Chapter 7

Circling around Rotational Motion and Orbits

In This Chapter

Working with centripetal acceleration

Feeling the pull of centripetal force

Incorporating angular displacement, velocity, and acceleration

Orbiting with Newton’s laws and gravity

Circular motion can include rockets’ moving around planets, race cars’ whizzing around a track, or bees’ buzzing around a hive. In this chapter, you look at the velocity and acceleration of objects that are moving in circles. This discussion leads to more general forms of rotational motion, where it’s useful to talk about motion in angular terms.

Angular equivalents exist for displacement, velocity, and acceleration. Instead of dealing with linear displacement as a distance, you deal with angular displacement as an angle. Angular velocity indicates what angle you sweep through in so many seconds, and angular acceleration gives you the rate of change in the angular velocity. All you have to do is take linear equations and substitute the angular equivalents: angular displacement for displacement, angular velocity for velocity, and angular acceleration for acceleration.

Centripetal Acceleration: Changing Direction to Move in a Circle

In order to keep an object moving in circular motion, its velocity constantly changes direction. Because velocity changes, you have acceleration. Specifically, you have centripetal acceleration — the acceleration needed to keep an object moving in circular motion. At any point, the velocity of the object is perpendicular to the radius of the circle.

If the string holding the ball in Figure 7-1 breaks at the top, bottom, left, or right moment you see in the illustration, which way would the ball go? If the velocity points to the left, the ball would fly off to the left. If the velocity points to the right, the ball would fly off to the right. And so on. That’s not intuitive for many people, but it’s the kind of physics question that may come up in introductory courses.

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Figure 7-1: Velocity constantly changes direction when an object is in circular motion.

Remember: The velocity of an object in circular motion is always at right angles to the radius of the object’s path. At any one moment, the velocity points along the tiny section of the circle’s circumference where the object is, so the velocity is tangential to the circle.

Keeping a constant speed with uniform circular motion

An object with uniform circular motion travels in a circle with a constant speed. Practical examples may be hard to come by, unless you see a race car driver on a perfectly circular track with his accelerator stuck, a clock with a seconds hand that’s in constant motion, or the moon orbiting the Earth.

Take a look at Figure 7-2, where a golf ball tied to a string is whipping around in circles. The golf ball is traveling at a uniform speed as it moves around in a circle, so you can say it’s traveling in uniform circular motion.

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Figure 7-2: A golf ball on a string traveling with constant speed.

Remember: An object in uniform circular motion doesn’t travel with a uniform velocity because its direction changes all the time.

Any object that travels in uniform circular motion always takes the same amount of time to move completely around the circle. That time is called its period, designated by T.

If you’re swinging a golf ball around on a string at a constant speed, you can easily relate the ball’s speed to its period. You know that the distance the ball must travel each time around the circle equals the circumference of the circle, which is (where r is the radius of the circle), so you can get the equation for finding an object’s period by first finding its speed:

Remember: If you solve for T, you get the equation for the period:

Another time measurement you’ll see in physics problems is frequency. Whereas the period is the time an object takes to go around in a circle, the frequency is the number of circles the object makes per second. The frequency, f, is connected to the period like this:

Example

Q. The moon’s orbital radius is 3.85 × 10⁸ meters, and its period is about 27.3 days. What is its speed as it goes around Earth?

A. The correct answer is 1,020 meters per second, when rounded for significant figures.

1. Convert 27.3 days to seconds:

   

2. Use the equation for the period to solve for speed:

   

3. Plug in the numbers:

   

Practice Questions

1. You have a ball on a string, and you’re whipping it around in a circle. If the radius of its circle is 1.0 meters and its period is 1.0 seconds, what is its speed?

2. You have a toy plane on a wire, and it’s traveling around in a circle. If the radius of its circle is 10.0 meters and its period is 0.75 seconds, what is its speed?

