Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 6

Planetary orbits The orbit in space of one body around another, such as a planet around the Sun, need not be circular. In general it takes the form of an ellipse, with the body sometimes closer in and sometimes further out. If you are given the distance \(\ell_{1}\) of closest approach that a planet makes to the Sun, also called its perihelion, and its linear velocity \(v_{1}\) at perihelion, then any other property of the orbit can be calculated from these two as follows. a) Kepler's second law tells us that the distance \(\ell_{2}\) and velocity \(v_{2}\) of the planet at its most distant point, or aphelion, satisfy \(\ell_{2} v_{2}=\ell_{1} v_{1}\). At the same time the total energy, kinetic plus gravitational, of a planet with velocity \(v\) and distance \(r\) from the Sun is given by $$ E=\frac{1}{2} m v^{2}-G \frac{m M}{r}, $$ where \(m\) is the planet's mass, \(M=1.9891 \times 10^{30} \mathrm{~kg}\) is the mass of the Sun, and \(G=6.6738 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\) is Newton's gravitational constant. Given that energy must be conserved, show that \(v_{2}\) is the smaller root of the quadratic equation $$ v_{2}^{2}-\frac{2 G M}{v_{1} \ell_{1}} v_{2}-\left[v_{1}^{2}-\frac{2 G M}{\ell_{1}}\right]=0 . $$ Once we have \(v_{2}\) we can calculate \(\ell_{2}\) using the relation \(\ell_{2}=\ell_{1} v_{1} / v_{2}\). b) Given the values of \(v_{1}, \ell_{1}\), and \(\ell_{2}\), other parameters of the orbit are given by simple formulas can that be derived from Kepler's laws and the fact that the orbit is an ellipse: $$ \begin{aligned} \text { Semi-major axis: } & a=\frac{1}{2}\left(\ell_{1}+\ell_{2}\right), \\ \text { Semi-minor axis: } & b=\sqrt{\ell_{1} \ell_{2}}, \\ \text { Orbital period: } & T=\frac{2 \pi a b}{\ell_{1} v_{1}}, \\ \text { Orbital eccentricity: } & e=\frac{\ell_{2}-\ell_{1}}{\ell_{2}+\ell_{1}} . \end{aligned} $$ Write a program that asks the user to enter the distance to the Sun and velocity at perihelion, then calculates and prints the quantities \(\ell_{2}, v_{2}, T\), and \(e\). c) Test your program by having it calculate the properties of the orbits of the Earth (for which \(\ell_{1}=1.4710 \times 10^{11} \mathrm{~m}\) and \(v_{1}=3.0287 \times 10^{4} \mathrm{~ms}^{-1}\) ) and Halley's comet \(\left(\ell_{1}=8.7830 \times 10^{10} \mathrm{~m}\right.\) and \(\left.v_{1}=5.4529 \times 10^{4} \mathrm{~m} \mathrm{~s}^{-1}\right)\). Among other things, you should find that the orbital period of the Earth is one year and that of Halley's comet is about 76 years.

Short Answer

Expert verified
Solve the quadratic for \(v_2\), use \(\ell_2 = \ell_1 v_1 / v_2\), then compute \(a, b, T, e\).

Step by step solution

01

Applying Kepler's Second Law

From Kepler's second law, the product of the distance \(\ell_2\) and the velocity \(v_2\) at aphelion is equal to the product of the distance \(\ell_1\) and the velocity \(v_1\) at perihelion: \[\ell_2 v_2 = \ell_1 v_1.\]
02

Expressing the Energy Conservation Equation

The total energy at perihelion and aphelion should be the same. Write down the energy equation for both points: \[E = \frac{1}{2} m v_1^2 - G \frac{m M}{\ell_1} = \frac{1}{2} m v_2^2 - G \frac{m M}{\ell_2}.\] Since energy is conserved, equate the two expressions and simplify.
03

Simplifying the Energy Equation

Substitute \(\ell_2 = \frac{\ell_1 v_1}{v_2}\). This allows the conservation of energy equation to become: \[\frac{1}{2} m v_1^2 - G \frac{m M}{\ell_1} = \frac{1}{2} m v_2^2 - G \frac{m M v_2}{\ell_1 v_1}.\]
04

Deriving the Quadratic Equation

Multiply both sides of the energy equation by 2 and \(m\) and simplify to get: \[v_2^2 - \frac{2 G M}{v_1 \ell_1} v_2 - \left[ v_1^2 - \frac{2 G M}{\ell_1} \right] = 0.\] This is the quadratic equation which needs to be solved for \(v_2\).
05

Solving the Quadratic Equation

Using the quadratic formula \[v_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] where \(a = 1, b = -\frac{2 G M}{v_1 \ell_1}, \text{and}\ c = - \left[ v_1^2 - \frac{2 G M}{\ell_1} \right].\) Solve for \(v_2\) and select the smaller root.
06

Calculate \(\ell_2\)

Once \(v_2\) is obtained, use the relation \( \ell_2 = \frac{\ell_1 \cdot v_1}{v_2} \) to find \(\ell_2\).
07

