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A satellite is in a circular Earth orbit of radius \(r\). The area \(A\) enclosed by the orbit depends on \(r^{2}\) because \(A=\pi r^{2} .\) Determine how the following properties of the satellite depend on \(r:\) (a) period, (b) kinetic energy, (c) angular momentum, and (d) speed.

Short Answer

Expert verified
(a) \(r^{3/2}\), (b) \(\frac{1}{r}\), (c) \(\sqrt{r}\), (d) \(r^{-1/2}\)

Step by step solution

01

Understand the Problem

We need to determine how various properties of a satellite—specifically its period, kinetic energy, angular momentum, and speed—depend on the radius of its circular orbit around the Earth. The given direct relationship is the area enclosed by the orbit, \(A = \pi r^2\).
02

Recall Relevant Formulas

For any satellite in a circular orbit, the period \(T\), kinetic energy \(K\), angular momentum \(L\), and speed \(v\) depend on physical laws and constants. Recall Kepler's third law: \(T^2 \propto r^3\), kinetic energy \(K = \frac{1}{2} mv^2\), angular momentum \(L = mvr\), and velocity \(v = \sqrt{\frac{GM}{r}}\). Here, \(G\) is the gravitational constant, \(M\) is the Earth’s mass, and \(m\) is the satellite’s mass.
03

Determine the Period's Dependence on r

From Kepler's third law, the period \(T\) is related to the radius \(r\) by \(T^2 \propto r^3\). Thus, \(T \propto r^{3/2}\). This indicates that as the radius increases, the satellite's orbital period also increases as a function of \(r^{3/2}\).
04

Determine the Kinetic Energy's Dependence on r

Using the formula \(v = \sqrt{\frac{GM}{r}}\), the kinetic energy expression becomes \(K = \frac{1}{2} m \left(\sqrt{\frac{GM}{r}}\right)^2 = \frac{GMm}{2r}\). Therefore, \(K \propto \frac{1}{r}\), indicating that as the radius increases, the kinetic energy decreases inversely with \(r\).
05

Determine the Angular Momentum's Dependence on r

Given \(L = mvr\) and substituting \(v = \sqrt{\frac{GM}{r}}\), we find \(L = m \left(\sqrt{\frac{GM}{r}}\right) r = m\sqrt{GMr}\). Thus, \(L \propto \sqrt{r}\), showing that angular momentum increases as a function of \(\sqrt{r}\) with increasing radius.
06

Determine the Speed's Dependence on r

We have the velocity formula \(v = \sqrt{\frac{GM}{r}}\). This shows that the speed is proportional to \(r^{-1/2}\). Therefore, as the radius increases, the speed decreases as a function of \(r^{-1/2}\).
07

Summarize the Dependencies

(a) The period \(T\) depends on the radius as \(T \propto r^{3/2}\). (b) The kinetic energy \(K\) depends on the radius as \(K \propto \frac{1}{r}\). (c) The angular momentum \(L\) depends on the radius as \(L \propto \sqrt{r}\). (d) The speed \(v\) depends on the radius as \(v \propto r^{-1/2}\).