Practice Answers

1. 6.3 m/s. Use the equation for the period to solve for speed:

Plug in the numbers:

2. 84 m/s. Use the equation for the period to solve for speed:

Plug in the numbers:

Finding the magnitude of the centripetal acceleration

When an object travels in uniform circular motion, its speed is constant, which means that the magnitude of the object’s velocity doesn’t change. Therefore, acceleration can have no component in the same direction as the velocity; if it did, the velocity’s magnitude would change.

However, the velocity’s direction is constantly changing — it always bends so that the object maintains movement in a constant circle. To make that happen, the object’s centripetal acceleration is always concentrated toward the center of the circle, perpendicular to the object’s velocity at any one time. The acceleration changes the direction of the object’s velocity while keeping the magnitude of the velocity constant.

In the ball’s case (refer to Figures 7-1 and 7-2), the string exerts a force on the ball to keep it going in a circle — a force that provides the ball’s centripetal acceleration. To provide that force, you have to constantly pull on the ball toward the center of the circle. (Picture what it feels like, force-wise, to whip an object around on a string.) You can see the centripetal acceleration vector, a_c, in Figure 7-2.

If you accelerate the ball toward the center of the circle to provide the centripetal acceleration, why doesn’t it hit your hand? The answer is that the ball is already moving at a high speed. The force, and therefore the acceleration, that you provide always acts at right angles to the velocity.

Remember: You always have to accelerate an object toward the center of the circle to keep it moving in circular motion. So can you find the magnitude of the acceleration you create? No doubt. If an object is moving in uniform circular motion at speed v and radius r, you can find the magnitude of the centripetal acceleration with the following equation:

For a practical example, imagine you’re driving around curves at a high speed. For any constant speed, you can see from the equation that the centripetal acceleration is inversely proportional to the radius of the curve. In other words, on tighter curves (as the radius decreases), your car needs to provide a greater centripetal acceleration (the acceleration increases).

Example

Q. Given that the moon goes around Earth about every 27.3 days and that its distance from the center of Earth is 3.85 × 10⁸ meters, what is the moon’s centripetal acceleration?

A. The correct answer is 2.7 × 10 ^–3 meters per second².

1. Start with this equation:

   

2. Find the speed of the moon. It goes 2πr in 27.3 days, so convert 27.3 to seconds:

   

3. Therefore, the speed of the moon is

   

4. Plug in the numbers:

   

Practice Questions

1. The tips of a helicopter’s blades are moving at 300 meters per second and have a radius of 7.0 meters. What is the centripetal acceleration of those tips?

2. Your ball on a string is revolving around in a circle. If it’s going 60 miles per hour at a radius of 2.0 meters, what is its centripetal acceleration?

Practice Answers

1. 1.3 × 10⁴ m/s². Use this equation:

Plug in the numbers:

2. 360 m/s². Use this equation:

Convert the 60 miles per hour speed to meters per second:

Plug in the numbers:

Getting Angular with Displacement, Velocity, and Acceleration

For objects moving in a circle, you can work with acceleration and velocity using the horizontal and vertical components, just as in previous chapters on motion. But when objects are undergoing rotational motion, using angular variables instead makes a lot of sense. With these variables, instead of specifying the horizontal and vertical components, you specify the radius and the angle of rotation.

In this section, you discover the angular equivalents of displacement, velocity, and acceleration. You can apply these variables to rotating objects and objects moving in a circle.

Measuring angles in radians

Remember: The natural unit of measurement of angles is the radian, not the degree. A full circle is made up of radians, which is also 360°, so radians. If you travel in a full circle, you go 360°, or radians. (If an object rotates one revolution, then the angle has magnitude of radians. Therefore, sometimes instead of radians per second, you see revolutions per second.) A half-circle is radians, and a quarter-circle is radians.

The radian is a natural measure of angle because a circular arc that has a length of one radius extends an angle of 1 radian (see Figure 7-3). So if you know the radius and the angle that an object has moved through in radians, you can easily find the distance that the object has moved in proportion to the radius. If the object moves radians in a circle of radius r, then the object travels a distance of along the circle.

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Figure 7-3: A circular arc extends an angle of one radian.

