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Chapter 8: Problem 7

(a) Using elementary Newtonian mechanics find the period of a mass \(m_{1}\) in a circular orbit of radius \(r\) around a fixed mass \(m_{2} .\) (b) Using the separation into \(\mathrm{CM}\) and relative motions, find the corresponding period for the case that \(m_{2}\) is not fixed and the masses circle each other a constant distance \(r\) apart. Discuss the limit of this result if \(m_{2} \rightarrow \infty\). (c) What would be the orbital period if the earth were replaced by a star of mass equal to the solar mass, in a circular orbit, with the distance between the sun and star equal to the present earth-sun distance? (The mass of the sun is more than 300,000 times that of the earth.)

Short Answer

Expert verified
(a) \(T = 2 \pi \sqrt{\frac{r^3}{G m_2}}\). (b) \(T = 2 \pi \sqrt{\frac{r^3}{G (m_1 + m_2)}}\). (c) Nearly Earth's orbit period.

Step by step solution

01

Understand the Problem

We need to analyze two scenarios: (a) a mass orbiting a fixed central mass, and (b) two masses orbiting their common center of mass. Finally, we will calculate an example using solar mass.
02

Analyze Fixed Mass System (Part a)

For mass \(m_1\) in a circular orbit of radius \(r\) around a fixed mass \(m_2\), use Newton’s second law and the gravitational force equation. The gravitational force provides the necessary centripetal force: \[ F = \frac{G m_1 m_2}{r^2} = \frac{m_1 v^2}{r} \] Solving for the orbital velocity \(v\): \[ v = \sqrt{\frac{G m_2}{r}} \] The period \(T\) is the time to complete one orbit: \[ T = \frac{2 \pi r}{v} = 2 \pi \sqrt{\frac{r^3}{G m_2}} \]
03

Analyze Relativistic System (Part b)

For two masses \(m_1\) and \(m_2\) orbiting each other, we use the concept of reduced mass \(\mu = \frac{m_1 m_2}{m_1 + m_2}\). The effective central force on each body is: \[ F = \frac{G m_1 m_2}{r^2} \] The center of mass system period is described as: \[ T = 2 \pi \sqrt{\frac{r^3}{G (m_1 + m_2)}} \]
04

Limiting Case of Infinite Mass (Part b)

In the limit as \(m_2 \to \infty\), the system resembles the fixed mass case, hence: \[ T = 2 \pi \sqrt{\frac{r^3}{G m_2}} \] This is identical to the result from part (a), confirming the consistency of the model when one mass is significantly larger than the other.
05

Orbital Period with Solar Mass (Part c)

Given the Earth replaced by a Star of Solar mass and using the sun as the other mass, the center of mass stays nearly at the Sun’s center. The period is: \[ T = 2 \pi \sqrt{\frac{r^3}{G (m_{\text{Sun}} + m_{\text{Star}})}} \] With \(m_{\text{Star}} = m_{\text{Sun}} \) and distance as Earth’s current distance from the sun, use: \[ T_T = 2 \pi \sqrt{\frac{r^3}{G 2m_{\text{Sun}}}} \]
06

Calculate with Given Values

If \(r = 1 \text{ AU}\), and considering \(m_{\text{Sun}}\) roughly more than 300,000 times Earth’s mass, the period \(T\) is nearly the same as Earth's current orbital period around the sun due to the minor mass of the Earth contributing little to the center of mass location.

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Key Concepts

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

Newtonian Mechanics
Newtonian mechanics forms the foundation of classical physics, explaining phenomena in our everyday macroscopic world. Central to this theory are Newton's Laws of Motion, particularly the first and second laws, essential for understanding motion in orbital mechanics. The first law, sometimes summarized as the law of inertia, states that an object will remain at rest or in uniform motion in a straight line unless acted upon by a force. This is important when considering objects in space, where the absence of significant resistive forces allows objects to maintain their paths unless another force intervenes.

The second law expresses the relationship between force, mass, and acceleration: \[ F = m imes a \].

In orbital mechanics scenarios, such as calculating the period of an orbiting mass, Newton's second law allows us to relate gravitational force (which acts as the centripetal force) to the movement of celestial bodies. Understanding that gravitational interactions can be considered forces helps explain the elegant and predictable orbits we observe in our solar system.
Gravitational Force
Gravitational force is the attractive force between two masses. It is a fundamental force of nature, acting over long distances, and is crucial for understanding the motion of planets, stars, and other celestial objects in space. The magnitude of this force is described by Newton's Law of Gravitation:\[F = \frac{G m_1 m_2}{r^2}\],where:
  • \( F \) is the gravitational force,
  • \( G \) is the gravitational constant,
  • \( m_1 \) and \( m_2 \) are the masses of the objects,
  • \( r \) is the distance between the centers of the two masses.

