Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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?

Short Answer

Expert verified
The "constant" in Kepler's third law varies by up to about 0.1% among the planets.

Step by step solution

01

Understanding Kepler's Original Third Law

Kepler's third law states that the square of the orbital period \( \tau \) of a planet is proportional to the cube of the semi-major axis \( a \) of its orbit. The original form is given by \[ \tau^2 = \left( \frac{4 \pi^2}{GM_s} \right) a^3 \] where \( M_s \) is the mass of the Sun and \( G \) is the gravitational constant.
02

Modify Kepler's Third Law to Account for Planet Mass

To account for the actual mass of the planets, \( m \), we modify the central gravitational force equation that led to Kepler's law. The modified equation then becomes: \[ \tau^2 = \left( \frac{4 \pi^2}{G(M_s + m)} \right) a^3 \] where the term \( (M_s + m) \) accounts for the combined gravitational attraction of both the Sun and the planet.
03

Calculate the Relative Difference in the Constant

Calculate the relative difference in the constant of proportionality for different planet masses. For Jupiter, the largest planet, \[ \frac{m}{M_s} = \frac{2 \times 10^{27}}{2 \times 10^{30}} = 0.001 \] This implies that the constant for Jupiter is actually 0.1% smaller.
04

Estimate Variation Among Planets

For the majority of planets, which have significantly smaller masses than Jupiter, the variation in the constant \[ \frac{m}{M_s} \] will be less than 0.1%. However, assuming that Jupiter has the largest effect, the variation among all planets' constants can be expected to be up to about 0.1%.

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.

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.

Orbital Mechanics
Orbital mechanics is the field of physics that focuses on the motion of objects in space. Specifically, it studies how celestial bodies, such as planets, moons, and satellites, move around each other under the influence of gravity. This concept is crucial for understanding how different bodies interact within a solar system. A key part of orbital mechanics is understanding the laws that govern these motions. One such law is Kepler's Third Law, which deals with the relationship between the orbital period of a planet and its distance from the Sun. In essence, orbital mechanics helps explain:
  • Why planets follow specific paths called orbits around the Sun.
  • How adjustments in mass and distance affect these orbits.
  • The relevance of gravitational forces in determining the shape and timing of these orbits.
In our solar system, planetary orbits are usually elliptical, with the Sun at one focus of the ellipse. Understanding orbital mechanics is fundamental not only for astronomy but also for mission planning in space exploration.
It provides insights into how spacecraft can travel between planetary bodies while conserving momentum and energy.
Gravitational Force
Gravitational force is what keeps the planets in orbit around the Sun. It is a fundamental force that acts between any two masses in the universe. According to Newton's law of universal gravitation, this force is proportional to the product of the two masses and inversely proportional to the square of the distance between their centers:\[F = G \frac{M_1 M_2}{r^2}\]Where:
  • \(F\) is the gravitational force.
  • \(G\) is the gravitational constant.
  • \(M_1\) and \(M_2\) are the masses of the two bodies.
  • \(r\) is the distance between the centers of the two masses.
Gravitational force is what keeps the planets in their elliptical orbits rather than moving off into space. Despite being a weak force on a small scale—for example, you can easily lift a book despite the Earth's gravitational pull—it is the dominant force on the scale of planets and stars. In the context of Kepler's Third Law, adjusting for the gravitational pull from both the Sun and the planet allows for a more accurate description of the planets' orbital mechanics.
It can slightly alter the proportionality constant, as seen from considering the mass of large planets like Jupiter.
Orbital Period
The orbital period is the time a planet takes to complete one full orbit around the Sun. In terms of days or years, this measurement provides crucial information about the speed and path the planet follows. Kepler's Third Law relates the square of the orbital period \( \tau \) to the cube of the average distance \( a \) from the Sun for planets in the solar system:\[\tau^2 = \left( \frac{4 \pi^2}{G(M_s + m)} \right) a^3\]Here:
  • \(\tau\) is the orbital period.
  • \(a\) is the semi-major axis of the orbit, or the average distance to the Sun.
  • \(M_s\) and \(m\) are the masses of the Sun and the planet, respectively.
This formula highlights that more massive planets exert a significant gravitational pull, affecting the orbital period.
However, for smaller planets, their masses are often negligible compared to the Sun, which is why the classic form assumes \(M_s\) only.
Understanding the orbital period is vital for predicting planetary positions and planning space missions.

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

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.

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.

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.]

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.

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 \(1 / r^{2}.\)

See all solutions

Recommended explanations on Physics Textbooks

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