This idea is useful in relating the angular velocity to the speed of an object moving in a circle. In addition, you can see why a full circle has an angle of radians: You know that the circumference of a circle is and that to go the whole way around the 360° of a circle, you need to travel times the radius. Therefore, there are radians to 360°.

Tip: How do you convert from degrees to radians and back again? Because radians (or 2 multiplied by 3.14, the rounded version of pi), you have an easy calculation.

Example

Q. Convert 180° into radians.

A. The correct answer is π.

1. Use the conversion factor π/180°.
2. Plug in the numbers:

   

Practice Questions

1. What is 23° in radians?

2. What is π/16 in degrees?

Practice Answers

1. 0.40 radians. Use the conversion factor π/180°.

Plug in the numbers:

2. 11.2°. Use the conversion factor 180°/π.

Plug in the numbers:

Relating linear and angular motion

Tip: The fact that you can think of the angle, , in rotational motion just as you think of the displacement, s, in linear motion is great, because it means you have an angular counterpart for many of the linear motion equations (see Chapter 3). Here are the variable substitutions you make to get the angular motion formulas:

• Displacement: Instead of s, which you use in linear travel, use , the angular displacement; is measured in radians.
• Velocity: In place of the velocity, v, use the angular velocity, ; angular velocity is the number of radians covered per second.
• Acceleration: Instead of acceleration, a, use the angular acceleration, ; the unit for angular acceleration is radians per second².

Table 7-1 compares the formulas for both linear and angular motion.

Table 7-1 Linear and Angular Motion Formulas

Type of Formula	Linear	Angular
Velocity
Acceleration
Displacement
Motion with time canceled out

Getting into Angular Velocity

The symbol for angular velocity is ω, so you can write the equation for angular velocity this way (this is actually a vector equation, of course, but we don’t get into the vector nature of angular motion until Chapter 11, so we look at this equation in scalar terms):

Figure 7-4 shows a line sweeping around in a circle. At a particular moment, it’s at angle θ, and if it took time t to get there, its angular velocity is ω = θ/t.

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Figure 7-4: Angular velocity in a circle.

So if the line in Figure 7-5 completes a full circle in 1.0 seconds, its angular velocity is 2π/1.0 s = 2π radians/s (because there are 2π radians in a complete circle). Technically speaking, radian isn’t a physical unit of measure (it’s a ratio), so the angular velocity can also be written 2π s^–1.

Given the angular velocity, you also can find the angle swept through in a number of seconds:

Example

Q. The moon goes around Earth in about 27.3 days. What is its angular velocity?

A. The correct answer is 2.66 × 10^–6 radians per second.

1. Convert 27.3 days to seconds:

   

2. Use the equation for angular velocity:

   

3. Plug in the numbers:

   

Practice Questions

1. You have a toy plane on a string that goes around three complete circles in 9.0 seconds. What is its angular velocity?

2. A satellite is orbiting Earth at 8.7 × 10^–4 radians per second. How long will it take to circle the entire world?

Practice Answers

1. 2.1 radians/s. Use the equation for the angular velocity:

Plug in the numbers:

2. 120 min. Start with the equation for the angular velocity:

Solve for Δt:

Plug in the numbers:

That’s about 120 minutes.

Whipping Around with Angular Acceleration

Just as with linear motion, you can have acceleration when you’re dealing with angular motion. For example, the line in Figure 7-5 may be sweeping around the circle faster and faster, which means it’s accelerating.

In linear motion, the following is the equation for acceleration, the rate at which the object’s velocity is changing:

As with all the equations of motion, you need only to substitute the correct angular quantities for the linear ones. In this case, v becomes ω. So the angular acceleration is Δω/Δt.

The symbol for linear acceleration is a, and the symbol for angular acceleration is α, which makes the equation for angular acceleration as follows:

The unit for angular acceleration is radians per second² (or, technically, just seconds^–2).

Example

Q. Your toy plane on a string accelerates from ω = 2.1 radians per second to 3.1 radians per second in 1.0 seconds. What is its angular acceleration?