This formula shows us that gravitational force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between them. In orbital mechanics, this gravitational force is what keeps planets in orbit around the sun or the moon orbiting Earth. Importantly, the gravitational force also provides the necessary centripetal force to keep the orbiting body moving in a circular path.
Centripetal Force
Centripetal force is the "center-seeking" force that keeps an object moving in a circular path. In the context of orbital mechanics, the gravitational force acts as the centripetal force required to keep a planet, moon, or satellite in its orbit. For an object of mass \( m \) moving with velocity \( v \) in a circle of radius \( r \), the necessary centripetal force is given by:\[F_c = \frac{m v^2}{r}.\]

This force is always directed towards the center of the circle. In our planetary systems, it is the gravitational force that acts as this centripetal force. By equating the gravitational force to the centripetal force, we can solve for orbital characteristics such as velocity and period, aiding in our understanding of how celestial bodies maintain their orbits without spiraling into or flying away from one another.
Center of Mass
The center of mass is a point representing the average position of the mass distribution in a system. In systems of orbiting bodies, this concept is pivotal. For a two-body system like the Earth and the Sun, both bodies actually orbit around their common center of mass, not just one around the other. The center of mass \( R \)of two masses \( m_1 \) and \( m_2 \) can be calculated using:\[R = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2},\]where \( x_1 \) and \( x_2 \) are the positions of the two masses.

If \( m_2 \) is significantly larger than \( m_1 \), the center of mass will be closer to \( m_2 \), often lying inside of it, which is typical for planetary systems where the star is much more massive than orbiting planets. This concept helps in visualizing the orbital paths and simplifies the understanding of orbital mechanics in systems with multiple bodies.

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Most popular questions from this chapter

Consider two particles interacting by a Hooke's law potential energy, \(U=\frac{1}{2} k r^{2},\) where \(\mathbf{r}\) is their relative position \(\mathbf{r}=\mathbf{r}_{1}-\mathbf{r}_{2},\) and subject to no external forces. Show that \(\mathbf{r}(t)\) describes an ellipse. Hence show that both particles move on similar ellipses around their common CM. [This is surprisingly awkward. Perhaps the simplest procedure is to choose the \(x y\) plane as the plane of the orbit and then solve the equation of motion (8.15) for \(x\) and \(y .\) Your solution will have the form \(x=A \cos \omega t+B \sin \omega t,\) with a similar expression for \(y .\) If you solve these for sin \(\omega t\) and \(\cos \omega t\) and remember that \(\sin ^{2}+\cos ^{2}=1,\) you can put the orbital equation in the form \(a x^{2}+2 b x y+c y^{2}=k\) where \(k\) is a positive constant. Now invoke the standard result that if \(a\) and \(c\) are positive and \(a c>b^{2}\) this equation defines an ellipse.]

Two particles whose reduced mass is \(\mu\) interact via a potential energy \(U=\frac{1}{2} k r^{2},\) where \(r\) is the distance between them. (a) Make a sketch showing \(U(r),\) the centrifugal potential energy \(U_{\mathrm{cf}}(r)\) and the effective potential energy \(U_{\text {eff }}(r) .\) (Treat the angular momentum \(\ell\) as a known, fixed constant.) (b) Find the "equilibrium" separation \(r_{\mathrm{o}},\) the distance at which the two particles can circle each other with constant \(r .\left[\text { Hint: This requires that } d U_{\text {eff }} / d r \text { be zero. }\right]\left(\text { c) By making a Taylor expansion of } U_{\text {eff }}(r)\right.\) about the equilibrium point \(r_{\mathrm{o}}\) and neglecting all terms in \(\left(r-r_{\mathrm{o}}\right)^{3}\) and higher, find the frequency of small oscillations about the circular orbit if the particles are disturbed a little from the separation \(r_{\mathrm{o}}.\)

An earth satellite is observed at perigee to be \(250 \mathrm{km}\) above the earth's surface and traveling at about \(8500 \mathrm{m} / \mathrm{s}\). Find the eccentricity of its orbit and its height above the earth at apogee. [Hint: The earth's radius is \(R_{e} \approx 6.4 \times 10^{6} \mathrm{m} .\) You will also need to know \(G M_{\mathrm{e}},\) but you can find this if you remember that \(\left.G M_{\mathrm{e}} / R_{\mathrm{e}}^{2}=g .\right]\)

Two particles of equal masses \(m_{1}=m_{2}\) move on a frictionless horizontal surface in the vicinity of a fixed force center, with potential energies \(U_{1}=\frac{1}{2} k r_{1}^{2}\) and \(U_{2}=\frac{1}{2} k r_{2}^{2} .\) In addition, they interact with each other via a potential energy \(U_{12}=\frac{1}{2} \alpha k r^{2},\) where \(r\) is the distance between them and \(\alpha\) and \(k\) are positive constants. (a) Find the Lagrangian in terms of the CM position \(\mathbf{R}\) and the relative position \(\mathbf{r}=\mathbf{r}_{1}-\mathbf{r}_{2} \cdot(\mathbf{b})\) Write down and solve the Lagrange equations for the \(\mathrm{CM}\) and relative coordinates \(X, Y\) and \(x, y .\) Describe the motion.

At time \(t_{\mathrm{o}}\) a comet is observed at radius \(r_{\mathrm{o}}\) traveling with speed \(v_{\mathrm{o}}\) at an acute angle \(\alpha\) to the line from the comet to the sun. Put the sun at the origin \(O\), with the comet on the \(x\) axis (at \(t_{\mathrm{o}}\) ) and its orbit in the \(x y\) plane, and then show how you could calculate the parameters of the orbital equation in the form \(r=c /[1+\epsilon \cos (\phi-\delta)] .\) Do so for the case that \(r_{0}=1.0 \times 10^{11} \mathrm{m}, v_{\mathrm{o}}=45 \mathrm{km} / \mathrm{s},\) and \(\left.\alpha=50 \text { degrees. [The sun's mass is about } 2.0 \times 10^{30} \mathrm{kg}, \text { and } G=6.7 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} .\right]\)
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