Calculate Other Orbital Parameters

With values of \(\ell_1\) and \(\ell_2\) known, calculate the remaining orbital parameters: \[a = \frac{1}{2} (\ell_1 + \ell_2), \ b = \sqrt{\ell_1 \ell_2}, \ T = \frac{2 \pi a b}{\ell_1 v_1}, \ e = \frac{\ell_2 - \ell_1}{\ell_2 + \ell_1}.\]
08

Implementing the Program

Write a program that prompts the user for \(\ell_1\) and \(v_1\), and implements the above steps to calculate and print \(\ell_2, v_2, T,\) and \(e\).
09

Testing the Program

Test the program with Earth's and Halley's comet values: \[\ell_1 = 1.4710 \times 10^{11} \text{m}, \ v_1 = 3.0287 \times 10^{4} \text{ms}^{-1}\] for Earth, and \[\ell_1 = 8.7830 \times 10^{10} \text{m}, v_1 = 5.4529 \times 10^{4} \text{m s}^{-1}\] for Halley's comet.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Kepler's Laws
Kepler's Laws were formulated by Johannes Kepler and are fundamental for understanding planetary motion. These laws describe how planets orbit the sun and can be summarized as follows:

1. **First Law (Law of Ellipses):** Each planet's orbit around the Sun is an ellipse with the Sun at one of the two foci.
2. **Second Law (Law of Equal Areas):** A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This implies that a planet moves faster when it is closer to the Sun and slower when it is farther from the Sun.
3. **Third Law (Harmonic Law):** The square of a planet's orbital period (T) is proportional to the cube of the semi-major axis (a) of its orbit, given by the formula \( T^2 \text{is proportional to} a^3 \).

Understanding these laws enables us to calculate various orbital parameters and understand the dynamics of celestial bodies.
Elliptical Orbits
Planets and other celestial bodies follow elliptical orbits around stars like our Sun. An ellipse is an elongated circle with two focal points, known as foci. The Sun occupies one of these foci in a planetary orbit.

The components of an ellipse include:
  • **Perihelion (\( \ell_1 \)):** The point in the orbit closest to the Sun.
  • **Aphelion (\( \ell_2 \)):** The point in the orbit farthest from the Sun.
  • **Semi-major axis (a):** Half the longest diameter of the ellipse.
  • **Semi-minor axis (b):** Half the shortest diameter of the ellipse.

Newton's laws of gravitation and motion explain why orbits are elliptical. The gravitational pull between a planet and the sun creates a balance of forces that results in this elliptical shape. Devices like telescopes and mathematical models based on Kepler's Laws and Newtonian physics help in predicting and understanding these orbits.
Energy Conservation
In a planetary orbit, energy conservation is crucial. The total mechanical energy (E) in an orbit is the sum of kinetic energy (K) and potential energy (U).

1. **Kinetic Energy (K):** Given by \( K = \frac{1}{2} mv^2 \), where m is the mass of the planet, and v is its velocity.
2. **Gravitational Potential Energy (U):** Given by \( U = -G \frac{mM}{r} \), where G is the gravitational constant, M is the Sun's mass, and r is the distance from the Sun.

The total energy equation therefore is:
\[ E = \frac{1}{2} mv^2 - G \frac{mM}{r} \]

Because energy is conserved, the total energy at any two points in the orbit (like perihelion and aphelion) must be equal. This principle allows us to derive the quadratic equation used to find the velocity at aphelion (\( v_2 \)). By solving this equation, we can understand and predict the orbital dynamics.
Orbital Parameters Calculation
Orbital parameters provide details about the shape and duration of an orbit. These parameters are crucial for predicting a planet’s position and movement over time. Using initial conditions like the perihelion distance (\( \ell_1 \)) and velocity (\( v_1 \)), we can compute:

*1. **Aphelion distance (\( \ell_2 \)):***

From Kepler's second law:
\[ \ell_2 = \frac{\ell_1 \cdot v_1}{v_2} \]

*2. **Semi-major axis (a):***

The semi-major axis is calculated as:
\[ a = \frac{1}{2} ( \ell_1 + \ell_2 ) \]

*3. **Semi-minor axis (b):***

Given by:
\[ b = \sqrt{\ell_1 \times \ell_2 } \]

*4. **Orbital period (T):***

The time it takes for a planet to complete one full orbit is:
\[ T = \frac{2 \pi a b}{\ell_1 v_1 } \]

*5. **Orbital eccentricity (e):***

This measures the deviation from a perfect circle:
\[ e = \frac{\ell_2 - \ell_1}{\ell_2 + \ell_1 } \]