A. The correct answer is 1.0 radians per second².

1. Use the equation for angular acceleration:

   

2. Plug in the numbers:

   

Practice Questions

1. You’re square dancing, turning your partner around at 1.0 radians per second. Then you speed up for 0.50 seconds at an angular acceleration of 10.0 radians per second². What is your partner’s final angular speed?

2. You’re trying a new yoga move, and, starting your arm at rest, you accelerate it at 15 radians per second² over 1.0 second. What’s your arm’s final angular velocity?

Practice Answers

1. 6.0 radians/s. Use the equation for angular acceleration:

Solve for Δω:

Plug in the numbers:

Add Δω to the initial angular velocity, ω_i:

2. 15 radians/s. Use the equation for angular acceleration:

Solve for Δω:

Plug in the numbers:

Add Δω to the initial angular velocity, ω_i:

Connecting Angular Velocity and Angular Acceleration to Angles

You can connect the distance traveled to the original velocity and linear acceleration like this:

And you can make the substitution from linear to angular motion by putting in the appropriate symbols:

Using this equation, you can connect angular velocity, angular acceleration, and time to the angle.

Example

Q. A marble is rolling around a circular track at 6.0 radians per second and then accelerates at 1.0 radian per second². How many radians has it gone through in 1 minute?

A. The correct answer is 2,200 radians.

1. Use this equation:

   

2. Plug in the numbers:

   

Practice Questions

1. Your model globe is spinning at 1.0 radians per second when you give it a push. If you accelerate it at 5.0 radians per second², how many radians has it turned through in 5.0 seconds?

2. A roulette wheel is slowing down, starting from 12.0 radians per second and going through 40.0 radians in 5.0 seconds. What was its angular acceleration?

Practice Answers

1. 68 radians. Use this equation:

Plug in the numbers:

2. –1.6 radians/s². Use this equation:

Solve for α, given that t_o = 0:

Plug in the numbers:

Connecting Angular Acceleration and Angle to Angular Velocity

You can connect angle, angular velocity, and angular acceleration. The corresponding equation for linear motion is

Substituting ω for v, α for a, and θ for s gives you:

Use this equation when you want to relate angle to angular velocity and angular acceleration.

Example

Q. A merry-go-round slows down from 6.5 radians per second to 2.5 radians per second, undergoing an angular acceleration of 1.0 radians per second². How many radians does the merry-go-round go through while this is happening?

A. The correct answer is 18 radians.

1. Start with the equation:

   

2. Solve for θ:

   

3. Plug in the numbers:

   

Practice Questions

1. A helicopter’s blades are speeding up. They go from 60 radians per second to 80 radians per second. If the angular acceleration is 10 radians per second², what is the total angle the blades have gone through?

2. Your ball on a string is traveling around in a circle. If it goes from 12 radians per second to 24 radians per second and the angular acceleration is 20 radians per second², what is the total angle the ball has gone through during this acceleration?

Practice Answers

1. 140 radians. Use this equation:

Solve for θ:

Plug in the numbers:

2. 11 radians. Use this equation:

Solve for θ:

Plug in the numbers:

Seeking the Center: Centripetal Force

When you’re driving a car around a bend, you create centripetal acceleration by the friction of your tires on the road. How do you know what force you need to create to turn the car at a given speed and turning radius? That depends on the centripetal force — the center-seeking, inward force needed to keep an object moving in uniform circular motion.

In this section, you discover how the centripetal force keeps the object moving in a circle and how the details of the circular motion, such as radius and velocity, depend upon the centripetal force.

Looking at the force you need

Centripetal force isn’t some new force that appears out of nowhere when an object travels in a circle; it’s the force the object needs to keep traveling in that circle.

As you know from Newton’s first law (see Chapter 5), if there’s no net force on a moving object, the object will continue to move uniformly in a straight line. If a force (or a component of a force) acts in the same direction as the object’s velocity, then the object begins to speed up, and if the force acts in the opposite direction to the velocity, then the object slows down. However, if the force always acts perpendicularly to the velocity while remaining of constant magnitude, then the magnitude of the velocity (the speed) doesn’t change; only its direction does — the object moves in a circle. In this case, the force is called centripetal force.

