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Suppose that a satellite goes around the Moon in an elliptical orbit. At its closest approach it has a speed \(v_{c}\) and a radius \(r_{c}\) from the center of the Moon. At its farthest distance, it has a speed \(v_{f}\) and a radius \(r_{f}\). Find the ratio \(v_{c} / v_{f}\). [Hint: Angular momentum is conserved, and, moreover, the satellite can be treated as a point mass.]

Short Answer

Expert verified
\(\frac{v_c}{v_f} = \frac{r_f}{r_c}\)

Step by step solution

01

Understand Conservation of Angular Momentum

The angular momentum of a satellite in orbit is conserved. This means the angular momentum at the closest approach is equal to the angular momentum at the farthest point. For a satellite treated as a point mass, the angular momentum \(L\) is given by \(L = mvr\), where \(m\) is the mass of the satellite, \(v\) is its speed, and \(r\) is the radius.
02

Express Angular Momentum at Closest Approach

At the closest approach, the satellite has a speed \(v_c\) and a radius \(r_c\). The angular momentum \(L_c\) at this point is \(L_c = m v_c r_c\).
03

Express Angular Momentum at Farthest Distance

At the farthest distance, the satellite has a speed \(v_f\) and a radius \(r_f\). The angular momentum \(L_f\) at this point is \(L_f = m v_f r_f\).
04

Set Angular Momenta Equal

Since angular momentum is conserved, we set \(L_c = L_f\). This yields the equation \(m v_c r_c = m v_f r_f\). Since the mass \(m\) is common on both sides, it cancels out, simplifying to \(v_c r_c = v_f r_f\).
05

Solve for the Ratio of Speeds

Rearrange the equation to find the ratio of the speeds: \(\frac{v_c}{v_f} = \frac{r_f}{r_c}\). Thus, the ratio of the speeds at the closest and farthest points is equal to the inverse ratio of their respective radii.

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Key Concepts

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

Conservation of Angular Momentum
In the realm of physics, the conservation of angular momentum is a fundamental principle. It states that if no external torque acts on a system, the total angular momentum of that system remains constant. For a satellite orbiting the Moon, this is crucial. Angular momentum, denoted by \( L \), depends on three elements: the mass \( m \), the velocity \( v \), and the radius \( r \) from the center of mass it is orbiting.

The equation for angular momentum is \( L = mvr \). In an elliptical orbit, at two specific points 'closest' (perigee) and 'farthest' (apogee) from the Moon, the angular momentum remains unchanged. This conservation allows us to relate velocities at these points directly.

In essence, conservation of angular momentum provides a balancing act: as a satellite moves closer to the Moon, its speed increases, and as it moves farther away, its speed decreases.
Elliptical Orbit
An elliptical orbit is one of Kepler's laws of planetary motion, describing the paths of moons and satellites. Unlike a perfect circle, an ellipse has two focal points, with celestial bodies orbiting in this stretched circle shape. When a satellite orbits in an ellipse, two main points are of interest: perigee, its closest point, and apogee, its farthest point.

This is relevant in satellite dynamics as it illustrates how the velocity and position of the satellite change throughout its orbit. At perigee, the satellite's speed is faster due to the gravitational pull of the Moon being stronger. When it reaches apogee, the speed notably drops as the gravitational pull weakens.

Understanding these changes and the behavior of a satellite in an elliptical orbit is crucial to predicting its motion and ensuring its operation aligns with its mission objectives.
Satellite Dynamics
Satellite dynamics delve into how and why satellites move in space. By treating a satellite as a point mass, we simplify understanding its motion around larger celestial bodies. Key to this is how forces like gravity influence its path, speed, and position.

A satellite's trajectory, especially in non-circular (elliptical) orbits, is influenced by gravitational pulls from the body it orbits, leading to changes in velocity and position. One practical outcome is the conservation of angular momentum, allowing predictions of speed changes as the satellite moves between perigee and apogee.

Engineers and scientists employ these principles when designing satellite paths to ensure precise deployment, position stability, and long-term success of satellite missions.
Physics Problem Solving
Physics problem solving involves applying theoretical principles to find solutions to real-world questions and scenarios. A common method used is breaking down complex problems into simpler steps, applying known laws, and utilizing mathematical relationships to find solutions.

In the exercise tackled, each step aims to harness the conservation of angular momentum. By equating angular momentum at the two critical points (perigee and apogee), and simplifying the equation by canceling common terms (like mass), we derive meaningful relationships like the speed ratio: \(\frac{v_c}{v_f} = \frac{r_f}{r_c}\).

This approach highlights the beauty of physics problem solving: from identifying applicable laws to simplifying equations, allowing us to understand complex behaviors through simple principles.

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Most popular questions from this chapter

A \(500-\mathrm{g}\) uniform sphere of \(7.0-\mathrm{cm}\) radius spins frictionlessly at \(30 \mathrm{rev} / \mathrm{s}\) on an axis through its center. Find its ( \(a\) ) \(\mathrm{KE}_{r},(b)\) angular momentum, and \((c)\) radius of gyration. We need the moment of inertia of a uniform sphere about an axis through its center. From Fig. 10-1, $$ I=\frac{2}{5} M r^{2}=(0.40)(0.50 \mathrm{~kg})(0.070 \mathrm{~m})^{2}=0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2} $$ (a) Knowing that \(\omega=30 \mathrm{rev} / \mathrm{s}=188 \mathrm{rad} / \mathrm{s}\), we have $$ \mathrm{KE}_{r}=\frac{1}{2} I \omega^{2}=\frac{1}{2}\left(0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(188 \mathrm{rad} / \mathrm{s})^{2}=0.017 \mathrm{~kJ} $$ Notice that \(\omega\) must be in \(\mathrm{rad} / \mathrm{s}\). (b) Its angular momentum is $$ L=I \omega=\left(0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(188 \mathrm{rad} / \mathrm{s})=0.18 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s} $$ (c) For any object, \(I=M k^{2}\), where \(k\) is the radius of gyration. Therefore, $$ k=\sqrt{\frac{I}{M}}=\sqrt{\frac{0.00098 \mathrm{~kg} \cdot \mathrm{m}^{2}}{0.50 \mathrm{~kg}}}=0.044 \mathrm{~m}=4.4 \mathrm{~cm} $$ Notice that this is a reasonable value in view of the fact that the radius of the sphere is \(7.0 \mathrm{~cm}\).

A tiny solid ball \(\left(I=2 M r^{2} / 5\right)\) rolls without slipping on the inside surface of a hemisphere as shown in Fig. 10-12. (The ball is much smaller than shown.) If the ball is released at \(A\), how fast is it moving as it passes ( \(a\) ) point- \(B\), and \((b)\) point- \(C\) ? Ignore friction losses. [Hint: Study the two previous questions. When it comes to the ball's descent, its own radius is negligible.]

When \(100 \mathrm{~J}\) of work is done on a stationary flywheel (that is otherwise free to rotate in place), its angular speed increases from 60 rev/min to 180 rev \(/ \mathrm{min}\). What is its moment of inertia?

The driving side of a belt has a tension of \(1600 \mathrm{~N}\), and the slack side has 500 - \(\mathrm{N}\) tension. The belt turns a pulley \(40 \mathrm{~cm}\) in radius at a rate of \(300 \mathrm{rpm} .\) This pulley drives a dynamo having 90 percent efficiency. How many kilowatts are being delivered by the dynamo?

A large horizontal disk is rotating on a vertical axis through its center. Its moment of inertia is \(I=4000 \mathrm{~kg} \cdot \mathrm{m}^{2}\). The disk is revolving freely at a rate of \(0.150 \mathrm{rev} / \mathrm{s}\) when a \(90.0-\mathrm{kg}\) person drops straight down onto it from an overhanging tree limb. The person lands and remains at a distance of \(3.00 \mathrm{~m}\) from the axis of rotation. What will be the rate of rotation after the person has landed?

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