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Key Concepts

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

Orbital Period
The orbital period is the time a satellite takes to complete one full orbit around the Earth. This concept is fundamentally linked to the size of the orbit. According to Kepler's third law of planetary motion, the square of the orbital period, denoted as \(T^2\), is directly proportional to the cube of the semi-major axis of the orbit, represented as \(r^3\). In simpler terms, \(T^2 \propto r^3\).
- This implies that the orbital period \(T\) is proportional to \(r^{3/2}\).- As a satellite orbits farther from Earth, it takes longer to complete an orbit.Understanding this principle helps in predicting how changes in orbital size affect the time it takes for a satellite to orbit once. It’s crucial for maintaining timing and synchronization for satellites used in communication and GPS systems.
Kinetic Energy
Kinetic energy in the context of orbital dynamics is the energy that a satellite possesses due to its motion. For a satellite in a stable, circular orbit, the formula for kinetic energy \(K\) is given by \(K = \frac{1}{2} mv^2\), where \(m\) is the mass of the satellite and \(v\) is its velocity. When you substitute the expression for velocity \(v = \sqrt{\frac{GM}{r}}\), the equation for kinetic energy becomes: - \(K = \frac{GMm}{2r}\)This shows that kinetic energy is inversely proportional to the orbital radius, i.e., \(K \propto \frac{1}{r}\).
- As the orbital radius \(r\) increases, the kinetic energy \(K\) decreases.- This is because a satellite further from Earth travels at slower velocities but still maintains its orbital path.This inverse relationship highlights how energy is distributed across different potential and kinetic forms as satellites orbit at various altitudes.
Angular Momentum
Angular momentum is a key concept in orbital mechanics. It describes the momentum of a satellite as it moves in its trajectory around a central body like the Earth. The angular momentum \(L\) for a satellite can be defined by the formula \(L = mvr\). When we further expand \(v\) with the relation \(v = \sqrt{\frac{GM}{r}}\), the expression becomes:- \(L = m\sqrt{GMr}\)This demonstrates that angular momentum is proportional to the square root of the orbital radius, \(L \propto \sqrt{r}\).
- As the radius \(r\) increases, the angular momentum \(L\) increases.- Conservation of angular momentum means any change in radius will adjust speed and path, maintaining a balance.Grasping angular momentum aids in understanding how satellites stabilize their orbits and how orbit transfers work in space missions.
Orbital Speed
Orbital speed refers to how fast a satellite travels along its orbit. The speed \(v\) of a satellite in a circular orbit is given by the expression \(v = \sqrt{\frac{GM}{r}}\), where \(G\) is the universal gravitational constant and \(M\) is the Earth's mass. From this, we derive:- \(v \propto r^{-1/2}\)This relationship indicates that orbital speed is inversely proportional to the square root of the radius.
- As the orbital radius \(r\) increases, the speed \(v\) decreases.- Satellites closer to the Earth travel faster than those in distant orbits.Understanding this dependency is essential for calculations involving satellite launches and orbital adjustments, as it impacts satellite stability in different orbits.

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Most popular questions from this chapter

Two Earth satellites, \(A\) and \(B\), each of mass \(m\), are to be launched into circular orbits about Earth's center. Satellite \(A\) is to orbit at an altitude of \(6370 \mathrm{~km} .\) Satellite \(B\) is to orbit at an altitude of \(19110 \mathrm{~km}\). The radius of Earth \(R_{E}\) is \(6370 \mathrm{~km}\). (a) What is the ratio of the potential energy of satellite \(B\) to that of satellite \(A\), in orbit? (b) What is the ratio of the kinetic energy of satellite \(B\) to that of satellite \(A\), in orbit? (c) Which satellite has the greater total energy if each has a mass of \(14.6 \mathrm{~kg} ?\) (d) By how much?

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In a certain binary-star system, each star has the same mass as our Sun, and they revolve about their center of mass. The distance between them is the same as the distance between Earth and the Sun. What is their period of revolution in years?

Two concentric spherical shells with uniformly distributed masses \(M_{1}\) and \(M_{2}\) are situated as shown in Fig. \(13-41\). Find the magnitude of the net gravitational force on a particle of mass \(m\), due to the shells, when the particle is located at radial distance (a) \(a\), (b) \(b\), and (c) \(c\).

In his 1865 science fiction novel From the Earth to the Moon, Jules Verne described how three astronauts are shot to the Moon by means of a huge gun. According to Verne, the aluminum capsule containing the astronauts is accelerated by ignition of nitrocellulose to a speed of \(11 \mathrm{~km} / \mathrm{s}\) along the gun barrel's length of \(220 \mathrm{~m} .\) (a) In \(g\) units, what is the average acceleration of the capsule and astronauts in the gun barrel? (b) Is that acceleration tolerable or deadly to the astronauts? A modern version of such gun-launched spacecraft (although without passengers) has been proposed. In this modern version, called the SHARP (Super High Altitude Research Project) gun, ignition of methane and air shoves a piston along the gun's tube, compressing hydrogen gas that then launches a rocket. During this launch, the rocket moves \(3.5 \mathrm{~km}\) and reaches a speed of \(7.0 \mathrm{~km} / \mathrm{s}\). Once launched, the rocket can be fired to gain additional speed. (c) In \(g\) units, what would be the average acceleration of the rocket within the launcher? (d) How much additional speed is needed (via the rocket engine) if the rocket is to orbit Earth at an altitude of \(700 \mathrm{~km} ?\)

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