If you’re spinning a ball on a string, then the centripetal force comes from the tension in the string. When the moon orbits the Earth, the centripetal force comes from gravity. And when you drive your car in a circle, the centripetal force comes from the friction of the tires against the road. The origin of the force isn’t important, only that it remains of constant magnitude and always acts perpendicularly to the velocity, toward the center of the circle.

Seeing how the mass, velocity, and radius affect centripetal force

Because force equals mass times acceleration, , and because centripetal acceleration is equal to (see the earlier section “Finding the magnitude of the centripetal acceleration”), you can determine the magnitude of the centripetal force needed to keep an object moving in uniform circular motion with the following equation:

This equation tells you the magnitude of the force that you need to move an object of a given mass, m, in a circle at a given radius, r, and speed, v. (Remember that the direction of the force is always toward the center of the circle.)

Think about how force is affected if you change one of the other variables. The equation shows that if you increase mass or speed, you’ll need a larger force; if you decrease the radius, you’re dividing by a smaller number, so you’ll also need a larger force. Here’s how these ideas play out in the real world:

• Increasing mass: You may have an easy time swinging a golf ball on a string in a circle, but if you replace the golf ball with a cannonball, watch out. You may now have to whip 10 kilograms around on the end of a 1.0-meter string every half-second. As you can tell, you need a heck of a lot more force.
• Increasing speed: Not interested in spinning cannonballs? Then imagine you’re driving your car around in a circle. If you’re going quite slowly around the circle, your tires have no problem generating enough frictional force to keep you going in the circle. But if you go too fast, then your tires can no longer generate the frictional force acting toward the center of the circle, so you start to skid.
• Decreasing the radius: You can see the effect of the radius in your car going around in a circle. If you drive your car at a fixed speed in a circle of smaller and smaller radius, eventually your tires won’t be able to supply enough centripetal force from the friction, and you’ll skid off the circular path.

The fictitious centrifugal force

You’ve probably heard of centrifugal force and most likely have felt it when a car you were in turned a corner. However, centrifugal force isn’t really a force as defined in Newton’s laws. It only appears to be a force. When you’re in a car turning a corner, your body has inertia and is naturally inclined to move at uniform speed in a straight line. But because the car is turning, it feels as though your body is being thrown outward, toward the car door.

Example

Q. The moon goes around Earth about every 27.3 days with a distance from Earth of 3.85 × 10⁸ meters. If the moon’s mass is 7.35 × 10²² kilograms, what is the centripetal force that Earth’s gravity exerts on it as it orbits Earth?

A. The correct answer is 2.0 × 10²⁰ newtons.

1. Start with this equation:

   

2. Find the speed of the moon. It goes 2πr in 27.3 days, so convert 27.3 days to seconds:

   

3. Therefore, the speed of the moon is

   

4. Plug in the numbers:

   

Practice Questions

1. You’re exerting a force on a string to keep a ball on a string going in a circle. If the ball has a mass of 0.10 kilograms and the speed of the ball is 16.0 meters per second at a distance of 2.0 meters, what is the centripetal force you need to apply to keep the ball going in a circle?

2. You have a 1.0-kilogram toy plane on the end of a 10-meter wire, and it’s going around at 60.0 meters per second. What is the force you have to apply to the wire to keep the plane going in a circle?

Practice Answers

1. 13 N. Use the equation for centripetal force:

Plug in the numbers:

2. 360 N. Use the equation for centripetal force:

Plug in the numbers:

Negotiating flat curves and banked turns

Imagine that you’re driving a car and you come to a curve. On a flat road, the centripetal force you need to negotiate the curve comes from the friction of the tires against the ground. If the surface is covered with a substance such as ice, you have less friction, and you can’t turn as safely at high speeds.

To make turns safer, engineers design roads so that curves are banked. With the road at an angle, there’s a component of the normal force of the road against your car, toward the center of the circle. This means that you don’t require as much friction from your tires to make the turn.