Understanding these parameters is essential for tasks like mission planning and predicting celestial events. They encapsulate the essence of a planet’s orbital characteristics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The semi-empirical mass formula In nuclear physics, the semi-empirical mass formula is a formula for calculating the approximate nuclear binding energy \(B\) of an atomic nucleus with atomic number \(Z\) and mass number \(A\) : $$ B=a_{1} A-a_{2} A^{2 / 3}-a_{3} \frac{Z^{2}}{A^{1 / 3}}-a_{4} \frac{(A-2 Z)^{2}}{A}+\frac{a_{5}}{A^{1 / 2}} $$ where, in units of millions of electron volts, the constants are \(a_{1}=15.8, a_{2}=18.3\), \(a_{3}=0.714, a_{4}=23.2\), and $$ a_{5}= \begin{cases}0 & \text { if } A \text { is odd, } \\ 12.0 & \text { if } A \text { and } Z \text { are both even, } \\ -12.0 & \text { if } A \text { is even and } Z \text { is odd. }\end{cases} $$ a) Write a program that takes as its input the values of \(A\) and \(Z\), and prints out the binding energy for the corresponding atom. Use your program to find the binding energy of an atom with \(A=58\) and \(Z=28\). (Hint: The correct answer is around \(500 \mathrm{MeV}\) ) b) Modify your program to print out not the total binding energy B, but the binding energy per nucleon, which is \(B / A\). c) Now modify your program so that it takes as input just a single value of the atomic number \(Z\) and then goes through all values of \(A\) from \(A=Z\) to \(A=3 Z\), to find the one that has the largest binding energy per nucleon. This is the most stable nucleus with the given atomic number. Have your program print out the value of \(A\) for this most stable nucleus and the value of the binding energy per nucleon d) Modify your program again so that, instead of taking \(Z\) as input, it runs through all values of \(Z\) from 1 to 100 and prints out the most stable value of \(A\) for each one. At what value of \(Z\) does the maximum binding energy per nucleon occur? (The true answer, in real life, is \(Z=28\), which is nickel.)

Quantum potential step A well-known quantum mechanics problem involves a particle of mass \(m\) that encounters a one-dimensional potential step, like this: The particle with initial kinetic energy \(E\) and wavevector \(k_{1}=\sqrt{2 m E} / \hbar\) enters from the left and encounters a sudden jump in potential energy of height \(V\) at position \(x=0\). By solving the Schrödinger equation, one can show that when \(E>V\) the particle may either (a) pass the step, in which case it has a lower kinetic energy of \(E-V\) on the other side and a correspondingly smaller wavevector of \(k_{2}=\sqrt{2 m(E-V)} / \hbar\), or \((b)\) it may be reflected, keeping all of its kinetic energy and an unchanged wavevector but moving in the opposite direction. The probabilities \(T\) and \(R\) for transmission and reflection are given by $$ T=\frac{4 k_{1} k_{2}}{\left(k_{1}+k_{2}\right)^{2}}, \quad R=\left(\frac{k_{1}-k_{2}}{k_{1}+k_{2}}\right)^{2} . $$ Suppose we have a particle with mass equal to the electron mass \(m=9.11 \times\) \(10^{-31} \mathrm{~kg}\) and energy \(10 \mathrm{eV}\) encountering a potential step of height \(9 \mathrm{eV}\). Write a Python program to compute and print out the transmission and reflection probabilities using the formulas above.

Catalan numbers The Catalan numbers \(C_{n}\) are a sequence of integers \(1,1,2,5,14,42,132 \ldots\) that play an important role in quantum mechanics and the theory of disordered systems. (They were central to Eugene Wigner's proof of the so- called semicircle law.) They are given by $$ C_{0}=1, \quad C_{n+1}=\frac{4 n+2}{n+2} C_{n} . $$ Write a program that prints in increasing order all Catalan numbers less than or equal to one billion.

A spaceship travels from Earth in a straight line at relativistic speed \(v\) to another planet \(x\) light years away. Write a program to ask the user for the value of \(x\) and the speed \(v\) as a fraction of the speed of light \(c\), then print out the time in years that the spaceship takes to reach its destination (a) in the rest frame of an observer on Earth and (b) as perceived by a passenger on board the ship. Use your program to calculate the answers for a planet 10 light years away with \(v=0.99 c\).

Altitude of a satellite A satellite is to be launched into a circular orbit around the Earth so that it orbits the planet once every \(T\) seconds. a) Show that the altitude \(h\) above the Earth's surface that the satellite must have is $$ h=\left(\frac{G M T^{2}}{4 \pi^{2}}\right)^{1 / 3}-R $$ where \(G=6.67 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2}\) is Newton's gravitational constant, \(M=\) \(5.97 \times 10^{24} \mathrm{~kg}\) is the mass of the Earth, and \(R=6371 \mathrm{~km}\) is its radius. b) Write a program that asks the user to enter the desired value of \(T\) and then calculates and prints out the correct altitude in meters. c) Use your program to calculate the altitudes of satellites that orbit the Earth once a day (so-called "geosynchronous" orbit), once every 90 minutes, and once every 45 minutes. What do you conclude from the last of these calculations? d) Technically a geosynchronous satellite is one that orbits the Earth once per sidereal day, which is \(23.93\) hours, not 24 hours. Why is this? And how much difference will it make to the altitude of the satellite?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Solid State Physics

Read Explanation

Work Energy and Power

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Energy Physics

Read Explanation

Waves Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free