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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 C.

Short Answer

Expert verified
Mars takes approximately 1.874 years to orbit the Sun, which matches the 1.88 years given in Appendix C.

Step by step solution

01

Understanding Kepler's Third Law

Kepler's Third Law states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. Mathematically, it is expressed as \( T^2 \propto a^3 \), where \( T \) is the orbital period, and \( a \) is the semi-major axis.
02

Expressing in Terms of Earth

Given that the orbital period and the semi-major axis for Earth are \(1\) year and \(1\) AU (astronomical unit) respectively, Kepler's law can be expressed as: \( T_{Earth}^2 = a_{Earth}^3 \). Since \(a_{Earth} = 1\), we have \(1^2 = 1^3\).
03

Expressing for Mars

For Mars, \( a_{Mars} = 1.52 \times a_{Earth} = 1.52 \). Using Kepler's law, we write: \( T_{Mars}^2 = a_{Mars}^3 = (1.52)^3 \). Solve this to find \(T_{Mars}\).
04

Calculating Mars's Orbital Period

Calculate \((1.52)^3\) which is approximately \(3.515648\). Then, \( T_{Mars} = \sqrt{3.515648} \). This calculation gives us \( T_{Mars} \approx 1.874\) years.
05

Comparing with Appendix C

The calculated period for Mars is approximately \(1.874\) years. Checking with Appendix C, you should find that the value given is about \(1.88\) years, which confirms our calculated answer.

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Key Concepts

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

Orbital Period
The orbital period is the time a planet takes to complete one full orbit around the Sun. It's crucial in understanding a planet’s motion and position over time. In Kepler's Third Law, the orbital period is denoted by the symbol \( T \). This law tells us that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. This relationship is incredibly useful because it allows us to predict how long a planet, like Mars, will take to orbit the Sun if we know its distance from the Sun.To provide a simple example, for Earth, the orbital period is about 365.25 days, or 1 year. This gives us a benchmark against which we can compare other planets. Using Kepler's Third Law, if Mars's semi-major axis is known, we can calculate its orbital period by understanding its proportional relationship with the semi-major axis.
Semi-Major Axis
The semi-major axis is a fundamental concept when we discuss planetary orbits, particularly in the application of Kepler's laws. It is half of the longest diameter of an elliptical orbit. The semi-major axis essentially measures the average distance of a planet from the Sun.In the exercise given, the semi-major axis of Mars is specified as \( 1.52 \times \) that of Earth’s semi-major axis. Since Earth's semi-major axis is 1 astronomical unit (AU), Mars' semi-major axis becomes \( 1.52 \) AU. This measurement helps in calculating Mars's orbital period.To connect this to Kepler's Third Law, if you know the semi-major axis of a planet's orbit, you can easily decipher its orbital period using the relationship \( T^2 \propto a^3 \), where \( T \) is the orbital period and \( a \) is the semi-major axis.
Astronomical Unit
The astronomical unit, often abbreviated as AU, is a key unit of measurement in astronomy. It is defined as the average distance from the Earth to the Sun, roughly about 149.6 million kilometers or 93 million miles. An astronomical unit provides a practical way to express larger distances in the solar system, which would otherwise involve cumbersome numbers.In the context of Kepler's Third Law, the Earth's orbit defines a baseline where both the semi-major axis and orbital period are normalized to 1 AU and 1 year, respectively. Hence, the semi-major axis of any other planet can be expressed in AU for simple comparison.For instance, Mars' semi-major axis is approximately \( 1.52 \) AU, which means Mars is about 1.52 times farther from the Sun than Earth is. Using the relationship provided by Kepler's law, this distance helps us to compute Mars’s orbital period in relation to Earth's year.

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Most popular questions from this chapter

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?

Two Earth satellites, \(A\) and \(B\), each of mass \(m\), are to be launched into circular orbits about Earth's center. Satellite \(A\) is to orbit at an altitude of \(6370 \mathrm{~km}\). Satellite \(B\) is to orbit at an altitude of \(19110 \mathrm{~km}\). The radius of Earth \(R_{E}\) is \(6370 \mathrm{~km}\). (a) What is the ratio of the potential energy of satellite \(B\) to that of satellite \(A\), in orbit? (b) What is the ratio of the kinetic energy of satellite \(B\) to that of satellite \(A\), in orbit? (c) Which satellite has the greater total energy if each has a mass of \(14.6 \mathrm{~kg}\) ? (d) By how much?

In 1610, Galileo used his telescope to discover four prominent moons around Jupiter. Their mean orbital radii \(a\) and periods \(T\) are as follows: $$\begin{array}{lcc} {\text { Name }} & a\left(10^{8} \mathrm{~m}\right) & T \text { (days) } \\ \hline \text { Io } & 4.22 & 1.77 \\ \text { Europa } & 6.71 & 3.55 \\ \text { Ganymede } & 10.7 & 7.16 \\ \text { Callisto } & 18.8 & 16.7 \\ \hline \end{array}$$ (a) Plot \(\log a(y\) axis \()\) against \(\log T(x\) axis \()\) and show that you get a straight line. (b) Measure the slope of the line and compare it with the value that you expect from Kepler's third law. (c) Find the mass of Jupiter from the intercept of this line with the \(y\) axis.

A comet that was seen in April 574 by Chinese astronomers on a day known by them as the Woo Woo day was spotted again in May 1994. Assume the time between observations is the period of the Woo Woo day comet and take its eccentricity as \(0.11\). What are (a) the semimajor axis of the comet's orbit and (b) its greatest distance from the Sun in terms of the mean orbital radius \(R_{p}\) of Pluto?

In deep space, sphere \(A\) of mass \(20 \mathrm{~kg}\) is located at the origin of an \(x\) axis and sphere \(B\) of mass \(10 \mathrm{~kg}\) is located on the axis at \(x=\) \(0.80 \mathrm{~m}\). Sphere \(B\) is released from rest while sphere \(A\) is held at the origin. (a) What is the gravitational potential energy of the twosphere system just as \(B\) is released? (b) What is the kinetic energy of \(B\) when it has moved \(0.20 \mathrm{~m}\) toward \(A\) ?

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