# Chapter 5

This chapter was copied with permission from Nick Strobel’s Astronomy Notes. Go to his site at www.astronomynotes.com for the updated and corrected version.

# Newton’s Law of Gravity

Newton’s laws of motion and his law of gravity are discussed. Applications of those laws (esp. gravity) are covered (e.g., measuring the masses of planets and stars, orbital motion, interplanetary trips, tides, etc.).

## 5.1 Introduction This chapter covers the revolutionary advancements due to probably the most brilliant scientist who ever lived: Isaac Newton (lived 1641–1727). His greatest contributions were in all branches of physics. Kepler’s discoveries about elliptical orbits and the planets’ non-uniform speeds made it impossible to maintain the idea of planetary motion as a natural one requiring no explanation. Newton had to answer some basic questions: What keeps the planets in their elliptical orbits? On our spinning Earth what prevents objects from flying away when they are thrown in the air? What keeps you from being hurled off the spinning Earth? Newton’s answer was that a fundamental force called “gravity” operating between all objects made them move the way they do.

Newton developed some basic rules governing the motion of all objects. He used these laws and Kepler’s laws to derive his unifying Law of Gravity. I will first discuss his three laws of motion and then discuss gravity. Finally, several applications in astronomy will be given. This chapter uses several math concepts that are reviewed in the mathematics review appendix. If your math skills are rusty, study the mathematics review appendix and don’t hesitate to ask for help. The vocabulary terms are in boldface.

I include images of world atlases from different time periods in this chapter and the previous one as another way to illustrate the advances in our understanding of our world and the universe. Links to the sites from which the photographs came are embedded in the images. Select the picture to go to the site (will display in another window). Nova orbis tabula [De Wit 1688]. Select the image to go to the Hargrett Library at the Univ. of Georgia from which this picture came. Note the curved path of the Sun between the Tropic of Cancer (at latitude 23.5° N) and the Tropic of Capricorn (at latitude 23.5° S).

## 5.2 Newton’s Laws of Motion

In order to accurately describe how things move, you need to be careful in how you describe the motion and the terms you use. Scientists are usually very careful about the words they use to explain something because they want to accurately represent nature. Language can often be imprecise and as you know, statements can often be misinterpreted. Because the goal of science is to find the single true nature of the universe, scientists try to carefully choose their words to accurately represent what they see. That is why scientific papers can look so “technical” (and even, introductory astronomy textbooks!)

When you think of motion, you may first think of something moving at a uniform speed. The speed = (the distance traveled)/(the time it takes). Because the distance is in the top of the fraction, there is a direct relation between the speed and the distance: the greater the distance traveled in a given time, the greater is the speed. However, there is an inverse relation between time and speed (time is in the bottom of the fraction): the smaller the time it takes to cover a given distance, the greater the speed must be.

To more completely describe all kinds of changes in motion, you also need to consider the direction along with the speed. For example, a ball thrown upward at the same speed as a ball thrown downward has a different motion. This inclusion of direction will be particularly important when you look at an object orbiting a planet or star. They may be moving at a uniform speed while their direction is constantly changing. The generalization of speed to include direction is called velocity. The term velocity includes both the numerical value of the speed and the direction something is moving. Galileo conducted several experiments to understand how something’s velocity can be changed. He found that an object’s velocity can be changed only if a force acts on the object. The philosopher René Descartes (lived 1596–1650, picture at left) used the idea of a greater God and an infinite universe with no special or privileged place to articulate the concept of inertia: a body at rest remains at rest, and one moving in a straight line maintains a constant speed and same direction unless it is deflected by a “force.” Newton took this as the beginning of his description of how things move, so this is now known as Newton’s 1st law of motion. A force causes a change in something’s velocity (an acceleration).

An acceleration is a change in the speed and/or direction of motion in a given amount of time: acceleration= (the velocity change)/(the time interval of the change). Something at rest is not accelerating and something moving at constant speed in a straight line is not accelerating. In common usage, acceleration usually means just a change in speed, but a satellite orbiting a planet is constantly being accelerated even if its speed is constant because its direction is constantly being deflected. The satellite must be experiencing a force since it is accelerating. That force turns out to be gravity. If the force (gravity) were to suddenly disappear, the satellite would move off in a straight line along a path tangent to the original circular orbit. A rock in your hand is moving horizontally as it spins around the center of the Earth, just like you and the rest of the things on the surface are. If you throw the rock straight up, there is no change in its horizontal motion because of its inertia. You changed the rock’s vertical motion because you applied a vertical force on it. The rock falls straight down because the Earth’s gravity acts on only the rock’s vertical motion. If the rock is thrown straight up, it does not fall behind you as the Earth rotates. Inertia and gravity also explain why you do not feel a strong wind as the Earth spins—as a whole, the atmosphere is spinning with the Earth.

Newton’s first law of motion is a qualitative one—it tells you when something will accelerate. Newton went on to quantify the amount of the change that would be observed from the application of a given force. In Newton’s second law of motion, he said that the force applied = mass of an object × acceleration. Mass is the amount of material an object has and is a way of measuring how much inertia the object has. For a given amount of force, more massive objects will have a smaller acceleration than less massive objects (a push needed to even budge a car would send a pillow flying!). For a given amount of acceleration, the more massive object requires a larger force than a less massive object. Newton also found that for every action force ON an object, there is an equal but opposite force BY the object (Newton’s third law of motion). For example, if Andre the Giant is stuck on the ice with Tom Thumb and he pushes Tom Thumb to the right, Andre will feel an equal force from Tom pushing him to the left. Tom will slide to the right with great speed and Andre will slide to the left with smaller speed since Andre’s mass is larger than Tom’s. Another example: an apple falls to the Earth because it is pulled by the force of the Earth’s gravity on the apple and the acceleration of the apple is large. The apple also exerts a gravitational force on the Earth of the same amount. However, the acceleration the Earth experiences is vastly smaller than the apple’s acceleration since the Earth’s mass is vastly larger than the apple’s—you will ordinarily refer to the apple falling to the Earth, rather than the Earth moving toward the apple or that they are falling toward each other.

