/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 The mean distance of Mars from t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 48

The mean distance of Mars from the Sun is \(1.52\) times that of Earth from the Sun. From Kepler's law of periods, calculate the number of years required for Mars to make one revolution around the Sun; compare your answer with the value given in Appendix \(\mathrm{C}\).

Short Answer

Expert verified
Mars takes approximately 1.87 years to orbit the Sun.

Step by step solution

01

Understand Kepler's Third Law

Kepler's Third Law of Planetary Motion states that the square of the sidereal period (orbital period) of a planet is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, it can be represented as \( T^2 \propto a^3 \), or \( \frac{T_1^2}{T_2^2} = \frac{a_1^3}{a_2^3} \) when comparing two planets, where \( T \) is the orbital period and \( a \) is the average distance from the sun.
02

Set Up Proportions

Let \( T_M \) be the orbital period of Mars and \( T_E \) be the orbital period of Earth (1 year). According to Kepler's law, \( \frac{T_M^2}{T_E^2} = \frac{a_M^3}{a_E^3} \). We know the mean distance of Mars from the Sun is \(1.52\) times that of Earth, so \( a_M = 1.52a_E \). Substitute this into the equation: \( \frac{T_M^2}{1^2} = \frac{(1.52a_E)^3}{a_E^3} \).
03

Simplify the Equation

Simplify the equation \( \frac{T_M^2}{1^2} = (1.52)^3 \). Since \( (1.52)^3 = 1.52 \times 1.52 \times 1.52 \), calculate this value to find \( 1.52^3 \).
04

Calculate \( 1.52^3 \)

Calculate \( 1.52^3 \): \( 1.52 \times 1.52 = 2.3104 \) and then \( 2.3104 \times 1.52 = 3.515328 \). So, \( 1.52^3 = 3.515328 \).
05

Solve for \( T_M \)

We have \( T_M^2 = 3.515328 \). Solve for \( T_M \) by taking the square root: \( T_M = \sqrt{3.515328} \).
06

Calculate \( T_M \)

Calculate \( \sqrt{3.515328} \) to find \( T_M \). The square root of \( 3.515328 \) is approximately \( 1.87 \). Hence, \( T_M \approx 1.87 \) 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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Orbital Periods
In the realm of astronomy, understanding the concept of Orbital Periods is crucial. An orbital period is the time a celestial object takes to complete one full orbit around another body. Such as a planet circling its star, like Mars orbiting the Sun. It helps us comprehend the dynamics of celestial motions, providing insights into how long planets take to return to the same position relative to their central star.
To calculate an object's orbital period, especially using Kepler's Third Law, we delve into the formula: \( T^2 \propto a^3 \). Here, \( T \) denotes the orbital period and \( a \) represents the semi-major axis, or the average distance from the object to the body it orbits. For example, Earth has an orbital period of one year.
Understanding this concept helps us calculate how long planets like Mars take to travel around the Sun by comparing their distances and periods to Earth's, providing a window into the cosmic clockwork.
  • Orbital Period: Time taken to complete one orbit
  • Kepler's Third Law: \( T^2 \propto a^3 \)
  • Comparison to Earth for calculations
Mars Revolution
The revolution of Mars around the Sun is a fascinating aspect of planetary motion. Just like Earth, Mars has its own orbital journey. The time Mars takes to complete one full revolution around the Sun is a significant measure known as its orbital period, which can be computed using Kepler's Third Law.
Since Mars is further from the Sun than Earth, its average distance, the semi-major axis of its orbit, is 1.52 times greater. This increased distance contributes to a longer revolution period. By leveraging the calculations from Kepler鈥檚 law, we determine Mars' orbital period as approximately 1.87 Earth years. This means it takes Mars about 687 Earth days to circle the Sun once.
Establishing Mars' orbital period not only aids in predicting its position in the solar system but also assists in planning space missions and understanding its seasonal changes.
  • Mars is 1.52 times further from the Sun than Earth
  • Orbital period for Mars: ~1.87 Earth years
  • Influences mission planning and scientific understanding
Sidereal Period
The Sidereal Period is an astronomical term used to describe the time taken for a planet to complete one full orbit relative to the stars, distinct from the solar time measured in relation to the Sun. This concept is critical to understanding the true orbital characteristics of planets.
For Mars, its sidereal period is about 1.87 Earth years, as calculated through Kepler's Third Law. This differs from Earth's sidereal period, which is exactly one year, or 365.25 days. The sidereal period provides a consistent framework to gauge orbital paths in space, free from the variances of solar timing.
Knowing a planet鈥檚 sidereal period enables astronomers to predict its exact position in the sky over extended periods and deepens our comprehension of celestial mechanics.
  • Sidereal Period: Time relative to fixed stars
  • For Mars, calculated as ~1.87 Earth years
  • Essential for accurate astronomical predictions

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

Moon effect. Some people believe that the Moon controls their activities. If the Moon moves from being directly on the opposite side of Earth from you to being directly overhead, by what percent does (a) the Moon's gravitational pull on you increase and (b) your weight (as measured on a scale) decrease? Assume that the Earth-Moon (center-to-center) distance is \(3.82 \times 10^{8} \mathrm{~m}\) and Earth's radius is \(6.37 \times 10^{6} \mathrm{~m}\).

A very early, simple satellite consisted of an inflated spherical aluminum balloon \(30 \mathrm{~m}\) in diameter and of mass \(20 \mathrm{~kg}\). Suppose a meteor having a mass of \(7.0 \mathrm{~kg}\) passes within \(3.0\) \(\mathrm{m}\) of the surface of the satellite. What is the magnitude of the gravitational force on the meteor from the satellite at the closest approach?

A mass \(M\) is split into two parts, \(m\) and \(M-m\), which are then separated by a certain distance. What ratio \(m / M\) maximizes the magnitude of the gravitational force between the parts?

An object lying on Earth's equator is accelerated (a) toward the center of Earth because Earth rotates, (b) toward the Sun because Earth revolves around the Sun in an almost circular orbit, and (c) toward the center of our galaxy because the Sun moves around the galactic center. For the latter, the period is \(2.5 \times 10^{8} \mathrm{y}\) and the radius is \(2.2 \times 10^{20} \mathrm{~m} .\) Calculate these three accelerations as multiples of \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\)

In a shuttle craft of mass \(m=3000 \mathrm{~kg}\), Captain Janeway orbits a planet of mass \(M=9.50 \times 10^{25} \mathrm{~kg}\), in a circular orbit of radius \(r=4.20 \times 10^{7} \mathrm{~m} .\) What are (a) the period of the orbit and (b) the speed of the shuttle craft? Janeway briefly fires a forwardpointing thruster, reducing her speed by \(2.00 \%\). Just then, what are (c) the speed, (d) the kinetic energy, (e) the gravitational potential energy, and (f) the mechanical energy of the shuttle craft? (g) What is the semimajor axis of the elliptical orbit now taken by the craft? (h) What is the difference between the period of the original circular orbit and that of the new elliptical orbit? (i) Which orbit has the smaller period?
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Electromagnetism

Read Explanation

Electrostatics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Wave Optics

Read Explanation

Atoms and Radioactivity

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.