vector

We use coordinates in the orbital plane with the body's center as the origin. At time \(t = 0\), the satellite is on the \(x\)-axis. The position of the satellite at any time \(t\) is described using polar coordinates, with the angle \(\theta\) given by the relation \(\theta = \frac{vt}{r_0}\). Therefore, the position vector \(\mathbf{r}(t)\) becomes: \[\mathbf{r}(t) = \left( r_0 \cos\left(\frac{vt}{r_0}\right)\right) \mathbf{i} + \left( r_0 \sin\left(\frac{vt}{r_0}\right)\right) \mathbf{j}\]

Step 2: Calculate the Acceleration

The position vector from Step 1 is already provided. To find the acceleration, we need to differentiate the position vector twice with respect to \(t\):1. Differentiate once to get the velocity:\[\mathbf{v}(t) = \frac{d}{dt}\mathbf{r}(t) = -v\sin\left(\frac{vt}{r_0}\right) \mathbf{i} + v\cos\left(\frac{vt}{r_0}\right) \mathbf{j}\]2. Differentiate the velocity to get acceleration:\[\mathbf{a}(t) = \frac{d}{dt}\mathbf{v}(t) = -\frac{v^2}{r_0}\cos\left(\frac{vt}{r_0}\right) \mathbf{i} - \frac{v^2}{r_0}\sin\left(\frac{vt}{r_0}\right) \mathbf{j}\] This shows that the acceleration is directed towards the center of the orbit.

Step 3: Apply Newton's Law of Gravitation

According to Newton's law of gravitation, the force on the satellite is:\[\mathbf{F} = \left(-\frac{GmM}{r_0^2}\right) \frac{\mathbf{r}}{r_0}\]From the second law \(\mathbf{F} = m\mathbf{a}\), substitute the acceleration:\[-m\frac{v^2}{r_0}\mathbf{r} = -\frac{GmM}{r_0^3}\mathbf{r}\]Cancelling common terms, we get:\[v^2 = \frac{GM}{r_0}\]

Step 4: Determine the Orbital Period Relation

The orbital period \(T\) is the time taken for a complete orbit. The circumference of the orbit is \(2\pi r_0\). At constant speed \(v\), for one orbit:\[vT = 2\pi r_0\]Solve for \(T\):\[T = \frac{2\pi r_0}{v}\]

Step 5: Derive the Period-Radius Relation

Substitute the expression for \(v\) from Step 3 into the relation from Step 4:\[T = \frac{2\pi r_0}{\sqrt{\frac{GM}{r_0}}}\]Simplify:\[T = 2\pi \sqrt{\frac{r_0^3}{GM}}\]Squaring both sides gives:\[T^2 = \frac{4\pi^2}{GM} r_0^3\]This supports Kepler's third law.

Key Concepts

Circular Orbit

In astronomy, and particularly when discussing satellites and planets, a circular orbit is one of the simplest forms of an orbit. It is defined as the path of an object whose centripetal force is provided by gravity, resulting in a constant speed around a larger body. In a circular orbit, the distance from the center of the larger body to the orbiting object remains constant, creating a perfect circle.
For a satellite in a circular orbit, its velocity and distance from the Earth's center give a precise balance between the gravitational pull and the object's tendency to move in a straight line. This balance keeps the satellite moving in a circular path. The velocity, at any point in this orbit, is tangent to the circle, and the gravitational force always points towards the center of this circle.
In our exercise, we noted that the satellite revolves around a massive body (like Earth) in a circular path, maintaining a constant radius. The radius will be critical later in understanding how the orbital period relates to the gravitational influences.

Newton's Laws of Motion

Newton's Laws of Motion are fundamental principles that explain how objects behave when forces are applied.

• First Law - Law of Inertia: An object at rest stays at rest, and an object in motion remains in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
• Second Law - Law of Acceleration: The acceleration of an object depends directly on the net force acting upon the object and inversely on the mass of the object. It can be formulated as: \( F = ma \).
• Third Law - Action and Reaction: For every action, there is an equal and opposite reaction.

For our satellite in circular orbit, Newton's Second Law is pivotal. The gravitational force acting as the centripetal force creates the centripetal acceleration necessary for the satellite to maintain its circular path. This equation, \( F = ma \), is directly used to balance the gravitational force and derive the velocity necessary for circular motion.

Gravitational Force

The gravitational force is a fundamental force of nature, governing the attraction between masses. It is described by Newton's Law of Universal Gravitation, which states: \[ F = -\frac{GmM}{r^2} \] Where:

• \( F \) is the force of attraction between two masses.
• \( G \) is the universal gravitational constant.
• \( m \) and \( M \) are the masses of the two objects.
• \( r \) is the distance between the centers of the two masses.

In the context of our satellite, the gravitational force acts as the centripetal force holding the satellite in its orbit. This force draws the satellite toward the larger mass (e.g., Earth) and is essential for calculating the orbital velocity. Using Newton's Second Law, we can relate this force to the required centripetal acceleration in the circular motion, helping to derive the satellite's velocity as \( v^2 = \frac{GM}{r_0} \). This is critical in determining the orbit's stability and the satellite's speed.

Kepler's Third Law

Kepler's Third Law is a vital principle in understanding orbits, stating that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. In mathematical terms, it can be expressed as: \[ T^2 \propto r^3 \] For a satellite in a circular orbit around Earth, as shown in our exercise, this law takes a specific form: \[ T^2 = \frac{4\pi^2}{GM} r_0^3 \] This equation identifies a direct relationship between the square of a satellite's orbital period and the cube of the orbit radius. It confirms that larger orbits take longer to complete. Kepler's Third Law, therefore, is not only useful in predicting orbital periods but also reveals consistency in our universe’s gravitational force. This relationship derives from the balance of the forces and motions we've discussed, providing a means to predict how orbital changes (i.e., moving to a higher orbit) will affect a satellite’s period, checking for compliance with natural laws.