### Vocabulary

• acceleration
• force
• inertia
• mass
• Newton’s 1st law
• Newton’s 2nd law
• Newton’s 3rd law
• velocity

### Formulae

Newton’s 2nd law: Force = mass × acceleration: F = m × a

### Review Questions 1

1. What 2 things can change for an acceleration?
2. If you give a bowling ball a push FAR away from any gravitational effects, what will it do? If you throw a feather (again far out in space) at the same speed as the bowling ball, how will its speed compare to the bowling ball after 5 minutes?
3. Let’s say you’re twirling a ball on a string and the string breaks. What path does the ball take and why is that?
4. How do you know gravity acts on an orbiting satellite?
5. How does a force exerted on an object relate to the object’s mass or acceleration? Given the same force will a boulder accelerate more than a regular marble? Why?
6. Why would you need to apply more force to a bowling ball than a feather (far out in space) so that they would be travelling at the same speed after 10 minutes?

## 5.3 Universal Law of Gravity

Using Kepler’s third law and his own second law, Newton found that the amount of the attractive force, called gravity, between a planet and Sun a distance d apart is Force = kp × (planet mass) / (d)2, where kp is a number that is the same for all the planets. In the same way he found that the amount of the gravity between the Sun and a planet is Force = ks × (Sun mass) / (d)2. Using his third law of motion, Newton reasoned that these forces must be the same (but acting in the opposite directions). He derived his Law of Gravity: the force of gravity = G × (mass #1) × (mass #2) / (distance between them)2 and this force is directed toward each object, so it is always attractive. The term G is a universal constant of nature. If you use the units of kilograms (kg) for mass and meters (m) for distance, G = 6.672 × 10-11 m3 /(kg sec2). If you need a refresher on exponents, square & cube roots, and scientific notation, then please study the math review appendix.

Spherically symmetric objects (eg., planets, stars, moons, etc.) behave as if all of their mass is concentrated at their centers. So when you use Newton’s Law of Gravity, you measure the distance between the centers of the objects. In a bold, revolutionary step, Newton stated that his gravity law worked for any two objects with mass—it applies for any motions on the Earth, as well as, any motions in space. He unified celestial and terrestrial physics and completed the process started by Copernicus of removing the Earth from a unique position or situation in the universe. His law of gravity also explained Kepler’s 1st and 2nd laws.

The UNL Astronomy Education program’s Planetary Orbit Simulator shows you how a planet’s velocity and acceleration relate to its orbits and Kepler’s laws (link will appear in a new window).

### 5.3.1 Characteristics of Gravity

Newton’s Law of Gravity says a lot about this force in a very compact, elegant way. It says that any piece of matter will feel it whether it is charged or not (this sets it apart from electrical and magnetic forces that affect only charged objects). Gravity depends only on the masses of the two attracting objects and their distance from each other. It does not depend on their chemical composition or density. A glob of peanut butter the mass of the Sun will have the same gravitational effect on the Earth as the Sun does. Gravity is always attractive, never repulsive (this is another way it is different from electrical and magnetic forces).

Because the masses are in the top of the fraction, more mass creates more gravity force. This also means that more massive objects produce greater accelerations than less massive objects. Since distance is in the bottom of the fraction, gravity has an inverse relation with distance: as distance increases, gravity decreases. However, gravity never goes to zero—it has an infinite range (in this respect it is like the electrical and magnetic forces). Stars feel the gravity from other stars, galaxies feel gravity from other galaxies, galaxy clusters feel gravity from other galaxies, etc. The always attractive gravity can act over the largest distances in the universe.

There is no way to get rid of the force of gravity. If you want to prevent a body from producing a gravitational acceleration on an object, you need to use a second body, with the same amount of gravity pull as the first body, in a way that its gravity pulling on the object is in the opposite direction. The resulting accelerations due to the forces from the two bodies will cancel each other out.

### Review Questions 2

1. What basic fundamental assumption did Newton make about the laws of nature on the Earth and in space?
2. Why is gravity often the most important force in astronomical interactions?
3. What things does gravity depend on?
4. How does gravity vary with distance between objects and with respect to what do you measure the distances?
5. What would happen to the orbit of Io (one of Jupiter’s moons) if all of the Hydrogen and Helium in Jupiter were converted to Silicon and Oxygen? Explain your answer.
6. What would happen to the Earth’s orbit if the Sun suddenly turned into a black hole (of the same mass)? Why?
7. How would antimatter respond to gravity? (Hint: antimatter has mass just like ordinary matter.)
8. What important laws of planet motion can be derived from Newton’s law of gravity?

## 5.4 Mass vs. Weight

Though the terms weight and mass are used interchangeably in common language, in science there is distinct difference between the two terms. The weight of an object = force of gravity felt by that object but the mass of an object is the amount of matter the object has. Mass is a measure of the object’s resistance to acceleration: a push on a skateboard will make it roll away quickly but the same push on a more massive car will barely budge it.

An object’s weight depends on the pull of the gravitating object but the object’s mass is independent of the gravity. For example, Joe Average weighs himself on the Earth’s surface and then on the Moon’s surface. His weight on the Moon will be about six times less than on the Earth but the number of atoms in his body has not changed so his mass is the same at the two places. In the old English unit system, there is a “pound” of force and “pound” of mass. On only the Earth’s surface, an object’s pound of mass = the number of pounds of force felt by the object due to the Earth’s gravity.

In the metric system there is no confusion of terms. A kilogram is a quantity of mass and a newton is a quantity of force. One kilogram (kg) = 2.205 pounds of mass and 4.45 newtons (N) = 1 pound of force. If someone uses “pounds,” be sure you understand if s/he means force or mass!