When you’re driving on a flat road, friction provides the centripetal force — toward the center of the circle — that allows you to make a turn.

Say you’re sitting in the passenger seat of the car, approaching a turn with a level, non-banked road surface. What’s the maximum speed the driver can go and still keep you safe? The frictional force needs to supply the centripetal force, so you come up with the following:

where m is the mass of the car, v is the velocity, r is the radius, is the coefficient of static friction (you use the coefficient of static friction if the wheels aren’t slipping), and g is the acceleration due to gravity, 9.8 meters per second².

If a curve is banked, then a component of the normal force of the road against the car contributes to the centripetal force, and so you can go around the curve much faster. Because you don’t have to rely on friction to supply the centripetal force, the question of whether you can safely make the turn no longer depends on road conditions.

Take a look at Figure 7-5, which shows a car banking around a turn. The engineers can make the driving experience enjoyable if they bank the turn so that drivers garner the centripetal force needed to go around the turn entirely by the component of the normal force of the road against the car acting toward the center of the turn’s circle. That component is (F_N is the normal force, the upward force perpendicular to the road; see Chapter 6), so

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Figure 7-5: The forces acting on a car banking around a turn.

To find the centripetal force, you need the normal force, F_N. If you look at Figure 7-3, you can see that F_N comes from a combination of the centripetal force due to the car’s banking around the turn and the car’s weight. The purely vertical component of F_N must equal mg, because no other forces are operating vertically, so

Plugging this result into the equation for centripetal force gives you

Because , you can also write this as

Solve for to find the angle of the road. The equation finally breaks down to

Tip: You don’t have to memorize this result, in case you’re panicking — this is the kind of equation used by highway engineers when they have to bank curves (notice that the mass of the car cancels out, meaning that it holds for vehicles regardless of weight). You can always derive this equation from your knowledge of Newton’s laws and circular motion.

Example

Q. What should the angle be if drivers go around a 200-meter-radius turn at 60.0 miles per hour?

A. The correct answer is 20°.

1. Use this equation:

   

2. Plug in the numbers: 60 miles per hour is about 27 meters per second, and the radius of the turn is 200 meters, so

   

   The designers should bank the turn at about 20° to give drivers a smooth experience.

Practice Questions

1. At what angle should a banked road be tilted if the magnitude of the force exerted by the ground on the car is the same as that exerted by gravity on the car? Give your answer in degrees.

2. What is the minimum velocity a 1,200-kilogram car must travel on an icy, banked road with a radius of curvature of 36 meters and an inclination of 9.0° to prevent it from slipping down the bank if the coefficient of static friction is 0.08?

Practice Answers

1. 8.1 m/s. The normal force — the force exerted by a surface on an object touching it — is equal to the force of gravity only when the surface is exactly horizontal. The angle of inclination is thus 0°.

2. 0°. Start by drawing the forces acting on the car:

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Because the car isn’t accelerating upward or downward, the sum of the vertical forces must equal 0:

Use ( is the coefficient of friction) and F_G=mg to solve for the normal force:

The sum of the horizontal forces provides the centripetal force keeping the car in its circular motion. Friction is pointing away from the “center of the circle” and is therefore antiparallel to the centripetal force:

Now use the result you got earlier for the normal force and the centripetal force equation, , where m is the mass of the object, v is the object’s velocity, and r is the radius (of the curve the object is traversing), to solve the equation.

Letting Gravity Supply Centripetal Force

You don’t have to tie objects to strings to observe travel in circular motion; larger bodies such as planets move in circular motion, too. Gravity provides the necessary centripetal force.

In this section, you discover Newton’s take on the gravitational forc e between two objects, and we show you how his theory relates to 9.8 meters per second², the value experimenters identified as the acceleration due to gravity near the surface of the Earth. Then you put Newton’s formula to use in looking at the orbits of satellites.

