kepler's law assignment

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Kepler’s Third Law

Kepler’s third law states “ The square of a planet’s orbital period is directly proportional to the cube of the orbit’s semi-major axis “.

Kepler’s third law establishes a relationship between the time taken by a planet to complete one revolution and its distance from the Sun. According to this law, the orbital period increases rapidly with the orbit’s radius.

The importance of the third law is that it has been successful in measuring the masses of the planets in the solar system. This law can also be applied to planets beyond the solar system, asteroids, comets, and artificial satellites. Another name for Kepler’s third law is the law of harmony.

German astronomer and mathematician Johannes Kepler published this law in 1619.

Kepler’s third law establishes the relationship between the orbital period and the distance of a planet’s orbit from the Sun. Kepler stated in his first law that the planets move around the Sun in elliptical orbits. The orbit’s size is represented by the semi-major axis a . Suppose T is the orbital period. Then, according to this law,

Where k is the constant of proportionality.

Newton showed in 1687 that Kepler’s third law was a direct consequence of his laws of motion and his law of gravity. To prove this and make the derivation easier, we make a few assumptions:

• The planets’ orbits are circular rather than elliptical.
• The Sun is at the center of the orbits rather than the focus.

Newton’s law of gravity states that the force of attraction between the Sun and a planet is given by

M s : Sun’s mass

M p :  Planet’s mass

R : Distance between the planet and Sun

G : Universal gravitational constant.

In this case, R is assumed to be the radius of the orbit.

When an object moves in a circle, there will be a centripetal force . When a planet moves with a velocity V , the centripetal force F C is given by

The gravitational force provides the centripetal force. Therefore,

However, the speed V is given by the distance the planet moves divided by the time it takes. For one complete revolution, this is given by

Where 2πR is the circumference and T is the time taken to complete one revolution, known as the orbital period. Substituting for V , we get

Replacing R with the semi-major axis a ,

This equation is the Kepler’s third law, where the constant of proportionality k is given by

Thus, we have derived Kepler’s third law from Newton’s laws of motion and Newton’s law of gravity.

T is expressed in Earth years

a is expressed in astronomical units (AU)

M s is expressed in solar mass or M s = 1

The following table shows the orbital period in years and the semi-major axes in astronomical units (A.U.) of the solar system planets. It is observed that the ratio T 2 /a 3 comes out to be close to 1.

Example Problems

Problem 1 : Kepler’s third law says that T 2 /a 3 is the same for all objects orbiting the Sun. Consider the dwarf planet Ceres in the asteroid belt that takes 4.6 years to orbit the Sun. Calculate the average Sun-Ceres distance.

From Kepler’s third law

Or, a = (T) 2/3

Or, a = (4.6) 2/3

Or, a = 2.77 A.U.

Problem 2 : Jupiter takes 11.86 years to complete one revolution. What is its distance from the Sun in km? (1 A.U. = 1.5 x 10 8 km)

T = 11.86 y

From Kepler’s Law

Or, a = (11.86) 2/3

Or, a = 5.2 A.U. = 5.2 A.U. x 1.5 x 10 8 km/A.U. = 7.6 x 10 8 km

• Kepler’s Third Law: The movement of solar system planets – Space.com
• Kepler’s Three Laws – Physicsclassroom.com
• Kepler’s Third Law – Spiff.rit.edu
• Kepler’s Law – Hyperphysics.phy-astr.gsu.edu
• Kepler’s Third Law – Farside.ph.utexas.edu

Article was last reviewed on Thursday, June 16, 2022

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Orbits and Kepler’s Laws

Planetary Physics: Kepler's Laws of Planetary Motion

Kepler's three laws describe how planetary bodies orbit the Sun. They describe how (1) planets move in elliptical orbits with the Sun as a focus, (2) a planet covers the same area of space in the same amount of time no matter where it is in its orbit, and (3) a planet’s orbital period is proportional to the size of its orbit (its semi-major axis).

Explore the process that Johannes Kepler undertook when he formulated his three laws of planetary motion.

The planets orbit the Sun in a counterclockwise direction as viewed from above the Sun's north pole, and the planets' orbits all are aligned to what astronomers call the ecliptic plane.

The story of our greater understanding of planetary motion could not be told if it were not for the work of a German mathematician named Johannes Kepler. Kepler lived in Graz, Austria during the tumultuous early 17th century. Due to religious and political difficulties common during that era, Kepler was banished from Graz on August 2nd, 1600.