### How do you do that?

To find something’s weight in newtons, you multiply the mass in kilograms by the acceleration of gravity in the units of meters/seconds2. For example: Joe Average has a mass of 63.5 kg and he feels a force of gravity on the Earth = 63.5 kg × 9.8 m/s2 = 623 kg m/s2 = 623 N. His weight is 623 N. The other value in the preceding equation, 9.8 m/s2, is the acceleration due to gravity close to the Earth’s surface. Joe Average’s weight at other places in the universe will be different but his mass will remain the same. • kilogram
• mass
• newton
• weight

### Review Questions 3

1. What is the difference between mass and weight?
2. When the astronauts landed on the Moon, how were they able to stay on the ground?
3. On the Moon, the astronauts weighed about six times less than they did on the Earth. Compare the amount of gravity on the Moon’s surface with that on the Earth’s surface. If objects fall with an acceleration of about 10 m/s2 on the Earth, how much would the acceleration be on the Moon’s surface? Explain your answer.
4. If Joe Astronaut has a mass of 40 kilograms on the Earth, how much mass would he have on an asteroid with 10 times less surface gravity than the Earth’s surface gravity? Explain your answer.

## 5.5 Inverse Square Law

Newton’s law of gravity describes a force that decreases with the SQUARE of the distance. For every factor of 2 the distance increases, the gravitational attraction decreases by a factor of 2 × 2 = 4; for every factor of 3 increase in distance, the gravity decreases by a factor of 3 × 3 = 9 (not by 3 + 3 = 6!); for every factor of 4 increase in distance, the gravity decreases by a factor of 4 × 4 = 16 (not by 4 + 4 = 8!), etc. See the mathematics review appendix for a review of “factor” and “times.” Some more examples are given in the table below. Notice how quickly an inverse square law gets very small.

A comparison of inverse and inverse square relations

distance inverse inverse square
1 1/1 = 1 1/12 = 1
2 1/2 = 0.5 1/22 = 1/4 = 0.25
3 1/3 = 0.33 1/32 = 1/9 = 0.11
4 1/4 = 0.25 1/42 = 1/16 = 0.0625
7 1/7 = 0.14 1/72 = 1/49 = 0.02
10 1/10 = 0.1 1/102 = 1/100 = 0.01
100 1/100 = 0.01 1/1002 = 1/10,000 = 0.0001

Example: Joe Average has a mass of 63.5 kilograms, so he weighs 623 newtons (=140 pounds) on the Earth’s surface. If he moves up 1 Earth radius (= 6378 kilometers) above the surface, he will be two times farther away from the Earth’s center (remember that distances are measured from center-to-center!), so his weight will be four times less, or 623/4 newtons = 155.8 newtons (= 140/4 pounds); NOT two times less, or 623/2 newtons = 311.5 newtons. If he moves up another Earth radius above the surface, he will be three times farther away than he was at the start, so his weight will drop by a factor of nine times, NOT 3 times. His weight will be 623/9 newtons = 69.22 newtons (= 140/9 pounds); NOT 623/3 newtons = 207.7 newtons. His mass will still be 63.5 kilograms. Figure below illustrates this.

Let us generalize this for any situation where the masses do not change: the force of gravity at distance A = (the force of gravity at distance B) × (distance B / distance A)2. Notice which distance is in the top of the fraction! To use this relation, have the gravity at distance A represent the unknown gravity force you are trying to find and the gravity at distance B represent the reference gravity force felt at the reference distance B. ### How do you do that?

Let’s find where the weight values in the inverse square law figure come from.

For Mr. Average’s case the reference weight is his weight on the surface of the Earth = 623 N. His weight at 6378 kilometers above the surface is gravity at A = 623 × [6378/(2 × 6378)]2 = 623 × 1/22 = 623 × 1/4 = 155.8 N.

When he is at two Earth radii above the surface, the gravity at A = 623 × [6378/(3 × 6378)]2 = 623 × 1/32 = 623 × 1/9 = 69.22 N.

### Formulae

Inverse Square Law: Gravity at A = gravity at B × (distance B / distance A)2.

### Review Questions 4

1. Why is gravity called an “inverse square law”?
2. What is the difference between a simple inverse relation and an inverse square relation?
3. If the Earth was 3 A.U. from the Sun (instead of 1 A.U.), would the gravity force between the Earth and the Sun be less or more than it is now? By how many times?
4. If Mercury was 0.2 A.U. from the Sun (instead of 0.4 A.U.), would the gravity force between Mercury and the Sun be less or more than it is now? By how many times?

## 5.6 Gravitational Acceleration

Galileo found that the acceleration due to gravity (called “g“) depends only on the mass of the gravitating object and the distance from it. It does not depend on the mass of the object being pulled. In the absence of air drag, a huge boulder will fall at the same rate as a small marble dropped from the same height as the boulder. A tiny satellite at the same distance from the Sun as Jupiter’s orbit from the Sun feels the same acceleration from the Sun as the large planet Jupiter does from the Sun. How is this possible? Most people would agree with Aristotle that the bigger object should fall faster than the smaller object, but experiments show they would be wrong.

A boulder falling toward the Earth is pulled by a stronger gravity force than the marble, since the boulder’s mass is greater than the marble, but the boulder also has greater resistance to a change in its motion because of its larger mass. The effects cancel each other out, so the boulder accelerates at the same rate as the marble. The same line of reasoning explains the equal acceleration experienced by Jupiter and the satellite.

You can use Newton’s second law of motion F = m × a (which relates the acceleration, a, felt by a object with mass m when acted on by a force F) to derive the acceleration due to gravity (here replace a with g) from a massive object:

The force of gravity = (G M m)d2 = mg,

so

g = (G M) / d2.