Using Newton’s law of universal gravitation

Remember: Sir Isaac Newton came up with one of the heavyweight laws in physics for you: the law of universal gravitation. This law says that every mass exerts an attractive force on every other mass. If the two masses are m₁ and m₂ and the distance between them is r, the magnitude of the force is

where G is a constant equal to .

Example

Q. Two people are sitting on a park bench, looking at each other and smiling. If the two lovebirds have masses of about 75 kilograms each, what’s the force of gravity pulling them together, assuming they start out 0.50 meters away?

A. The correct answer is .

Use Newton’s law of universal gravitation:

Practice Questions

1. What is the gravitational force between the sun and the Earth? The sun has a mass of about kilograms, and the Earth has a mass of about kilograms. A distance of about meters separates the two bodies.

2. Calculate the force of gravity between Earth and its moon. Approximate the distance between the center of Earth and the center of the moon as . The moon has a mass of about , and the Earth has a mass of about kilograms.

Practice Answers

1. . Plug the numbers into Newton’s equation:

2. . Convert the Earth to moon distance to meters to match the units in the gravitational constant, G:

Plug the numbers into your calculator:

Deriving the force of gravity on the Earth’s surface

The equation for the force of gravity — — holds true no matter how far apart two masses are. But you also come across a special gravitational case (which most of the work on gravity in this book is about): the force of gravity near the surface of the Earth.

The gravitational force between a mass and the Earth is the object’s weight. Mass is considered a measure of an object’s inertia, and its weight is the force exerted on the object in a gravitational field. On the surface of the Earth, the two forces are related by the acceleration due to gravity: . Kilograms and slugs are units of mass; newtons and pounds are units of weight.

You can use Newton’s law of gravitation to get the acceleration due to gravity, g, on the surface of the Earth just by knowing the gravitational constant G, the radius of the Earth, and the mass of the Earth. The force on an object of mass m₁ near the surface of the Earth is

This force is provided by gravity between the object and the Earth, according to Newton’s gravity formula, so you can write

The radius of the Earth, r_e, is about meters, and the mass of the Earth is kilograms. Putting in the numbers, you have

Dividing both sides by m₁ gives you the acceleration due to gravity:

Newton’s law of gravitation gives you the acceleration due to gravity near the surface of the Earth: 9.8 meters per second².

Of course, you can measure g by letting an apple drop and timing it, but what fun is that when you can calculate it in a roundabout way that requires you to first measure the mass of the Earth?

Using the law of gravitation to examine circular orbits

In space, bodies are constantly orbiting other bodies due to gravity. Satellites (including the moon) orbit the Earth, the Earth orbits the sun, the sun orbits around the center of the Milky Way, the Milky Way orbits around the center of its local group of galaxies. This is big-time stuff. In the case of orbital motion, gravity supplies the centripetal force that causes the orbits.

The force of gravity between orbiting bodies is quite a bit different from small-time orbital motion — such as when you have a ball on a string — because for a given distance and two masses, the gravitational force is always going to be the same. You can’t increase the force to increase the speed of an orbiting planet as you can with a ball. The following sections examine the speed and period of orbiting bodies in space.

A particular satellite can have only one speed when in orbit around a particular body at a given distance because the force of gravity doesn’t change. So what’s that speed? You can calculate it with the equations for centripetal force and gravitational force. You know that for a satellite of a particular mass, m₁, to orbit, you need a corresponding centripetal force (see the section “Seeking the Center: Centripetal Force”):

This centripetal force has to come from the force of gravity, so

You can rearrange this equation to get the speed:

This equation represents the speed that a satellite at a given radius must have in order to orbit if the orbit is due to gravity. The speed can’t vary as long as the satellite has a constant orbital radius — that is, as long as it’s going around in circles. This equation holds for any orbiting object where the attraction is the force of gravity, whether it’s a human-made satellite orbiting the Earth or the Earth orbiting the sun. If you want to find the speed for satellites that orbit the Earth, for example, you use the mass of the Earth in the equation:

Remember: Here are a few details you should note on reviewing the orbiting speed equation:

• You have to use the distance from the center of the Earth, not the distance above Earth’s surface, as the radius. Therefore, the distance you use in the equation is the distance between the two orbiting bodies. In this case, you add the distance from the center of the Earth to the surface of the Earth, meters, to the satellite’s height above the Earth.
• The equation assumes that the satellite is high enough off the ground that it orbits out of the atmosphere. That assumption isn’t really true for artificial satellites; even at 400 miles above the surface of the Earth, satellites do feel air friction. Gradually, the drag of friction brings them lower and lower, and when they hit the atmosphere, they burn up on re-entry. When a satellite is less than 100 miles above the surface, its orbit decays appreciably each time it circles the Earth. (Look out below!)
• The equation is independent of mass. If the moon rather than the artificial satellite orbited at 400 miles and you could ignore air friction and collisions with the Earth, it would have to go at the same speed as the satellite in order to preserve its close orbit (which would make for some pretty spectacular moonrises).

You can think of a satellite in motion around the Earth as always falling. The only thing that keeps it from striking the Earth is that its velocity points over the horizon. The satellite is falling, but its velocity takes it over the horizon — that is, over the curve of the world as it falls — so it doesn’t get any closer to the Earth. (The same is true of the astronauts inside. They only have the appearance of being weightless, but they’re continuously falling, too.)

Sometimes it’s more important to know the period of an orbit rather than the speed, such as when you’re counting on a satellite to come over the horizon before communication can take place. The period of a satellite is the time it takes it to make one full orbit around an object. The period of the Earth as it travels around the sun is one year. If you know the satellite’s speed and the radius at which it orbits (see the preceding section), you can figure out its period. The satellite travels around the entire circumference of the circle — which is if r is the radius of the orbit — in the period, T. This means the orbital speed must be , giving you

If you solve this for the period of the satellite, you get

Understanding Kepler’s laws of orbiting bodies

Johannes Kepler (1571–1630), a German national born in the Holy Roman Empire, came up with three laws that helped explain a great deal about orbits before Newton came up with his law of universal gravitation. Here are Kepler’s laws:

1. Law 1: Planets orbit in ellipses. An ellipse is a shape like a squashed circle, and the degree of squashedness is called the eccentricity of the ellipse. The orbits allowed can have any degree of eccentricity. When the eccentricity is zero, you have a circular orbit.
2. Law 2: Planets move so that a line between the sun and the planet sweeps out the same area in the same time, independent of where they are in their orbits. This means that when the planet is in that part of its orbit where it is close to the sun, it has to travel faster to sweep out the same area that it does when it’s farther away.
3. Law 3: The square of a planet’s orbital period (the time it takes the planet to make one complete orbit) is proportional to its average distance from the sun cubed.

The third law can be derived from Newton’s laws. It takes the form of this equation: . Although Kepler’s third law says that T ² is proportional to r ³, you can get the exact constant relating these quantities by using Newton’s law of gravitation.

Example

Q. Human-made satellites typically orbit at heights of 400 miles from the surface of the Earth (about 640 kilometers, or meters). What’s the speed of such a satellite?

A. The correct answer is meters per second or 16,800 miles per hour.

1. All you have to do is put in the numbers into the satellite speed equation:

   

2. This converts to about 16,800 miles per hour.

Practice Questions

1. How far above the surface of the Earth is a satellite whose period is the same as the Earth’s 24-hour period?

2. If the International Space Station (ISS) is located 420 kilometers above Earth’s surface, how many hours does it take to make a complete orbit?

Practice Answers

1. meters. In cases of stationary satellites, the period, T, is 24 hours, or about 86,400 seconds.

Solve the equation for periods for the radius:

Plug in the numbers:

Subtract the Earth’s radius of meters to get meters. At this distance, geosynchronous satellites orbit the Earth at the same rate the Earth is turning, which means that they stay put over the same piece of real estate.

2. 1.55. Use the equation relating orbital period to orbital position, .

Find the distance in meters between the ISS and the center of the Earth:

Substitute that into the orbital equation:

Convert to hours: