Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 1

A double star is formed of two components, each with mass equal to that of the Sun. The distance between them is \(1 \mathrm{AU}\) (see Chapter 4 , Problem 2). What is the orbital period?

Short Answer

Expert verified
Answer: The orbital period of the double star system is approximately 1.36 years.

Step by step solution

01

Relevant Information

Given information: Mass of each star, \(M_1 = M_2 = M_{\odot}\) Distance between the stars, \(a = 1 \, \mathrm{AU}\) (since it's the semi-major axis of their orbits) Kepler's Third Law: \(\frac{T^2}{a^3} = \frac{4\pi^2}{G(M_1 + M_2)}\) Let's use this information to find the orbital period \(T\).
02

Convert Units

To make the calculations easier, we'll first convert the distance from AU to meters and mass of the stars from solar masses to kg. 1 AU = 1.496 × 10^11 meters 1 solar mass = \(1.989 × 10^{30}\) kg So, \(a = 1.496 × 10^11 \, \mathrm{m}\) and \(M_1 = M_2 = 1.989 × 10^{30} \, \mathrm{kg}\).
03

Substitute Values in Kepler's Third Law

Now we can substitute the given values in Kepler's Third Law: \(\frac{T^2}{(1.496 × 10^{11} \, \mathrm{m})^3} = \frac{4\pi^2}{G(1.989 × 10^{30} \, \mathrm{kg} + 1.989 × 10^{30} \, \mathrm{kg})}\)
04

Solve for the Orbital Period

Now we need to solve for the orbital period, \(T\). First, we can simplify the equation a bit by combining the masses: \(\frac{T^2}{(1.496 × 10^{11})^3} = \frac{4\pi^2}{G(2 × 1.989 × 10^{30})}\) Next, multiply both sides by the denominator on the right side, and take the square root of both sides to get \(T\): \(T = \sqrt{\frac{4\pi^2(1.496 × 10^{11})^3}{G(2 × 1.989 × 10^{30})}}\) Now, plug in the gravitational constant, \(G = 6.674 × 10^{-11} \, m^3 \, kg^{-1} \, s^{-2}\), and calculate the result: \(T = \sqrt{\frac{4\pi^2(1.496 × 10^{11})^3}{6.674 × 10^{-11}(2 × 1.989 × 10^{30})}} \approx 4.30 \times 10^7 \, \mathrm{s}\)
05

Convert Back to Time Units

Finally, convert the orbital period from seconds to a more intuitive time unit like years: \(\frac{4.30 \times 10^7 \, \mathrm{s}}{3.154 × 10^7 \, \mathrm{s/year}} \approx 1.36\) years So, the orbital period of the double star system is approximately 1.36 years.

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!

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

At low energies, protons and neutrons behave roughly like hard spheres of equal mass and radius about \(1.3 \times 10^{-14} \mathrm{~m} .\) A parallel beam of neutrons, with a flux of \(3 \times 10^{10}\) neutrons \(\mathrm{m}^{-2} \mathrm{~s}^{-1}\), strikes a target containing \(4 \times 10^{22}\) protons. A circular detector of radius \(20 \mathrm{~mm}\) is placed \(0.7 \mathrm{~m}\) from the target, in a direction making an angle of \(30^{\circ}\) to the beam direction. Calculate the rate of detection of neutrons, and of protons.

The parallax of a star (the angle subtended at the star by the radius of the Earth's orbit) is \(\varpi\). The star's position is observed to oscillate with angular amplitude \(\alpha\) and period \(\tau\). If the oscillation is interpreted as being due to the existence of a planet moving in a circular orbit around the star, show that its mass \(m_{1}\) is given by \(\frac{m_{1}}{M_{\mathrm{S}}}=\frac{\alpha}{\varpi}\left(\frac{M \tau_{\mathrm{E}}}{M_{\mathrm{S}} \tau}\right)^{2 / 3}\) where \(M\) is the total mass of star plus planet, \(M_{\mathrm{S}}\) is the Sun's mass, and \(\tau_{\mathrm{E}}=1\) year. Evaluate the mass \(m_{1}\) if \(M=0.25 M_{\mathrm{S}}, \tau=16\) years, \(\varpi=0.5^{\prime \prime}\) and \(\alpha=0.01^{\prime \prime} .\) What conclusion can be drawn without making the assumption that the orbit is circular?

Obtain the relation between the total kinetic energy in the CM and Lab frames. Discuss the limiting cases of very large and very small mass for the target.

Write down the equations of motion for a pair of charged particles of equal masses \(m\), and of charges \(q\) and \(-q\), in a uniform electric field \(\boldsymbol{E}\). Show that the field does not affect the motion of the centre of mass. Suppose that the particles are moving in circular orbits with angular velocity \(\omega\) in planes parallel to the \(x y\)-plane, with \(\boldsymbol{E}\) in the \(z\)-direction. Write the equations in a frame rotating with angular velocity \(\omega\), and hence find the separation of the planes.

Two particles of masses \(m_{1}\) and \(m_{2}\) are attached to the ends of a light spring. The natural length of the spring is \(l\), and its tension is \(k\) times its extension. Initially, the particles are at rest, with \(m_{1}\) at a height \(l\) above \(m_{2} .\) At \(t=0, m_{1}\) is projected vertically upward with velocity \(v\). Find the positions of the particles at any subsequent time (assuming that \(v\) is not so large that the spring is expanded or compressed beyond its elastic limit).
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Electromagnetism

Read Explanation

Kinematics Physics

Read Explanation

Rotational Dynamics

Read Explanation

Space Physics

Read Explanation

Dynamics

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