The gravitational acceleration depends on only the mass of the gravitating object M and the distance d from it. Notice that the mass of the falling object m has been cancelled out. This explains why astronauts orbiting the Earth feel “weightless.” In orbit they are continually “falling” toward the Earth because of gravity (the Earth’s surface curves away from them at the same rate they are moving forward). If Jane Astronaut drops a pen in the space shuttle, it accelerates toward the Earth, but she accelerates by the same amount so the pen remains at the same position relative to her.In fact the entire shuttle and its contents are accelerating toward the Earth at the same rate, so Jane and her companions “float” around inside! This is because all of them are at very nearly the same distance from the Earth. The acceleration decreases with the SQUARE of the distance (inverse square law). To compare gravity accelerations due to the same object at different distances, you use the gravity acceleration g at distance A = (the gravity acceleration g at distance B) × (distance B / distance A)2. Notice which distance is in the top of the fraction. An example of using the inverse square law is given in the “How do you do that?” box below.

### How do you do that?

Find how many times more gravitational acceleration the Galileo atmosphere probe felt at 100,000 miles from Jupiter’s center than the orbiter felt at 300,000 miles. You have

 probe’s g = orbiter’s g × (300,000/100,000)2 = orbiter’s g × (3/1)2 = orbiter’s g × 9.

The probe accelerated by an amount nine times greater than the orbiter.

### 5.6.1 Measuring the Mass of the Earth

Measuring the acceleration of an object dropped to the ground enables you to find the mass of the Earth. You can rearrange the gravity acceleration relation to solve for the mass M to find M = g d2/G. Close to the Earth’s surface at a distance of 6.4 × 106 meters from the center, g = 9.8 meters/second2. The distance is given in meters to match the units of the gravity acceleration — when you do a calculation, you must be sure you check that your units match up or you will get nonsense answers. The big G is the universal gravitational constant, approximately 6.7×10-11 m3/(kg sec2). Plugging in the values, you will find the Earth’s mass = 9.8 × (6.4×106)2 / (6.7 × 10-11) kilograms = 6.0 × 1024 kilograms. If you are unsure of how to work with scientific notation, read the scientific notation section in the mathematics review appendix (pay close attention to the part describing how to enter scientific notation on your calculator!).

You can determine masses of stars and planets in a similar way: by measuring the acceleration of objects orbiting them and the distance between the star or planet and the object. A small object falling to the Earth has mass and, therefore, has a gravitational acceleration associated with it: the Earth is accelerated toward the falling object (an example of Newton’s third law)! However, if you plug some typical masses of terrestrial objects (less than, say, 1000 kilograms) into the acceleration formula, you will see that the amount the Earth is accelerated is vastly smaller than the falling object’s acceleration. You can ignore the Earth’s acceleration.

A side note: determining the mass of the Earth also depends on knowing the value of the gravitational constant G. The constant was first measured by Henry Cavendish in 1798. After discussing his experimental results, he then applied his measurement to the subject of his paper’s title: “Weighing the Earth.”

### Formulae

1. Gravitational Acceleration: g = (G × Mass)/(distance from the center)2.
2. Comparing gravitational accelerations: acceleration at position A = acceleration at position B × (distance B/distance A)2.
3. Calculating mass: Mass = (g × distance2)/G.

### Review Questions 5

1. What did Galileo discover about how objects of different masses fall to the Earth?
2. If you dropped a hammer and feather from the same height above the Earth’s surface, which would actually hit the ground first? Why would it be different than what Galileo said about falling objects? Explain why if you let the feather fall quill end first, the result is closer to what Galileo said.
3. If you dropped a hammer and feather from the same height above the airless Moon’s surface, which would actually hit the ground first? Explain why your answer is different than for the previous question.
4. How many times less/more gravity acceleration due to the Sun does the Ulysses spacecraft feel at 2.3 A.U. above the Sun than the solar gravity acceleration it felt at Jupiter (5.2 A.U. from the Sun)? Is it accelerated more or less at 2.3 A.U. than when it was at 5.2 A.U.?
5. Why do astronauts in orbit around the Earth feel “weightless” even though the Earth’s gravity is still very much present?
6. Put the following in order of their acceleration around the Earth: a 200-ton space station 6580 kilometers from the center, a 60-kilogram astronaut 6580 kilometers from the center, a 1-ton satellite 418,000 kilometers from the center, and the 7.4×1019-ton Moon 384,000 kilometers from the center. Explain your answer.
7. How can you find the mass of the Earth using ordinary objects in your house?

## 5.7 A Closer Look at Newton’s Gravity

Newton found that his gravity law is obeyed everywhere in the universe and could explain Kepler’s three laws of orbital motion. Newton’s development of the unifying law of gravity was also the culmination of a process of Occam’s Razor in action. From Ptolemy to Newton, the theories of how the planets move got simpler and more powerful as time went on. Ptolemy’s model had become extremely complicated by the time of the Renaissance and Copernicus reduced the number of circular motions to around 50 so it was simpler to use. Kepler vastly simplified the theory of planet motion by reducing the number of essential parts to just three laws. Newton unified all of those laws to the ONE unifying law of gravity. This law was so simple and elegant that it could also explain motions on the Earth.

But what is gravity? Newton understood how the gravity force affected the motion of objects but not why gravity worked the way it did. Recognizing the limits of his knowledge, he adopted an instrumentalist view: the scientist’s job is to capture observations in precise mathematical equations; explain the “how” not the “why.” Only things verified by our experience of the world are admissible into science. Though the “why” question is intriguing and a few scientists will spend years trying to answer it, most scientists share Newton’s instrumentalist view.

With Newton, there was no longer a hierarchical-teleological universe (one designed by God for some purpose with man playing a crucial role in the plan). The universe was now a perfect machine, based on mathematics, set in motion by God long ago. God is the reference point for absolute space and time. Newtonian mechanics requires an absolute coordinate system to keep things sensible (according to Newton this also gave God something to do).

With the success of Newton’s ideas, a major change occurred in how people viewed the world around them. Reality was completely reduced to material objects. Ideas, thought, feelings, and values were secondary. Newtonism undercut the role of God and religion and the validity of science: science became just a subjective perspective of the machine universe.

