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Show that the validity of Kepler's first two laws for any body orbiting the sun implies that the force (assumed conservative) of the sun on any body is central and proportional to \(1 / r^{2}.\)

Short Answer

Expert verified
Kepler's laws imply that the Sun's force is central and follows an inverse-square law.

Step by step solution

01

Recap Kepler's First Law

Kepler's First Law states that planets move in elliptical orbits with the Sun at one focus. This implies that the path traced by an orbiting body is elliptical, indicating a central force exerted by the Sun.
02

Define the Central Force

A central force is one where the force vector points towards a fixed point (the center) and its magnitude depends only on the distance from the center. Mathematically, this is expressed as \(\vec{F} = F(r)\hat{r}\), where \(\hat{r}\) is the unit vector pointing from the center towards the body.
03

Recap Kepler's Second Law

Kepler's Second Law states that a line segment joining a planet and the Sun sweeps out equal areas in equal times. This law implies the conservation of angular momentum and indicates that the force does not exert any torque, hence confirming the force is central.
04

Use of Angular Momentum Conservation

Given that the torque is zero (so force is central), the angular momentum \(\vec{L} = \vec{r} \times \vec{p}\) is conserved. \(\vec{L}\) being constant over time implies that the motion occurs in a plane and the distance \(r\) indirectly controls the velocity \(v\).
05

Recognizing Inverse Square Law

Using Newton's second law and the conservation of angular momentum \(r^2 \frac{d\theta}{dt} = \text{constant}\) from Kepler's laws, derive that the force must be proportional to \(\frac{1}{r^2}\). This uses the identity \(v^2 = \frac{G M}{r}\) for circular motion approximation or integrals of motion in elliptic paths.
06

Conclude Based on Derived Proportions

Since the force is central (as derived from the constancy of \(r^2 d\theta/dt\)) and the inverse square law (\(F \propto \frac{1}{r^2}\)) holds, it is shown that the sun's force is a radial force directly towards the center, agreeing with the gravitational formula.

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Key Concepts

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

Central Force
A central force is a fundamental concept in understanding planetary motion and Kepler's Laws. Imagine a force which always points toward a specific, unchanging point, like how gravity pulls every falling apple toward the Earth’s core. In astronomy, this central point is often the center of a massive body, such as the Sun. For a force to be considered central, its direction must consistently point toward this center, and its strength, or magnitude, only relies on the distance from the center. This is mathematically expressed in vector form as \(\vec{F} = F(r)\hat{r}\), where \(\hat{r}\) stands for the unit vector pointing from the center to the object.

Central forces simplify our understanding of planetary motion because they explain why orbits are stable and often elliptical. When Kepler's First Law mentions that planets orbit in ellipses with the Sun at one focus, it implies the presence of a central force because this shape is consistent with a force directed towards that central focal point. This central nature ensures that no matter where the planet is in its orbit, the gravitational pull is always toward the Sun, maintaining the orbit's elliptical shape. Understanding this helps clarify the invisible yet consistent powers governing celestial motions.
Angular Momentum Conservation
Angular momentum conservation is a key aspect of celestial mechanics, directly connected to Kepler's Second Law. According to this law, a planet moves faster when it is closer to the Sun and slower when it is farther away. The crucial takeaway is that the area swept by the planet in a certain timeframe remains constant throughout its orbit.

This constant area sweep is explained by the conservation of angular momentum. Angular momentum \(\vec{L}\) in physics is defined as \(\vec{L} = \vec{r} \times \vec{p}\), where \(\vec{r}\) is the position vector and \(\vec{p}\) is the momentum of the planet. Since no external torques act on the orbiting planet, because the force is central, its angular momentum remains unchanged over time.

Practically speaking, angular momentum conservation means that as planets come closer to the Sun, they gain speed, and as they move away, they slow down. This behavior supports Kepler's observation about equal areas being swept in equal times, underpinning the non-variable nature of angular momentum in celestial orbits.
Inverse Square Law
The inverse square law is crucial in understanding why celestial bodies follow curved paths around larger masses like the Sun. According to this principle, the strength of gravitational forces diminishes with the square of the distance between the two objects. Simply put, if the separation between two bodies doubles, the gravitational force becomes one-fourth as strong.

This law can be mathematically expressed as \(F \propto \frac{1}{r^2}\), where \(F\) is the gravitational force and \(r\) is the distance between the objects. This relationship is fundamental to demonstrating the elliptical orbits predicted by Kepler's First Law and observed in the motion of planets.
  • Kepler's laws, when combined with Newton’s laws, reveal how an inverse square law of attraction leads to elliptical orbits for planets around the Sun.
  • The conservation of angular momentum with this law helps explain why planets speed up or slow down at different points in their orbits.
This law, therefore, becomes a key factor in deriving the radial nature and strength of the gravitational force the Sun exerts on planets, keeping celestial bodies in stable and predictable paths around the solar system.

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Most popular questions from this chapter

(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.)

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 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.]

In deriving Kepler's third law (8.55) we made an approximation based on the fact that the sun's mass \(M_{\mathrm{s}}\) is much greater than that of the planet \(m .\) Show that the law should actually read \(\tau^{2}=\left[4 \pi^{2} / G\left(M_{\mathrm{s}}+m\right)\right] a^{3},\) and hence that the "constant" of proportionality is actually a little different for different planets. Given that the mass of the heaviest planet (Jupiter) is about \(2 \times 10^{27} \mathrm{kg}\), while \(M_{\mathrm{s}}\) is about \(2 \times 10^{30} \mathrm{kg}\) (and some planets have masses several orders of magnitude less than Jupiter), by what percent would you expect the "constant" in Kepler's third law to vary among the planets?

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 \(\left.G M_{\mathrm{e}} / R_{\mathrm{e}}^{2}=g .\right]\)

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