Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 23

The planet Mars has a satellite, Phobos, which travels in an orbit of radius \(9400 \mathrm{~km}\) with a period of 7 h 39 min. Calculate the mass of Mars from this information. (The mass of Phobos is negligible compared with that of Mars.)

Short Answer

Expert verified
After simplifying and calculating the above expression, the mass of Mars is approximately \(6.427 \times 10^{23} \mathrm{kg}\).

Step by step solution

01

Write down the given parameters

Here are the given parameters: The radius \(r = 9400 \mathrm{~km} = 9400 \times 10^3 \mathrm{~m}\) (as we will work in SI units) and the period \(T = 7 \mathrm{~h} 39 \mathrm{~min} = 7.65 \mathrm{~h} = 7.65 \times 3600 \mathrm{~s}\).
02

Use the formula for the period of revolution

The formula for the period of revolution of a body around a celestial object is \(T = 2\pi \sqrt{\frac{r^3}{GM}}\), where \(T\) - period, \(r\) - radius, \(G\) - the gravitational constant (\(6.674 \times 10^{-11} \mathrm{Nm}^2/\mathrm{kg}^2\)) and \(M\) - mass of the celestial object (in this case, Mars).
03

Solve for the mass of Mars

You want to solve for \(M\), so rearrange the equation in Step 2 to solve for \(M\): \(M = \frac{r^3}{G(T/2\pi)^2}\). Plug in the known values and solve for \(M\). Mars' mass \(M = \frac{(9400 \times 10^3) ^3}{6.674 \times 10^{-11} \times (7.65 \times 3600/2\pi)^2}\).

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

A typical neutron star may have a mass equal to that of the Sun but a radius of only \(10.0 \mathrm{~km} .(a)\) What is the gravitational acceleration at the surface of such a star? \((b)\) How fast would an object be moving if it fell from rest through a distance of \(1.20 \mathrm{~m}\) on such a star?

A rocket is accelerated to a speed of \(v=2 \sqrt{g R_{\mathrm{E}}}\) near the Earth's surface and then coasts upward. (a) Show that it will escape from the Earth. ( \(b\) ) Show that very far from the Earth its speed is \(v=\sqrt{2 g R_{\mathrm{E}}}\).

Show that the velocity of escape from the Sun at the Earth's distance from the Sun is \(\sqrt{2}\) times the speed of the Earth in its orbit, assumed to be a circle. (This is a specific case of a general result for circular orbits: \(v_{\text {esc }}=\sqrt{2} v_{\text {orb }} .\) )

The orbit of the Earth about the Sun is almost circular. The closest and farthest distances are \(1.47 \times 10^{8} \mathrm{~km}\) and \(1.52 \times\) \(10^{8} \mathrm{~km}\), respectively. Determine the maximum variations in ( \(a\) ) potential energy, (b) kinetic energy, ( \(c\) ) total energy, and (d) orbital speed that result from the changing Earth-Sun distance in the course of 1 year. (Hint: Use conservation of energy and angular momentum.)

If \(g\) is to be determined by dropping an object through a distance of (exactly) \(10 \mathrm{~m}\), how accurately must the time be measured to obtain a result good to \(0.1 \%\) ? Calculate a percent error and an absolute error, in milliseconds.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Translational Dynamics

Read Explanation

Physical Quantities And Units

Read Explanation

Electricity and Magnetism

Read Explanation

Quantum Physics

Read Explanation

Electrostatics

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