Descartes saw the need to rescue thoughts, ideas and values. He developed a mind-body dualism: a world of thought and spirit exists independent of, but parallel to, the material world. There is a correspondence between the God-inaugurated, mathematical thoughts of scientists and the motions in the physical world. Descartes said that mathematical ideas work so well because there is a pre-established parallelism between the physical world and the human mind. What is real does NOT depend on us — this is probably the actual completion of the Copernican revolution and was soon so widely accepted that it became “common sense” (how about that for a paradigm shift!).

### Review Questions 6

1. What important discoveries and ideas did Newton make?
2. How does Occam’s Razor relate to the progress of planet motion theory from Ptolemy to Newton?
3. How can Newton’s work be considered the completion of the process started by Copernicus almost 120 years earlier?

## 5.8 Orbits

Now I will apply Newton’s laws of motion and gravity to topics more astronomical: objects moving around other objects. What kinds of things can you find out about celestial objects from just observing their motions?

### 5.8.1 Centripetal Force Newton’s first law of motion says that an object’s inertia will keep it from changing its speed and/or direction unless some force acts on it. This means that satellites orbiting the Earth must be feeling some force that constantly deflects them toward the center of the Earth. If there was no force, they would travel in a straight line at a constant speed.

If you whirl a ball attached to string around your head, it moves in a circular path around you because the string is always pulling the ball directly toward the hand grabbing the string. The ball wants to move in a straight line and the string is pulling it directly inward. The resulting deflection is a compromise: a circular path. The string is applying a centripetal force to the ball: an inward force. If you let go of the string, there is no centripetal force and the ball will fly off in a straight line because of its inertia. If you do not whirl the ball fast enough it will move inward to a smaller non-circular path around you. If you whirl the ball too fast, you may not be able to give it enough centripetal force to keep it in a circular path around you. The amount of centripetal force needed to balance an object’s inertia and keep it in a circular path of radius r is found from Newton’s second law: the centripetal force = m v2 / r, where v and m are the object’s speed and mass, respectively. The radius of the orbit r is the same as the distance between the moving object and the central body.

### 5.8.2 Measuring Planet and Star Masses

Now for orbits! Satellites are not being deflected by strings but by gravity. Gravity provides the centripetal force needed to keep the satellites in orbit. If you focus on the simple case of circular orbits, you can use the centripetal force formula above with the law of gravity to determine the mass of a planet or star. Simply set the force of gravity equal to the centripetal force and solve for the mass of the planet or star, M.

(G M m) / r2 = m v2 / r.

The satellite mass m cancels out from both sides and if you put M on one side and the rest on the other side of the equation, you get

M = (r2 / G) × (v2 / r)

M = r v2 / G.

This assumes that the satellite’s mass, m, is much less than the central object’s mass so you can ignore the acceleration of the central object toward the satellite!

### How do you do that?

Let’s use this result to get an estimate of the mass of the Sun. You need to use something orbiting with a known radius and speed. The Earth’s orbit is roughly circular with radius = 1.5 × 1011 meters and the Earth moves with a speed 30,000 meters/second (= 30 km/s) in its orbit. The distance is given in meters to match the units of the speed. The distance unit of a meter is used because you will be using the gravitational constant in your calculation and it uses the meter unit. When you do a calculation, you must be sure you check that your units match up or you will get nonsense answers.

Plug the values into the mass relation:

the Sun’s mass = (30,000)2 × (1.5 × 1011)/(6.7 × 10-11) = 2 × 1030 kilograms. This is much larger than the Earth’s mass so it was okay to ignore the Sun’s movement toward the Earth. Using no approximations (ie., not assuming a circular orbit and including the Sun’s motion toward the Earth) gives a value for the Sun’s mass that is very close to this. Your answer does not depend on which planet you choose to use (here you used the Earth’s orbit). You would get the same value for the mass of the Sun if you had used any of the other planets orbital speeds and sizes.

This relation tells you what you need to know in order to measure a planet’s or star’s mass: the orbital speed of a satellite and the distance it is from the center of the planet or star. Because the velocity is on top of the fraction, satellites are made to move faster if the mass of the central object is greater. At the same distance, a massive planet will exert more gravity force than a low-mass planet, so the massive planet will produce greater inward accelerations on satellites orbiting it. The satellites will, therefore, orbit at faster speeds.

Sometimes the orbital period P is measured instead of the orbital velocity. The orbital period is the time it takes the object to travel the circumference of its orbit (for a circle, the circumference = 2πr, where π is approximately 3.1416). Recall that speed = (distance traveled)/time, so the speed v = the circumference of the orbit/orbital period. When you substitute this for the speed in the mass relation above, you get

M = (4π2 / G) × (r3 / P2)

This may look familiar to you—it is Kepler’s third law! There is a distance cubed, an orbital period squared, and some other constant factors. Newton found that when Kepler used the motions of the planets to formulate his third law, Kepler was actually measuring the mass of the Sun. If you use the convenient set of units of an astronomical unit for the distance, a year for the time, and a “solar mass” (mass relative to the Sun) for the mass, the complicated term (4π2)/G becomes the simple value of 1. For the planets orbiting the Sun, the mass relation becomes 1 = r3 / P2, or r3 = P2, just what Kepler found. ### 5.8.3 Orbital Speed

The mass formula above tells you that satellites orbiting massive planets must move faster than satellites orbiting low-mass planets at the same distance. Massive planets have stronger gravity than low-mass planets so a satellite orbiting a massive planet is accelerated by a greater amount than one going around a lesser mass planet at the same distance. To balance the stronger inward gravitational pull of the massive planet, the satellite must move faster in its orbit than if it was orbiting a lesser mass planet. Of course, this also applies to planets orbiting stars, stars orbiting other stars, etc.

If you solve for the orbit speed, v, in the mass formula, you can find how fast something needs to move to balance the inward pull of gravity:

v2 = (G M)/r .

Taking the square root of both sides (you want just v not v2), you get

v = Sqrt[(G M)/r].

### How do you do that?

Find the orbital speed of Jupiter around the Sun. Jupiter’s distance from the Sun is 5.2 A.U., or 7.8×1011 meters and the Sun’s mass is 2×1030 kilograms. The orbital speed of Jupiter around the Sun is Sqrt[6.7×10-11 × (2×1030)/(7.8×1011)] = Sqrt[1.718×108] = 1.3×104 meters/second, or 13 kilometers/second. What do you think you would find if you used one of the Trojan asteroids, millions of times less massive than Jupiter, that orbits the Sun at 5.2 A.U.?

Notice that the orbital speed v (and therefore, the orbital period P) does not depend on the mass of the satellite! The mass M in the formula above is the mass of the central object. If we consider the Sun-Jupiter-Trojan asteroid example in the “How do you do that?” box above (and recall the discussion in the Gravitational Acceleration section), the Sun-Jupiter gravity force is MUCH greater than the Sun-Trojan asteroid gravity force but Jupiter has much greater inertia (resistance to change = mass). The greater gravity force between the Sun and Jupiter is compensated for by the greater inertia of Jupiter so the acceleration of Jupiter will be the same as the acceleration of the Trojan asteroid—Jupiter and the Trojan asteroid will have the same orbital velocity and same orbital period.

### 5.8.4 Escape Velocity

If an object moves fast enough it can escape a massive object’s gravity and not be drawn back toward the massive object. The critical speed needed to do this is the escape velocity. More specifically, this is the initial speed something needs to escape the object’s gravity and assumes that there is no other force acting on the object besides gravity after the initial boost. Rockets leaving the Earth do not have the escape velocity at the beginning but the engines provide thrust for an extended period of time, so the rockets can eventually escape. The concept of escape velocity applies to anything gravitationally attracted to anything else (gas particles in planet atmospheres, comets orbiting the Sun, light trying to escape from black holes, galaxies orbiting each other, etc.). Using Newton’s laws of motion and law of gravity, you can find that the escape velocity vesc looks very similar to the orbital speed:

vesc = Sqrt[(2 G M)/r].This is a factor Sqrt larger than the circular orbital speed. Since the mass M is on top of the fraction, the escape velocity increases as the mass increases. More massive bodies exert greater gravity force, so escaping objects have to move faster to overcome the greater gravity. Also, the distance from the center of the object r is in the bottom of the fraction, so the escape velocity DEcreases as the distance increases. Gravity decreases with greater distance, so objects farther from a massive body do not need to move as quickly to escape it than those closer to it.

### How do you do that?

Find the escape velocity from the surface of the Earth. Using the acceleration of gravity, you can find that the Earth has a mass of 6.0×1024 kilograms. The Earth’s radius is 6.4×106meters. Since the mass and distance from the center are in the standard units, you just need to plug their values into the escape velocity relation.The Earth’s surface escape velocity is Sqrt[2× (6.7×10-11) × (6.0×1024)/ (6.4×106)] = Sqrt[1.256×108] = 1.1×104 meters/second (= 11 km/s). Here are some other surface escape velocities: Moon = 2.4 km/s, Jupiter = 59.6 km/s, Sun = 618 km/s.

### Vocabulary

• centripetal force
• escape velocity

### Formulae

1. Mass of central object = [(orbital speed)2 × distance)]/G
M = v2 r / G
2. Mass of central object (Kepler’s 3rd law) = (4π2)/G × [(distance)3/(orbital period)2]
M = (4 π2) / G (r3 / P2)
3. Orbital speed = Sqrt[G × Mass / distance] (orbital speed)
v = Sqrt[ G M / r ]
4. Escape velocity = Sqrt[2G × Mass / distance]
v = Sqrt[ 2 G M / r ]

### Review Questions 7

1. What keeps satellites orbiting the Earth moving along their curved paths?
2. What two things must be determined first in order to calculate the mass of a planet or a star?
3. Jupiter’s moon Io has about the same mass as the Moon and orbits Jupiter at about the same distance that the Moon orbits the Earth (center to center). Then why does Io take only 1.8 days to orbit Jupiter but our Moon takes 27.3 days to orbit the Earth?
4. Astronomers were able to accurately measure the orbital periods of the moons of Jupiter since the time of Galileo, so why was an accurate value for Jupiter’s mass not found for over 300 years until the astronomical unit was measured accurately?
5. Which would have a shorter orbital period, a planet orbiting a massive star at 3 A.U. or a planet orbiting a low-mass star at 3 A.U.? Explain your answer.
6. If a planet orbiting a massive star has the same orbital period as a planet orbiting a low-mass star, which of the planets orbits at a greater distance from its star? Explain your answer.
7. What two things does the escape velocity depend on?
8. Why does the planet Saturn with over 95 times the Earth’s mass have a smaller escape velocity at its cloudtops than the Earth has at its cloudtops?
9. Why is Jupiter’s escape velocity at its cloudtops over two times higher than the Earth’s surface escape velocity, even though Jupiter has a much larger diameter than the Earth?

## 5.9 Kepler’s Third Law

Kepler’s third law of planetary motion says that the average distance of a planet from the Sun cubed is directly proportional to the orbital period squared. Newton found that his gravity force law could explain Kepler’s laws. Since Newton’s law of gravity applies to any object with mass, Kepler’s laws can be used for any object orbiting another object. Let’s look at satellites orbiting a planet.

If you have two satellites (#1 and #2) orbiting a planet, Kepler’s third law says:

(period #1/period #2)2 = (distance #1/distance #2)3,where the distance is the average distance of the satellite from the planet—the orbit’s semimajor axis. The satellites must be orbiting the same planet in order to use Kepler’s third law! Kepler found this law worked for the planets because they all orbit the same star (the Sun).

If you have measured the orbital period of one satellite around a planet, you can then easily find how long it would take any other satellite to orbit the planet in any size orbit. Kepler’s third law can be simplified down to

period #1 = period #2 × Sqrt[(distance #1/distance #2)3]

or

period #1 = period #2 × (distance #1/distance #2)3/2.

