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Chapter 8: Problem 29

What would become of the earth's orbit (which you may consider to be a circle) if half of the sun's mass were suddenly to disappear? Would the earth remain bound to the sun? [Hints: Consider what happens to the earth's KE and PE at the moment of the great disappearance. The virial theorem for the circular orbit (Problem 4.41) helps with this one.] Treat the sun (or what remains of it) as fixed.

Short Answer

Expert verified
Earth's orbit becomes elliptical, but it remains bound to the sun.

Step by step solution

01

Understanding the Earth's Initial Orbit

Initially, the earth orbits the sun in a nearly circular path. The gravitational force provides the necessary centripetal force for this circular motion. According to the virial theorem, for a stable circular orbit, the kinetic energy (KE) is related to the gravitational potential energy (PE) by \( KE = -\frac{1}{2} PE \).
02

Analyze the Changes in Gravitational Force

When half of the sun's mass suddenly disappears, the gravitational force experienced by the earth is halved. The new gravitational force \( F' \) becomes \( F' = \frac{G \cdot (M/2) \cdot m}{r^2} \), where \( M \) is the original mass of the sun, \( m \) is the earth's mass, and \( r \) is the radius of the orbit.
03

Change in Potential Energy (PE')

The potential energy in the gravitational field is given by \( PE = -\frac{G \cdot M \cdot m}{r} \). After the mass disappears, the potential energy becomes \( PE' = -\frac{G \cdot (M/2) \cdot m}{r} = \frac{1}{2} PE \).
04

Conservation of Kinetic Energy (KE)

The kinetic energy of the earth does not change instantly when the mass of the sun changes because KE depends on the velocity of the earth which remains constant immediately after mass reduction. Thus, immediately after the disappearance, \( KE' = KE \).
05

Determine the New Nature of Earth's Orbit

With the sudden loss of mass, the gravitational attraction drops, affecting the force balance necessary for circular motion. As the centripetal force decreases and KE remains initially unchanged, the orbit becomes elliptical, with the earth possessing greater kinetic energy than required for the new, smaller gravitational pull, leading the earth to move away from the sun.

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy (PE) is the energy an object possesses due to its position in a gravitational field. In the context of celestial bodies, it is essential for understanding how objects like planets and stars interact. For a planet orbiting nearly circular around the sun, the gravitational potential energy is determined by the equation: \[ PE = -\frac{G \cdot M \cdot m}{r} \]where:- \( G \) is the gravitational constant,- \( M \) is the mass of the sun,- \( m \) is the mass of the Earth,- \( r \) is the distance between the Earth and the Sun.This formula shows that potential energy is negative, indicating that an energy input would be needed to separate the two bodies. After half of the sun's mass disappears, the gravitational potential energy reduces to half its original value, aligning with the new mass \( M/2 \). This change is crucial because it alters the force that maintains the Earth's orbit.
Kinetic Energy
Kinetic Energy (KE) refers to the energy an object possesses due to its motion. In the scenario of Earth's orbit around the sun, the kinetic energy is derived from Earth's velocity as it moves through space. The formula for kinetic energy is:\[ KE = \frac{1}{2} mv^2 \]where:- \( m \) is the mass of the Earth,- \( v \) is the velocity of the Earth in its orbit.Immediately after a drastic change like the reduction of the Sun’s mass, Earth's velocity remains constant for an instant, meaning its kinetic energy stays the same irrespective of the sun's mass change. This unchanging kinetic energy becomes significant when the gravitational force diminishes. It results in an orbit shift because the remaining gravitational pull isn't enough for a circular path, causing the Earth to transition into a more elliptical orbit.
Virial Theorem
The virial theorem is a key principle in astrophysics that relates the average total kinetic energy and the average total potential energy of a stable, bound system in equilibrium. For planets in stable orbit, the theorem is expressed as:\[ 2 \langle KE \rangle = -\langle PE \rangle \]Simplifying, for a circular orbit:\[ KE = -\frac{1}{2} PE \]This relationship implies that, in a stable orbit, the planet's kinetic energy is half the absolute value of its potential energy. When half of the sun’s mass vanishes, this delicate balance is disrupted. The gravitational potential energy drops, but the kinetic energy does not instantly change, disturbing the initial equilibrium. The Virial theorem explains why Earth would not maintain its circular orbit when the gravitational potential energy drops; the balance required for a circular trajectory is lost, causing the Earth to start moving in an elliptical path.

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Most popular questions from this chapter

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.

Verify that the positions of two particles can be written in terms of the CM and relative positions as \(\mathbf{r}_{1}=\mathbf{R}+m_{2} \mathbf{r} /\mathbf{M}\) and \(\mathbf{r}_{2}=\mathbf{R}-m_{1} \mathbf{r} / M .\) Hence confirm that the total \(\mathrm{KE}\) of the two particles can be expressed as \(T=\frac{1}{2} \mathbf{M} \dot{\mathbf{R}}^{2}+\frac{1}{2} \mu \dot{\mathbf{r}}^{2},\) where \(\mu\) denotes the reduced mass \(\mu=\mathbf{m}_{1} \mathbf{m}_{2} / \mathbf{M}.\)

Consider two particles of equal masses, \(m_{1}=m_{2},\) attached to each other by a light straight spring (force constant \(k\), natural length \(L\) ) and free to slide over a frictionless horizontal table. (a) Write down the Lagrangian in terms of the coordinates \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\), and rewrite it in terms of the \(\mathrm{CM}\) and relative positions, \(\mathbf{R}\) and \(\mathbf{r},\) using polar coordinates \((r, \phi)\) for \(\mathbf{r} .\) (b) Write down and solve the Lagrange equations for the CM coordinates \(X, Y\). (c) Write down the Lagrange equations for \(r\) and A. Solve these for the two special cases that \(r\) remains constant and that \(\phi\) remains constant. Describe the corresponding motions. In particular, show that the frequency of oscillations in the second case is \(\omega=\sqrt{2 k / m_{1}}\).

Consider a particle of reduced mass \(\mu\) orbiting in a central force with \(U=k r^{n}\) where \(k n>0\). (a) Explain what the condition \(k n>0\) tells us about the force. Sketch the effective potential energy \(U_{\mathrm{eff}}\) for the cases that \(n=2,-1,\) and \(-3 .\) (b) Find the radius at which the particle (with given angular momentum \(\ell\) ) can orbit at a fixed radius. For what values of \(n\) is this circular orbit stable? Do your sketches confirm this conclusion? (c) For the stable case, show that the period of small oscillations about the circular orbit is \(\tau_{\mathrm{osc}}=\tau_{\mathrm{orb}} / \sqrt{n+2} .\) Argue that if \(\sqrt{n+2}\) is a rational number, these orbits are closed. Sketch them for the cases that \(n=2,-1,\) and 7.

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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