Fortunately, an opportunity to work as an assistant for the famous astronomer Tycho Brahe presented itself and the young Kepler moved his family from Graz 300 miles across the Danube River to Brahe's home in Prague. Tycho Brahe is credited with the most accurate astronomical observations of his time and was impressed with the studies of Kepler during an earlier meeting. However, Brahe mistrusted Kepler, fearing that his bright young intern might eclipse him as the premier astronomer of his day. He, therefore, led Kepler to see only part of his voluminous planetary data.

He set Kepler, the task of understanding the orbit of the planet Mars, the movement of which fit problematically into the universe as described by Aristotle and Ptolemy. It is believed that part of the motivation for giving the Mars problem to Kepler was Brahe's hope that its difficulty would occupy Kepler while Brahe worked to perfect his own theory of the solar system, which was based on a geocentric model, where the earth is the center of the solar system. Based on this model, the planets Mercury, Venus, Mars, Jupiter, and Saturn all orbit the Sun, which in turn orbits the earth. As it turned out, Kepler, unlike Brahe, believed firmly in the Copernican model of the solar system known as heliocentric, which correctly placed the Sun at its center. But the reason Mars' orbit was problematic was because the Copernican system incorrectly assumed the orbits of the planets to be circular.

After much struggling, Kepler was forced to an eventual realization that the orbits of the planets are not circles, but were instead the elongated or flattened circles that geometers call ellipses, and the particular difficulties Brahe hand with the movement of Mars were due to the fact that its orbit was the most elliptical of the planets for which Brahe had extensive data. Thus, in a twist of irony, Brahe unwittingly gave Kepler the very part of his data that would enable Kepler to formulate the correct theory of the solar system, banishing Brahe's own theory.

Since the orbits of the planets are ellipses, let us review three basic properties of ellipses. The first property of an ellipse: an ellipse is defined by two points, each called a focus, and together called foci. The sum of the distances to the foci from any point on the ellipse is always a constant. The second property of an ellipse: the amount of flattening of the ellipse is called the eccentricity. The flatter the ellipse, the more eccentric it is. Each ellipse has an eccentricity with a value between zero, a circle, and one, essentially a flat line, technically called a parabola.

The third property of an ellipse: the longest axis of the ellipse is called the major axis, while the shortest axis is called the minor axis. Half of the major axis is termed a semi-major axis. Knowing then that the orbits of the planets are elliptical, johannes Kepler formulated three laws of planetary motion, which accurately described the motion of comets as well.

Kepler's First Law: each planet's orbit about the Sun is an ellipse. The Sun's center is always located at one focus of the orbital ellipse. The Sun is at one focus. The planet follows the ellipse in its orbit, meaning that the planet to Sun distance is constantly changing as the planet goes around its orbit.

Kepler's Second Law: the imaginary line joining a planet and the Sun sweeps equal areas of space during equal time intervals as the planet orbits. Basically, that planets do not move with constant speed along their orbits. Rather, their speed varies so that the line joining the centers of the Sun and the planet sweeps out equal parts of an area in equal times. The point of nearest approach of the planet to the Sun is termed perihelion. The point of greatest separation is aphelion, hence by Kepler's Second Law, a planet is moving fastest when it is at perihelion and slowest at aphelion.

Kepler's Third Law: the squares of the orbital periods of the planets are directly proportional to the cubes of the semi-major axes of their orbits. Kepler's Third Law implies that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit. Thus we find that Mercury, the innermost planet, takes only 88 days to orbit the Sun. The earth takes 365 days, while Saturn requires 10,759 days to do the same. Though Kepler hadn't known about gravitation when he came up with his three laws, they were instrumental in Isaac Newton deriving his theory of universal gravitation, which explains the unknown force behind Kepler's Third Law. Kepler and his theories were crucial in the better understanding of our solar system dynamics and as a springboard to newer theories that more accurately approximate our planetary orbits.

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7.1 Kepler's Laws of Planetary Motion

Section learning objectives.

By the end of this section, you will be able to do the following:

• Explain Kepler’s three laws of planetary motion
• Apply Kepler’s laws to calculate characteristics of orbits

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.

In this section students will apply Kepler’s laws of planetary motion to objects in the solar system.

[BL] [OL] Discuss the historical setting in which Kepler worked. Most people still thought Earth was the center of the universe, and yet Kepler not only knew that the planets circled the sun, he found patterns in the paths they followed. What would it be like to be that far ahead of almost everyone? A fascinating description of this is given in the program Cosmos with Carl Sagan (Episode 3, Harmony of the Worlds).

[AL] Explain that Kepler’s laws were laws and not theories. Laws describe patterns in nature that always repeat themselves under the same set of conditions. Theories provide an explanation for the patterns. Kepler provided no explanation.

Section Key Terms

Concepts related to kepler’s laws of planetary motion.