Those of you with a scientific calculator (one that does powers, trig functions, scientific notation, etc.) will want to use the formula on the last line (remember that 3/2 = 1.5). Those with a calculator that just has a square root button will want to use the formula on the second-to-last line.

If the satellite is orbiting the Sun, then the relation can be greatly simplified with an appropriate choice of units: the unit of years for the orbit period and the distance unit of astronomical units. In this case, the reference “satellite” is the Earth and Kepler’s third law becomes period = distance3/2. Let’s use this to find out how long it takes to explore the solar system.

### 5.9.1 Interplanetary Trips

The simplest way to travel between the planets is to let the Sun’s gravity do the work and take advantage of Kepler’s laws of orbital motion. A fuel efficient way to travel is to put the spacecraft in orbit around the Sun with the Earth at one end of the orbit at launch and the other planet at the opposite end at arrival. These orbits are called “Hohmann orbits” after Walter Hohmann who developed the theory for transfer orbits. The spacecraft requires only an acceleration at the beginning of the trip and a deceleration at the end of the trip to put it in orbit around the other planet. Let’s go to Mars! The relative positions of Earth and Mars must be just right at launch so that Mars will be at the right position to greet the spacecraft when it arrives several months later. These good positionings happen once every 780 days (the synodic period of Mars). The spacecraft must be launched within a time interval called the “launch window” that is just few of weeks long to use a Hohmann orbit for the spacecraft’s path. The Earth is at the perihelion (point closest to the Sun) of the spacecraft orbit (here, 1.0 A.U.) and Mars is at the aphelion (point farthest from the Sun—here, 1.52 A.U.).

Kepler’s third law relates the semi-major axis of the orbit to its sidereal period. The major axis is the total length of the long axis of the elliptical orbit (from perihelion to aphelion). For the Mars journey, the major axis = 1.52 + 1.0 A.U. = 2.52 A.U. The semi-major axis is one-half of the major axis, so divide the major axis by two: 2.52/2 = 1.26 A.U. Now apply Kepler’s third law to find the orbital period of the spacecraft = 1.263/2 = 1.41 years. This is the period for a full orbit (Earth to Mars and back to Earth), but you want to go only half-way (just Earth to Mars). Traveling from Earth to Mars along this path will take (1.41 / 2) years = 0.71 years or about 8.5 months.

When the craft is launched, it already has the Earth’s orbital velocity of about 30 kilometers/second. Since this is the speed for a circular orbit around the Sun at 1.0 A.U., a reduction in the spacecraft’s speed would make it fall closer to the Sun and the Hohmann orbit would be inside the Earth’s orbit. Since you want to go beyond the Earth’s orbit, the spacecraft needs an increase in its speed to put it in an orbit that is outside the Earth’s orbit. It will slow down gradually as it nears aphelion.

At aphelion the spacecraft will not be traveling fast enough to be in a circular orbit at Mars’ distance (1.52 A.U.) so it will need to arrive at aphelion slightly before Mars does. Mars will then catch up to it. But the spacecraft will be moving much too fast to be in a circular orbit around Mars, so it will need to slow down to go in orbit around Mars.

On its journey to Mars, the spacecraft’s distance from the Sun is continuously monitored to be sure the craft is on the correct orbit. Though the spacecraft responds mostly to the Sun’s gravity, the nine planets’ gravitational pulls on the spacecraft can affect the spacecraft’s path as it travels to Mars, so occasional minor firings of on-board thrusters may be required to keep the craft exactly on track.

• aphelion
• perihelion

### Formulae

1. Kepler’s third law: period #1 = period #2 × Sqrt[(distance #1/distance #2)3]
P1 = P2 Sqrt[ (r1 / r2)3 ]
2. Kepler’s third law: period #1 = period #2 × (distance #1/distance #2)3/2
P1 = P2 (r1 / r2)3/2
3. If considering objects orbiting the Sun, measure the orbit period in years and the distance in A.U. With these units, Kepler’s third law is simply: period = distance3/2.
P = r3/2

### Review Questions 8

1. How can you predict the orbital period of Jupiter’s satellite Europa from observations of the other Jovian moon Io?
2. If Io takes 1.8 days to orbit Jupiter at a distance of 422,000 kilometers from its center, find out how long it would take Europa to orbit Jupiter at 671,000 kilometers from its center.
3. If the Moon were twice as far from the Earth as it is now, how long could a solar eclipse last? (Solar eclipses currently last up to about two hours from the start of the cover-up to when the Moon no longer blocks the Sun at all.)
4. The Hubble Space Telescope orbits the Earth 220 kilometers above the surface and takes about 1.5 hours to complete one orbit. How can you find out how far up to put a communication satellite, so that it takes 24 hours to circle the Earth? (Such an orbit is called a “geosynchronous orbit” because the satellite remains above a fixed point on the Earth.)
5. Why does NASA not launch interplanetary spacecraft when the planets are at opposition (closest to the Earth)?
6. Find out how long it will take the Cassini spacecraft to travel to Saturn 9.5 A.U. from the Sun.

## 5.10 Tides

When you look in the paper at the section containing the tide tables, you will often see the phase of the moon indicated as well. That is because the ocean tides are caused by different strengths of the Moon’s gravity at different points on the Earth. The side of the Earth facing the Moon is about 6400 kilometers closer to the Moon than the center of the Earth is, and the Moon’s gravity pulls on the near side of the Earth more strongly than on the Earth’s center. This produces a tidal bulge on the side of the Earth facing the Moon. The Earth rock is not perfectly rigid; the side facing the Moon responds by rising toward the Moon by a few centimeters on the near side. The more fluid seawater responds by flowing into a bulge on the side of the Earth facing the Moon. That bulge is the high tide.

At the same time the Moon exerts an attractive force on the Earth’s center that is stronger than that exerted on the side away from the Moon. The Moon pulls the Earth away from the oceans on the far side, which flow into a bulge on the far side, producing a second high tide on the far side. These tidal bulges are always along the Earth-Moon line and the Earth rotates beneath the tidal bulge. When the part of the Earth where you are located sweeps under the bulges, you experience a high tide; when it passes under one of the depressions, you experience a low tide. An ideal coast should experience the rise and fall of the tides twice a day. In reality, the tidal cycle also depends on the latitude of the site, the shape of the shore, winds, etc.

The Sun’s gravity also produces tides that are about half as strong as the Moon’s and produces its own pair of tidal bulges. They combine with the lunar tides. At new and full moon, the Sun and Moon produce tidal bulges that add together to produce extreme tides. These are called spring tides (the waters really spring up!). When the Moon and Sun are at right angles to each other (1st & 3rd quarter), the solar tides reduce the lunar tides and you have neap tides.  ### 5.10.1 Tides Slow Earth Rotation

As the Earth rotates beneath the tidal bulges, it attempts to drag the bulges along with it. A large amount of friction is produced which slows down the Earth’s spin. The day has been getting longer and longer by about 0.0016 seconds each century.

Over the course of time this friction can have a noticeable effect. Astronomers trying to compare ancient solar eclipse records with their predictions found that they were off by a significant amount. But when they took the slowing down of the Earth’s rotation into account, their predictions agreed with the solar eclipse records. Also, growth rings in ancient corals about 400 hundred million years old show that the day was only 22 hours long so that there were over 400 days in a year. In July 1996 a research study reported evidence, from several sedimentary rock records providing an indicator of tidal periods, that the day was only 18 hours long 900 million years ago.

Eventually the Earth’s rotation will slow down to where it keeps only one face toward the Moon. Gravity acts both ways so the Earth has been creating tidal bulges on the Moon and has slowed it’s rotation down so much that it rotates once every orbital period. The Moon keeps one face always toward the Earth.

Here is a list of references about the evidence for the slowing down of the Earth’s rotation:

1. Growth Rhythms and the History of the Earth’s rotation, edited by G.D. Rosenberg and S.K. Runcorn (Wiley: New York, 1975). An excellent source on the eclipse records and the biology of coral and their use as chronometers.
2. Tidal Friction and the Earth’s Rotation, edited by P. Brosche and J. Sundermann (Springer Verlag, 1978). The second volume put out in 1982 does not talk about eclipse records or the use of coral but, instead, goes into the astrophysics of the Earth-Moon dynamics and geophysics of internal Earth processes effects on the Earth’s rotation.
3. Earth’s Rotation from Eons to Days, edited by P. Brosche and J. Sundermann (Springer Verlag, 1990). Has several articles about the use of ancient Chinese observations.
4. Richard Monastersky 1994, Ancient tidal fossils unlock lunar secrets in Science News vol. 146, no. 11, p. 165 of the 10 Sept 1994 issue.
5. C. P. Sonett, E. P. Kvale, A. Zakharian, Marjorie A. Chan, T. M. Demko 1996, Late Proterozoic and Paleozoic Tides, Retreat of the Moon, and Rotation of the Earth in Science vol 273, no. 5271, p. 100 of the 05 July 1996 issue. ### 5.10.2 Tides Enlarge Moon Orbit

Friction with the ocean beds drags the tidal bulges eastward out of a direct Earth-Moon line and since these bulges contain a lot of mass, their gravity pulls the moon forward in its orbit. The increase in speed enlarges the Moon’s orbit. Currently, the Moon’s distance from the Earth is increasing by about 3 centimeters per year. Astronomers have been able to measure this slow spiraling out of the Moon by bouncing laser beams off reflectors left by the Apollo astronauts on the lunar surface.

The consequence of the Moon’s recession from the Earth because of the slowing down of the Earth’s rotation is also an example of the conservation of angular momentum. Angular momentum is the amount of spin motion an object or group of objects has. It depends on the geometric size of the object or group of objects, how fast the object (or group of objects) is moving, and the mass of the object (or the group). Since the Earth’s angular momentum is decreasing, the Moon’s angular momentum must increase to keep the overall angular momentum of the Earth-Moon system the same. The concept of angular momentum is discussed further in the Angular Momentum appendix.

The slow spiraling out of the Moon means that there will come a time in the future when the angular size of the Moon will be smaller than the Sun’s and we will not have any more total solar eclipses! Fifty billion years in the future the Earth day will equal 47 of our current days and the Moon will take 47 of our current days to orbit the Earth. Both will be locked with only one side facing the other — people on one side of the Earth will always see the Moon while people on the other side will only have legends about the Moon that left their pleasant sky.

### 5.10.3 Tidal Effects Elsewhere

Tidal effects are larger for more massive objects and at closer distances. The Sun produces a tidal bulge on the planet Mercury (the planet closest to the Sun) and has slowed that planet’s rotation period so it rotates three times for every two times it orbits the Sun (a “3 – 2 spin-orbit resonance”). Jupiter’s moon, Io, orbits at about the same distance from Jupiter’s center as the Earth’s moon. Jupiter is much more massive than the Earth, so Jupiter’s tidal effect on Io is much greater than the Earth’s tidal effect on the Moon. Io is stretched by varying amounts as it orbits Jupiter in its elliptical orbit. This tidal flexing of the rock material creates huge amounts of heat from friction in Io’s interior which in turn is released in many volcanic eruptions seen on Io. Galaxies passing close to each other can be severely stretched and sometimes pulled apart by mutual tidal effects.

### Vocabulary

• conservation of angular momentum
• neap tide
• spring tide

### Review Questions 9

1. What causes the tides?
2. How are tides related to the position of the Moon and Sun with respect to the Earth?
3. Why are there two high tides roughly every 12.5 hours? Explain why there are two tidal bulges AND why they are over 12 hours apart.
4. At what phases do spring tides occur?
5. At what phases do neap tides occur?
6. How are tides responsible for the slowing down of the Earth’s spin and the Moon’s spiraling away from us?
7. Where are some other places that tides play a significant role in the appearance and motion of objects?