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A satellite of mass \(m\) is in an elliptical orbit (that satisfies Kepler's laws) about a body of mass \(M,\) with \(m\) negligible compared to \(M\) a) Find the total energy of the satellite as a function of its speed, \(v\), and distance, \(r\), from the body it is orbiting. b) At the maximum and minimum distance between the satellite and the body, and only there, the angular momentum is simply related to the speed and distance. Use this relationship and the result of part (a) to obtain a relationship between the extreme distances and the satellite's energy and angular momentum. c) Solve the result of part (b) for the maximum and minimum radii of the orbit in terms of the energy and angular momentum per unit mass of the satellite. d) Transform the results of part (c) into expressions for the semimajor axis, \(a\), and eccentricity of the orbit, \(e\), in terms of the energy and angular momentum per unit mass of the satellite.

Short Answer

Expert verified
Question: In terms of energy and angular momentum per unit mass, provide expressions for the semimajor axis and eccentricity of an elliptical orbit of a satellite around a massive body. Answer: Semimajor axis, \(a\): \(a = \frac{1}{2} \left(\frac{L}{2m}(E + \sqrt{E^2 - 4 \left( \frac{1}{2m} \right) (-\frac{GMm}{r_{max}}) \left( \frac{L^2}{4m^2} \right) }) + \frac{L}{2m}(E - \sqrt{E^2 - 4 \left( \frac{1}{2m} \right) (-\frac{GMm}{r_{min}}) \left( \frac{L^2}{4m^2} \right) })\right)\) Eccentricity, \(e\): \(e = \frac{\frac{L}{2m}(E + \sqrt{E^2 - 4 \left( \frac{1}{2m} \right) (-\frac{GMm}{r_{max}}) \left( \frac{L^2}{4m^2} \right) }) - \frac{L}{2m}(E - \sqrt{E^2 - 4 \left( \frac{1}{2m} \right) (-\frac{GMm}{r_{min}}) \left( \frac{L^2}{4m^2} \right) })}{\frac{L}{2m}(E + \sqrt{E^2 - 4 \left( \frac{1}{2m} \right) (-\frac{GMm}{r_{max}}) \left( \frac{L^2}{4m^2} \right) }) + \frac{L}{2m}(E - \sqrt{E^2 - 4 \left( \frac{1}{2m} \right) (-\frac{GMm}{r_{min}}) \left( \frac{L^2}{4m^2} \right) }))}\)

Step by step solution

01

Find the energy function

Total energy of the satellite at any given moment is the sum of its kinetic energy and potential energy: \(E = T + U\), where \(T\) is kinetic energy and \(U\) is potential energy. Since \(m << M\), we can write the kinetic energy as \(T = \frac{1}{2} mv^2\). The gravitational potential energy is given by \(U = -\frac{GMm}{r}\), where \(G\) is the gravitational constant. The total energy function is thus: \(E = \frac{1}{2} mv^2 -\frac{GMm}{r}\).
02

Maximum and minimum distance

At the maximum and minimum distance between the satellite and the body, the angular momentum is simply related to speed and distance. The angular momentum, \(L\), is given by \(L = mvr\). From step 1, we have the energy function \(E = \frac{1}{2} mv^2 -\frac{GMm}{r}\). We want to obtain a relationship between the extreme distances and the satellite's energy and angular momentum. Divide the angular momentum equation by the satellite's mass to get \(\frac{L}{m} = vr\). We will call the extreme distances \(r_{max}\) and \(r_{min}\).
03

Maximum and minimum radii in terms of energy and angular momentum

Using the result of part (b), let's solve for \(r_{max}\) and \(r_{min}\) based on energy and angular momentum per unit mass: First, let's rewrite the energy function to be in terms of \(\frac{L}{m}\): \(E = \frac{1}{2m} (\frac{L}{m}r)^2 -\frac{GMm}{r}\). Now, let's manipulate the equation to solve for \(r_{max}\): \(r_{max} = \frac{L}{2m}(E + \sqrt{E^2 - 4 \left( \frac{1}{2m} \right) (-\frac{GMm}{r_{max}}) \left( \frac{L^2}{4m^2} \right) })\) And similarly for \(r_{min}\): \(r_{min} = \frac{L}{2m}(E - \sqrt{E^2 - 4 \left( \frac{1}{2m} \right) (-\frac{GMm}{r_{min}}) \left( \frac{L^2}{4m^2} \right) })\)
04

Find semimajor axis and eccentricity expressions

Now let's transform our results in step 3 to find the semimajor axis, \(a\), and eccentricity, \(e\), in terms of the energy and angular momentum per unit mass of the satellite. Semimajor axis: \(a = \frac{1}{2} (r_{max} + r_{min})\) Eccentricity: \(e = \frac{r_{max} - r_{min}}{r_{max} + r_{min}}\) By plugging in our expressions for \(r_{max}\) and \(r_{min}\) from step 3, we can obtain expressions for the semimajor axis and eccentricity in terms of energy and angular momentum per unit mass of the satellite.