Examples of orbits abound. Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris. The moon’s orbit around Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets around the sun are no less interesting. If we look farther, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through gravity.

All these motions are governed by gravitational force. The orbital motions of objects in our own solar system are simple enough to describe with a few fairly simple laws. The orbits of planets and moons satisfy the following two conditions:

• The mass of the orbiting object, m , is small compared to the mass of the object it orbits, M .
• The system is isolated from other massive objects.

[OL] Ask the students to explain the criteria to see if they understand relative mass and isolated systems.

Based on the motion of the planets about the sun, Kepler devised a set of three classical laws, called Kepler’s laws of planetary motion , that describe the orbits of all bodies satisfying these two conditions:

• The orbit of each planet around the sun is an ellipse with the sun at one focus.
• Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times.
• The ratio of the squares of the periods of any two planets about the sun is equal to the ratio of the cubes of their average distances from the sun.

These descriptive laws are named for the German astronomer Johannes Kepler (1571–1630). He devised them after careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho Brahe (1546–1601). Such careful collection and detailed recording of methods and data are hallmarks of good science. Data constitute the evidence from which new interpretations and meanings can be constructed. Let’s look closer at each of these laws.

[BL] Relate orbit to year and rotation to day. Be sure that students know that an object rotates on its axis and revolves around a parent body as it follows its orbit.

[OL] See how many levels of orbital motion the students know and fill in the ones they don’t. For example, moons orbit around planets; planets around stars; stars around the center of the galaxy, etc.

[AL] From the point of view of Earth, which objects appear (incorrectly) to be orbiting Earth (stars, the sun, galaxies) and which can be seen to be orbiting parent bodies (the moon, moons of other planets, stars in other galaxies)?

Kepler’s First Law

The orbit of each planet about the sun is an ellipse with the sun at one focus, as shown in Figure 7.2 . The planet’s closest approach to the sun is called perihelion and its farthest distance from the sun is called aphelion .

[AL] Ask for a definition of planet. Prepare to discuss Pluto’s demotion if it comes up. Discuss the first criterion in terms of center of rotation of a moon-planet system. Explain that for all planet-moon systems in the solar system, the center of rotation is within the planet. This is not true for Pluto and its largest moon, Charon, because their masses are similar enough that they rotate around a point in space between them.

If you know the aphelion ( r a ) and perihelion ( r p ) distances, then you can calculate the semi-major axis ( a ) and semi-minor axis ( b ).

[AL] If any students are interested and proficient in algebra and geometry, ask them to derive a formula that relates the length of the string and the distance between pins to the major and minor axes of an ellipse. Explain that this is a real world problem for workers who design elliptical tabletops and mirrors.

[BL] [OL] Impress upon the students that Kepler had to crunch an enormous amount of data and that all his calculations had to be done by hand. Ask students to think of similar projects where scientists found order in a daunting amount of data (the periodic table, DNA structure, climate models, etc.).

Teacher Demonstration

Demonstrate the pins and string method of drawing an ellipse, as shown in Figure 7.3 , or have the students try it at home or in class.

Ask students: Why does the string and pin method create a shape that conforms to Kepler's second law? That is, why is the shape an ellipse?

Explain that the pins are the foci and explain what each of the three sections of string represents. Note that the pencil represents a planet and one of the pins represents the sun.

Kepler’s Second Law

Each planet moves so that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal times, as shown in Figure 7.4 .

Ask the students to imagine how complicated it would be to describe the motion of the planets mathematically, if it is assumed that Earth is stationary. And yet, people tried to do this for hundreds of years, while overlooking the simple explanation that all planets circle the sun.

[OL] Ask students to use this figure to understand why planets and comets travel faster when they are closer to the sun. Explain that time intervals and areas are constant, but both velocity and distance from the sun vary.

Tips For Success

Note that while, for historical reasons, Kepler’s laws are stated for planets orbiting the sun, they are actually valid for all bodies satisfying the two previously stated conditions.

Kepler’s Third Law

The ratio of the periods squared of any two planets around the sun is equal to the ratio of their average distances from the sun cubed. In equation form, this is

where T is the period (time for one orbit) and r is the average distance (also called orbital radius). This equation is valid only for comparing two small masses orbiting a single large mass. Most importantly, this is only a descriptive equation; it gives no information about the cause of the equality.

[BL] See if students can rearrange this equation to solve for any one of the variables when the other three are known.

[AL] Show a solution for one of the periods T or radii r and ask students to interpret the fractional powers on the right hand side of the equation.

[OL] Emphasize that this approach only works for two satellites orbiting the same parent body. The parent body must be the same because r 2 / T 2 = G M / ( 4 π 2 ) r 2 / T 2 = G M / ( 4 π 2 ) and M is the mass of the parent body. If M changes, the ratio r 3 / T 2 also changes.

Links To Physics

History: ptolemy vs. copernicus.

Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth as shown in Figure 7.5 (a) . This is called the Ptolemaic model , named for the Greek philosopher Ptolemy who lived in the second century AD. The Ptolemaic model is characterized by a list of facts for the motions of planets, with no explanation of cause and effect. There tended to be a different rule for each heavenly body and a general lack of simplicity.

Figure 7.5 (b) represents the modern or Copernican model . In this model, a small set of rules and a single underlying force explain not only all planetary motion in the solar system, but also all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling.

Nicolaus Copernicus (1473–1543) first had the idea that the planets circle the sun, in about 1514. It took him almost 20 years to work out the mathematical details for his model. He waited another 10 years or so to publish his work. It is thought he hesitated because he was afraid people would make fun of his theory. Actually, the reaction of many people was more one of fear and anger. Many people felt the Copernican model threatened their basic belief system. About 100 years later, the astronomer Galileo was put under house arrest for providing evidence that planets, including Earth, orbited the sun. In all, it took almost 300 years for everyone to admit that Copernicus had been right all along.

Grasp Check

Explain why Earth does actually appear to be the center of the solar system.

• Earth appears to be the center of the solar system because Earth is at the center of the universe, and everything revolves around it in a circular orbit.
• Earth appears to be the center of the solar system because, in the reference frame of Earth, the sun, moon, and planets all appear to move across the sky as if they were circling Earth.
• Earth appears to be at the center of the solar system because Earth is at the center of the solar system and all the heavenly bodies revolve around it.
• Earth appears to be at the center of the solar system because Earth is located at one of the foci of the elliptical orbit of the sun, moon, and other planets.

Introduce the historical debate around the geocentric versus the heliocentric view of the universe. Stress how controversial this debate was at the time. Explain that this was important to people because their world view and cultural beliefs were at stake.

Virtual Physics

Acceleration.

This simulation allows you to create your own solar system so that you can see how changing distances and masses determines the orbits of planets. Click Help for instructions.

• The orbiting object moves fastest when it is closest to the central object and slowest when it is farthest away.
• The orbiting object moves slowest when it is closest to the central object and fastest when it is farthest away.
• The orbiting object moves with the same speed at every point on the circumference of the elliptical orbit.
• There is no relationship between the speed of the object and the location of the planet on the circumference of the orbit.

Give the students ample time to manipulate this animation. It may take some time to get the parameters adjusted so that they can see how mass and eccentricity affect the orbit. Initially, the planet is likely to disappear off the screen or crash into the sun.

Calculations Related to Kepler’s Laws of Planetary Motion

Refer back to Figure 7.2 (a) . Notice which distances are constant. The foci are fixed, so distance f 1 f 2 ¯ f 1 f 2 ¯ is a constant. The definition of an ellipse states that the sum of the distances f 1 m ¯ + m f 2 ¯ f 1 m ¯ + m f 2 ¯ is also constant. These two facts taken together mean that the perimeter of triangle Δ f 1 m f 2 Δ f 1 m f 2 must also be constant. Knowledge of these constants will help you determine positions and distances of objects in a system that includes one object orbiting another.

Refer back to Figure 7.4 . The second law says that the segments have equal area and that it takes equal time to sweep through each segment. That is, the time it takes to travel from A to B equals the time it takes to travel from C to D, and so forth. Velocity v equals distance d divided by time t : v = d / t v = d / t . Then, t = d / v t = d / v , so distance divided by velocity is also a constant. For example, if we know the average velocity of Earth on June 21 and December 21, we can compare the distance Earth travels on those days.

The degree of elongation of an elliptical orbit is called its eccentricity ( e ). Eccentricity is calculated by dividing the distance f from the center of an ellipse to one of the foci by half the long axis a .

When e = 0 e = 0 , the ellipse is a circle.

The area of an ellipse is given by A = π a b A = π a b , where b is half the short axis. If you know the axes of Earth’s orbit and the area Earth sweeps out in a given period of time, you can calculate the fraction of the year that has elapsed.

[OL] Review the definitions of major and minor axes, semi-major and semi-minor axes, and distance f . The major axis is the length of the ellipse and passes through both foci. The minor axis is the width of the ellipse and is perpendicular to the major axis. The semi-major and semi-minor axes are half of the major and minor axes, respectively.

Worked Example

At its closest approach, a moon comes within 200,000 km of the planet it orbits. At that point, the moon is 300,000 km from the other focus of its orbit, f 2 . The planet is focus f 1 of the moon’s elliptical orbit. How far is the moon from the planet when it is 260,000 km from f 2 ?

Show and label the ellipse that is the orbit in your solution. Picture the triangle f 1 m f 2 collapsed along the major axis and add up the lengths of the three sides. Find the length of the unknown side of the triangle when the moon is 260,000 km from f 2 .

Perimeter of f 1 m f 2 = 200 , 000 km + 100,000 km + 300,000 km = 600,000 km. f 1 m f 2 = 200 , 000 km + 100,000 km + 300,000 km = 600,000 km.

m f 1 = 600,000 km − ( 100,000 km + 260,000 km ) = 240,000 km. m f 1 = 600,000 km − ( 100,000 km + 260,000 km ) = 240,000 km.

The perimeter of triangle f 1 mf 2 must be constant because the distance between the foci does not change and Kepler’s first law says the orbit is an ellipse. For any ellipse, the sum of the two sides of the triangle, which are f 1 m and mf 2 , is constant.

Walk the students through the process of mentally collapsing the f 1 mf 2 at the end of the major axis to reveal what the three sides of the triangle f 1 mf 2 are equal to. Picture the sections of the string as the pencil approaches the major axis. This distance f 1 f 2 remains constant, f 1 m is the distance from f 1 to the end of the major axis, and mf 2 is f 1 m + f 1 f 2 .

[OL] Have students relate eccentricity, distance between foci, and shape of orbit.

[AL] Ask for examples of orbits with high eccentricity (comets, Pluto) and low eccentricity (moon, Earth).

Figure 7.6 shows the major and minor axes of an ellipse. The semi-major and semi-minor axes are half of these, respectively.

Earth’s orbit is very slightly elliptical, with a semi-major axis of 1.49598 × 10 8 km and a semi-minor axis of 1.49577 × 10 8 km. If Earth’s period is 365.26 days, what area does an Earth-to-sun line sweep past in one day?

Each day, Earth sweeps past an equal-sized area, so we divide the total area by the number of days in a year to find the area swept past in one day. For total area use A = π a b A = π a b . Calculate A , the area inside Earth’s orbit and divide by the number of days in a year (i.e., its period).

The area swept out in one day is thus 1.93 × 10 14 km 2 1.93 × 10 14 km 2 .

The answer is based on Kepler’s law, which states that a line from a planet to the sun sweeps out equal areas in equal times.

Explain that this formula is easy to remember because it is similar to A = π r 2 . A = π r 2 . Discuss Earth’s eccentricity. Compare it with that of other planets, asteroids, or comets to further emphasize what defines a planet. Note that Earth has one of the least eccentric orbits and Mercury has the most eccentric orbit of the planets.

[BL] Have the students memorized the value of π ? π ?

[OL] [AL] What is the formula when a = b ? Is the formula familiar?

[OL] Can the student verify this statement by rearranging the equation?

Kepler’s third law states that the ratio of the squares of the periods of any two planets ( T 1 , T 2 ) is equal to the ratio of the cubes of their average orbital distance from the sun ( r 1 , r 2 ). Mathematically, this is represented by

From this equation, it follows that the ratio r 3 /T 2 is the same for all planets in the solar system. Later we will see how the work of Newton leads to a value for this constant.

Given that the moon orbits Earth each 27.3 days and that it is an average distance of 3.84 × 10 8 m 3.84 × 10 8 m from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1,500 km above Earth’s surface.

The period, or time for one orbit, is related to the radius of the orbit by Kepler’s third law, given in mathematical form by T 1 2 T 2 2 = r 1 3 r 2 3 T 1 2 T 2 2 = r 1 3 r 2 3 . Let us use the subscript 1 for the moon and the subscript 2 for the satellite. We are asked to find T 2 . The given information tells us that the orbital radius of the moon is r 1 = 3.84 × 10 8 m r 1 = 3.84 × 10 8 m , and that the period of the moon is T 1 = 27.3 days T 1 = 27.3 days . The height of the artificial satellite above Earth’s surface is given, so to get the distance r 2 from the center of Earth we must add the height to the radius of Earth (6380 km). This gives r 2 = 1500 km + 6380 km = 7880 km r 2 = 1500 km + 6380 km = 7880 km . Now all quantities are known, so T 2 can be found.

To solve for T 2 , we cross-multiply and take the square root, yielding

This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will complete one orbit in the same amount of time.

Remind the students that this only works when the satellites are small compared to the parent object and when both satellites orbit the same parent object.

Practice Problems

A planet with no axial tilt is located in another solar system. It circles its sun in a very elliptical orbit so that the temperature varies greatly throughout the year. If the year there has 612 days and the inhabitants celebrate the coldest day on day 1 of their calendar, when is the warmest day?

• 2.75 × 10 3 km
• 1.96 × 10 4 km
• 1.40 × 10 5 km
• 1.00 × 10 6 km

Check Your Understanding

• Kepler’s laws are purely descriptive.
• Kepler’s laws are purely causal.
• Kepler’s laws are descriptive as well as causal.
• Kepler’s laws are neither descriptive nor causal.

True or false—According to Kepler’s laws of planetary motion, a satellite increases its speed as it approaches its parent body and decreases its speed as it moves away from the parent body.

• One focus is the parent body, and the other is located at the opposite end of the ellipse, at the same distance from the center as the parent body.
• One focus is the parent body, and the other is located at the opposite end of the ellipse, at half the distance from the center as the parent body.
• One focus is the parent body and the other is located outside of the elliptical orbit, on the line on which is the semi-major axis of the ellipse.
• One focus is on the line containing the semi-major axis of the ellipse, and the other is located anywhere on the elliptical orbit of the satellite.

Use the Check Your Answers questions to assess whether students master the learning objectives for this section. If students are struggling with a specific objective, the Check Your Answers will help identify which objective is causing the problem and direct students to the relevant content.

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Chapter 6 Uniform Circular Motion and Gravitation

6.6 Satellites and Kepler’s Laws: An Argument for Simplicity

• State Kepler’s laws of planetary motion.
• Derive the third Kepler’s law for circular orbits.
• Discuss the Ptolemaic model of the universe.

Examples of gravitational orbits abound. Hundreds of artificial satellites orbit Earth together with thousands of pieces of debris. The Moon’s orbit about Earth has intrigued humans from time immemorial. The orbits of planets, asteroids, meteors, and comets about the Sun are no less interesting. If we look further, we see almost unimaginable numbers of stars, galaxies, and other celestial objects orbiting one another and interacting through gravity.

All these motions are governed by gravitational force, and it is possible to describe them to various degrees of precision. Precise descriptions of complex systems must be made with large computers. However, we can describe an important class of orbits without the use of computers, and we shall find it instructive to study them. These orbits have the following characteristics:

• A small mass [latex]{m}[/latex] orbits a much larger mass [latex]{M}.[/latex] This allows us to view the motion as if [latex]{M}[/latex] were stationary—in fact, as if from an inertial frame of reference placed on [latex]{M}[/latex] —without significant error. Mass [latex]{m}[/latex] is the satellite of [latex]{M},[/latex] if the orbit is gravitationally bound.
• The system is isolated from other masses . This allows us to neglect any small effects due to outside masses.

The conditions are satisfied, to good approximation, by Earth’s satellites (including the Moon), by objects orbiting the Sun, and by the satellites of other planets. Historically, planets were studied first, and there is a classical set of three laws, called Kepler’s laws of planetary motion, that describe the orbits of all bodies satisfying the two previous conditions (not just planets in our solar system). These descriptive laws are named for the German astronomer Johannes Kepler (1571–1630), who devised them after careful study (over some 20 years) of a large amount of meticulously recorded observations of planetary motion done by Tycho Brahe (1546–1601). Such careful collection and detailed recording of methods and data are hallmarks of good science. Data constitute the evidence from which new interpretations and meanings can be constructed.

Kepler’s Laws of Planetary Motion

Kepler’s First Law

The orbit of each planet about the Sun is an ellipse with the Sun at one focus.

Kepler’s Second Law

Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times (see Figure 2 ).

Kepler’s Third Law

The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun. In equation form, this is

where [latex]{T}[/latex] is the period (time for one orbit) and [latex]{r}[/latex] is the average radius. This equation is valid only for comparing two small masses orbiting the same large one. Most importantly, this is a descriptive equation only, giving no information as to the cause of the equality.

Note again that while, for historical reasons, Kepler’s laws are stated for planets orbiting the Sun, they are actually valid for all bodies satisfying the two previously stated conditions.

Example 1: Find the Time for One Orbit of an Earth Satellite

Given that the Moon orbits Earth each 27.3 d and that it is an average distance of [latex]{3.84\times10^8\text{ m}}[/latex] from the center of Earth, calculate the period of an artificial satellite orbiting at an average altitude of 1500 km above Earth’s surface.

The period, or time for one orbit, is related to the radius of the orbit by Kepler’s third law, given in mathematical form in [latex]{\frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}}.[/latex] Let us use the subscript 1 for the Moon and the subscript 2 for the satellite. We are asked to find [latex]{T_2}.[/latex] The given information tells us that the orbital radius of the Moon is [latex]{r_1=3.84\times10^8\text{ m}},[/latex] and that the period of the Moon is [latex]{T_1=27.3\text{ d}}.[/latex] The height of the artificial satellite above Earth’s surface is given, and so we must add the radius of Earth (6380 km) to get [latex]{r_2=(1500+6380)\text{ km}=7880\text{ km}}.[/latex] Now all quantities are known, and so [latex]{T_2}[/latex] can be found.

Kepler’s third law is

To solve for [latex]{T_2},[/latex] we cross-multiply and take the square root, yielding

Substituting known values yields

Discussion This is a reasonable period for a satellite in a fairly low orbit. It is interesting that any satellite at this altitude will orbit in the same amount of time. This fact is related to the condition that the satellite’s mass is small compared with that of Earth.

People immediately search for deeper meaning when broadly applicable laws, like Kepler’s, are discovered. It was Newton who took the next giant step when he proposed the law of universal gravitation. While Kepler was able to discover what was happening, Newton discovered that gravitational force was the cause.

Derivation of Kepler’s Third Law for Circular Orbits

We shall derive Kepler’s third law, starting with Newton’s laws of motion and his universal law of gravitation. The point is to demonstrate that the force of gravity is the cause for Kepler’s laws (although we will only derive the third one).

Let us consider a circular orbit of a small mass [latex]{m}[/latex] around a large mass [latex]{M},[/latex] satisfying the two conditions stated at the beginning of this section. Gravity supplies the centripetal force to mass [latex]{m}.[/latex] Starting with Newton’s second law applied to circular motion,

The net external force on mass [latex]{m}[/latex] is gravity, and so we substitute the force of gravity for [latex]{F_{\text{net}}}:[/latex]

The mass [latex]{m}[/latex] cancels, yielding

The fact that [latex]{m}[/latex] cancels out is another aspect of the oft-noted fact that at a given location all masses fall with the same acceleration. Here we see that at a given orbital radius [latex]{r},[/latex] all masses orbit at the same speed. (This was implied by the result of the preceding worked example.) Now, to get at Kepler’s third law, we must get the period [latex]{T}[/latex] into the equation. By definition, period [latex]{T}[/latex] is the time for one complete orbit. Now the average speed [latex]{v}[/latex] is the circumference divided by the period—that is,

Substituting this into the previous equation gives

Solving for [latex]{T^2}[/latex] yields

Using subscripts 1 and 2 to denote two different satellites, and taking the ratio of the last equation for satellite 1 to satellite 2 yields

This is Kepler’s third law. Note that Kepler’s third law is valid only for comparing satellites of the same parent body, because only then does the mass of the parent body [latex]{M}[/latex] cancel.

Now consider what we get if we solve [latex]{T^2=\frac{4\pi^2}{GM}r^3}.[/latex] We obtain a relationship that can be used to determine the mass [latex]{M}[/latex] of a parent body from the orbits of its satellites:

If [latex]{r}[/latex] and [latex]{T}[/latex] are known for a satellite, then the mass [latex]{M}[/latex] of the parent can be calculated. This principle has been used extensively to find the masses of heavenly bodies that have satellites. Furthermore, the ratio [latex]{r^3{/}T^2}[/latex] should be a constant for all satellites of the same parent body (because [latex]{r^3{/}T^2=GM{/}4\pi^2}[/latex] ). (See Table 2 ).

It is clear from Table 2 that the ratio of [latex]{r^3{/}T^2}[/latex] is constant, at least to the third digit, for all listed satellites of the Sun, and for those of Jupiter. Small variations in that ratio have two causes—uncertainties in the [latex]{r}[/latex] and [latex]{T}[/latex] data, and perturbations of the orbits due to other bodies. Interestingly, those perturbations can be—and have been—used to predict the location of new planets and moons. This is another verification of Newton’s universal law of gravitation.

MAKING CONNECTIONS

Newton’s universal law of gravitation is modified by Einstein’s general theory of relativity, as we shall see in Chapter 33 Particle Physics . Newton’s gravity is not seriously in error—it was and still is an extremely good approximation for most situations. Einstein’s modification is most noticeable in extremely large gravitational fields, such as near black holes. However, general relativity also explains such phenomena as small but long-known deviations of the orbit of the planet Mercury from classical predictions.

The Case for Simplicity

The development of the universal law of gravitation by Newton played a pivotal role in the history of ideas. While it is beyond the scope of this text to cover that history in any detail, we note some important points. The definition of planet set in 2006 by the International Astronomical Union (IAU) states that in the solar system, a planet is a celestial body that:

• is in orbit around the Sun,
• has sufficient mass to assume hydrostatic equilibrium and
• has cleared the neighborhood around its orbit.

A non-satellite body fulfilling only the first two of the above criteria is classified as “dwarf planet.”

In 2006, Pluto was demoted to a ‘dwarf planet’ after scientists revised their definition of what constitutes a “true” planet.

The universal law of gravitation is a good example of a physical principle that is very broadly applicable. That single equation for the gravitational force describes all situations in which gravity acts. It gives a cause for a vast number of effects, such as the orbits of the planets and moons in the solar system. It epitomizes the underlying unity and simplicity of physics.

Before the discoveries of Kepler, Copernicus, Galileo, Newton, and others, the solar system was thought to revolve around Earth as shown in Figure 3 (a). This is called the Ptolemaic view, for the Greek philosopher who lived in the second century AD. This model is characterized by a list of facts for the motions of planets with no cause and effect explanation. There tended to be a different rule for each heavenly body and a general lack of simplicity.

Figure 3 (b) represents the modern or Copernican model. In this model, a small set of rules and a single underlying force explain not only all motions in the solar system, but all other situations involving gravity. The breadth and simplicity of the laws of physics are compelling. As our knowledge of nature has grown, the basic simplicity of its laws has become ever more evident.

Section Summary

Kepler’s first law

Kepler’s second law

Each planet moves so that an imaginary line drawn from the Sun to the planet sweeps out equal areas in equal times.

Kepler’s third law

The ratio of the squares of the periods of any two planets about the Sun is equal to the ratio of the cubes of their average distances from the Sun:

where [latex]{T}[/latex] is the period (time for one orbit) and rr size 12{m} {} is the average radius of the orbit.

Conceptual Questions

1: In what frame(s) of reference are Kepler’s laws valid? Are Kepler’s laws purely descriptive, or do they contain causal information?

Problems & Exercises

1: A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). Calculate the radius of such an orbit based on the data for the moon in Table 2 .

2: Calculate the mass of the Sun based on data for Earth’s orbit and compare the value obtained with the Sun’s actual mass.

3: Find the mass of Jupiter based on data for the orbit of one of its moons, and compare your result with its actual mass.

4: Find the ratio of the mass of Jupiter to that of Earth based on data in Table 2 .

5: Astronomical observations of our Milky Way galaxy indicate that it has a mass of about [latex]{8.0\times10^{11}}[/latex] solar masses. A star orbiting on the galaxy’s periphery is about [latex]{6.0\times10^4}[/latex] light years from its center. (a) What should the orbital period of that star be? (b) If its period is [latex]{6.0\times10^7}[/latex] instead, what is the mass of the galaxy? Such calculations are used to imply the existence of “dark matter” in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies.

6: Integrated Concepts

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a satellite in an orbit 900 km above Earth’s surface. (b) Suppose a loose rivet is in an orbit of the same radius that intersects the satellite’s orbit at an angle of [latex]{90^0}[/latex] relative to Earth. What is the velocity of the rivet relative to the satellite just before striking it? (c) Given the rivet is 3.00 mm in size, how long will its collision with the satellite last? (d) If its mass is 0.500 g, what is the average force it exerts on the satellite? (e) How much energy in joules is generated by the collision? (The satellite’s velocity does not change appreciably, because its mass is much greater than the rivet’s.)

7: Unreasonable Results

(a) Based on Kepler’s laws and information on the orbital characteristics of the Moon, calculate the orbital radius for an Earth satellite having a period of 1.00 h. (b) What is unreasonable about this result? (c) What is unreasonable or inconsistent about the premise of a 1.00 h orbit?

8: Construct Your Own Problem

On February 14, 2000, the NEAR spacecraft was successfully inserted into orbit around Eros, becoming the first artificial satellite of an asteroid. Construct a problem in which you determine the orbital speed for a satellite near Eros. You will need to find the mass of the asteroid and consider such things as a safe distance for the orbit. Although Eros is not spherical, calculate the acceleration due to gravity on its surface at a point an average distance from its center of mass. Your instructor may also wish to have you calculate the escape velocity from this point on Eros.

[latex]{1.98\times10^{30}\text{ kg}}[/latex]

[latex]{\frac{M_{\text{J}}}{M_{\text{E}}}=316}[/latex]

a) [latex]{7.4\times10^3\text{ m/s}}[/latex]

b) [latex]{1.05\times10^3\text{ m/s}}[/latex]

c) [latex]{2.86\times10^{-7}\text{ s}}[/latex]

d) [latex]{1.84\times10^7\text{ N}}[/latex]

e) [latex]{2.76\times10^4\text{ J}}[/latex]

a) [latex]{5.08\times10^3\text{ km}}[/latex]

b) This radius is unreasonable because it is less than the radius of earth.

c) The premise of a one-hour orbit is inconsistent with the known radius of the earth.

College Physics Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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