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Key Concepts

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

Kepler's Laws of Planetary Motion
Understanding Kepler's laws of planetary motion can bring light to many phenomena related to celestial bodies. These three fundamental laws describe the motion of planets around a star and are integral to satellite orbital mechanics.

The First Law

The first of Kepler's laws, known as the 'Law of Ellipses', states that a planet orbits the sun in an elliptical path, with the sun at one of the two foci. Similarly, a satellite orbits around its parent body, also following an elliptical trajectory.

The Second Law

Kepler's second law, or the 'Law of Equal Areas', indicates that a line segment joining a planet to the sun sweeps out equal areas during equal intervals of time. This means that satellites move faster when they're closer to the object they're orbiting and slower when they're farther away.

The Third Law

The third law, the 'Harmonic Law', establishes the relationship between the orbital period of a planet and its semimajor axis, the largest distance from the center of its elliptical orbit to its perimeter. It tells us that the square of a planet's orbital period is proportional to the cube of the semimajor axis of its orbit.

These laws are pivotal in the determination of the character of the orbit of a satellite and influence calculations including orbital period, velocity, and position at any given time.
Gravitational Potential Energy
Gravitational potential energy (GPE) is the energy an object possesses by virtue of its position in a gravitational field. For satellites, GPE is crucial as it contributes to the total energy governing its orbit.

In the context of orbital mechanics, the GPE of a satellite is always negative because work must be done to overcome the gravitational pull and move a satellite to an infinite distance away. The formula for GPE in the vicinity of a celestial body, such as Earth, is given by \( U = -\frac{GMm}{r} \) where \( G \) is the gravitational constant, \( M \) is the mass of the larger body, \( m \) is the mass of the satellite, and \( r \) is the distance from the center of the larger body to the satellite. It's important for students to understand that GPE allows us to evaluate the satellite's energy at various points in its orbit, and it plays a part in determining the shape and dynamics of the orbit.
Angular Momentum
In orbital mechanics, angular momentum is a measure of the motion of the satellite around the body it is orbiting. It is a conserved quantity, meaning it remains constant unless an external torque is applied. For a satellite, angular momentum \(L\) is given by the product of its mass \(m\), velocity \(v\), and the distance \(r\) from the center of the object it is orbiting: \( L = mvr \).

Conservation of angular momentum explains why satellites in elliptical orbits move faster at the points closest to the orbital focus (periapsis) and slower at the furthest points (apoapsis). This concept is directly connected to Kepler's second law and is used in calculating the changing velocity of the satellite as it moves along its orbit.
Orbit Eccentricity and Semimajor Axis
The shape of a satellite orbit is defined in terms of its eccentricity and semimajor axis. These two elements are crucial in describing not only the form but also the orientation and scale of the orbit.

Eccentricity \(e\) indicates how much an orbit deviates from being circular. An eccentricity of 0 represents a perfect circle, while an eccentricity close to 1 suggests a highly elongated ellipse. The formula obtained from the exercise relates eccentricity to the maximum and minimum radii, showing how energy and angular momentum influence the orbit's shape.

The semimajor axis \(a\) is half the longest diameter of the elliptical orbit. It can be thought of as the average distance between the satellite and the body it orbits around. The semimajor axis is directly tied to Kepler’s third law, as it is used to determine the orbital period of the satellite. Understanding these elements gives insight into not just the current orbit, but also how the orbit might change in response to factors like thrust from satellite thrusters or gravitational influences from other bodies.

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Most popular questions from this chapter

A planet with a mass of \(7.00 \cdot 10^{21} \mathrm{~kg}\) is in a circular orbit around a star with a mass of \(2.00 \cdot 10^{30} \mathrm{~kg} .\) The planet has an orbital radius of \(3.00 \cdot 10^{10} \mathrm{~m}\). a) What is the linear orbital velocity of the planet? b) What is the period of the planet's orbit? c) What is the total mechanical energy of the planet?

Is the orbital speed of the Earth when it is closest to the Sun greater than, less than, or equal to the orbital speed when it is farthest from the Sun? Explain.

Halley's comet orbits the Sun with a period of 76.2 yr. a) Find the semimajor axis of the orbit of Halley's comet in astronomical units ( \(1 \mathrm{AU}\) is equal to the semimajor axis of the Earth's orbit). b) If Halley's comet is \(0.56 \mathrm{AU}\) from the Sun at perihelion, what is its maximum distance from the Sun, and what is the eccentricity of its orbit?

Consider the Sun to be at the origin of an \(x y\) coordinate system. A telescope spots an asteroid in the \(x y\) -plane at a position given by \(\left(2.0 \cdot 10^{11} \mathrm{~m}, 3.0 \cdot 10^{11} \mathrm{~m}\right)\) with a velocity given by \(\left(-9.0 \cdot 10^{3} \mathrm{~m} / \mathrm{s},-7.0 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\right) .\) What will the asteroid's speed and distance from the Sun be at closest approach?

For two identical satellites in circular motion around the Earth, which statement is true? a) The one in the lower orbit has less total energy. b) The one in the higher orbit has more kinetic energy. c) The one in the lower orbit has more total energy. d) Both have the same total energy.

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