## 5 Steps to a 5: AP Physics C (2016)

### STEP __4__

### Review the Knowledge You Need to Score High

### CHAPTER 15

### Gravitation and Circular Motion

**IN THIS CHAPTER**

**Summary:** When an object moves in a circle, it is accelerating toward the center of the circle. Any two massive objects attract each other due to gravity.

**Key Ideas**

Centripetal (center-seeking) acceleration is equal to .

Circular motion (and gravitation) problems still require a free-body diagram and the four-step problem-solving process.

The gravitational force between two objects is bigger for bigger masses, and smaller for larger separations between the objects.

Kepler”s laws apply to the orbits of planets.

**Relevant Equations**

Centripetal acceleration:

Gravitational force between any two masses:

Gravitational potential energy a long way from a planet:

It might seem odd that we”re covering gravitation and circular motion in the same chapter. After all, one of these topics relates to the attractive force exerted between two massive objects, and the other one relates to, well, swinging a bucket over your head.

However, as you”re probably aware, the two topics have plenty to do with each other. Planetary orbits, for instance, can be described only if you understand both gravitation and circular motion. And, sure enough, the AP exam frequently features questions about orbits.

So let”s look at these two important topics, starting with circular motion.

**Velocity and Acceleration in Circular Motion**

Remember how we defined acceleration as an object”s change in velocity in a given time? Well, velocity is a vector, and that means that an object”s velocity can change either in magnitude or in direction (or both). In the past, we have talked about the magnitude of an object”s velocity changing. Now, we discuss what happens when the direction of an object”s velocity changes.

When an object maintains the same speed but turns in a circle, the magnitude of its acceleration is constant and directed **toward the center** of the circle. This means that the acceleration vector is perpendicular to the velocity vector at any given moment, as shown in __Figure 15.1__ .

**Figure 15.1 Velocity and acceleration of an object traveling in uniform circular motion.**

The velocity vector is always directed tangent to the circle, and the acceleration vector is always directed toward the center of the circle. There”s a way to prove that statement mathematically, but it”s complicated, so you”ll just have to trust us. (You can refer to your textbook for the complete derivation.)

**Centripetal Acceleration**

On to a few definitions.

**Centripetal acceleration:** The acceleration keeping an object in uniform circular motion, abbreviated *a* _{c}

We know that the net force acting on an object is related to the object”s acceleration by *F* _{net} = *ma* . And we know that the acceleration of an object in circular motion points toward the center of the circle. So we can conclude that the centripetal force acting on an object also points toward the center of the circle.

The formula for centripetal acceleration is

In this equation, *v* is the object”s velocity, and *r* is the radius of the circle in which the object is traveling.

**Centrifugal acceleration:** As far as you”re concerned, nonsense. Acceleration in circular motion is always **toward** , not away from, the center.

Centripetal acceleration is real; centrifugal acceleration is nonsense, unless you”re willing to read a multipage discussion of “non-inertial reference frames” and “fictitious forces.” So for our purposes, there is no such thing as a centrifugal (center-fleeing) acceleration. *When an object moves in a circle, the acceleration (and also the net force) must point to the center of the circle* .

The main thing to remember when tackling circular motion problems is that a *centripetal force is simply whatever force is directed toward the center of the circle in which the object is traveling* . So, first label the forces on your free-body diagram, and then find the net force directed toward the center of the circle. That net force is the centripetal force. But NEVER label a free-body diagram with “*F _{c} *.”

**Exam tip from an AP Physics veteran:**

On a free-response question, do not label a force as “centripetal force,” even if that force does act toward the center of a circle; you will not earn credit. Rather, label with the actual source of the force; i.e., tension, friction, weight, electric force, etc.

*—Mike, high school junior*

**Mass on a String**

A block of mass *M* = 2 kg is swung on a rope in a vertical circle of radius *r* constantspeed *v* . When the block is the circle, the tension in the rope is measued to be 10 N. What is the tension in the rope when the block is at the bottom of the circle?

Let”s begin by drawing a free-body diagram of the block at the top of the circle and another of the block at the bottom of the circle.

Next, we write Newton”s second law for each diagram. Acceleration is always toward the center of the circle.

The acceleration is centripetal, so we can plug in *v* ^{2} /*r* for both accelerations.

At both top and bottom, the speed *v* and the radius *r* are the same. So *Mv* ^{2} /*r* has to be the same at both the top and bottom, allowing us to set the left side of each equation equal to one another.

With *M* = 2 kg and *T* _{top} = 10 N, we solve to get *T* _{bottom} = 50 N.

**Car on a Curve**

This next problem is a bit easier than the last one, but it”s still good practice.

A car of mass *m* travels around a flat curve that has a radius of curvature *r* . What is the necessary coefficient of friction such that the car can round the curve with a velocity *v* ?

Before we draw our free-body diagram, we should consider how friction is acting in this case. Imagine this: what would it be like to round a corner quickly while driving on ice? You would slide off the road, right? Another way to put that is to say that without friction, you would be unable to make the turn. Friction provides the centripetal force necessary to round the corner. Bingo! The force of friction must point in toward the center of the curve.

We can now write some equations and solve for *μ* , the coefficient of friction.

The net force in the horizontal direction is *F* _{f}_{ }, which can be set equal to mass times (centripetal) acceleration.

We also know that *F* _{f}_{ }= *μ* _{N}_{ }. So,

Furthermore, we know that the car is in vertical equilibrium—it is neither flying off the road nor being pushed through it—so *F* _{N}_{ }= *mg* .

Solving for *μ* we have

Note that this coefficient doesn”t depend on mass. Good—if it did, we”d need tires made of different materials depending on how heavy the car is.

**Newton”s Law of Gravitation**

We now shift our focus to gravity. Gravity is an amazing concept—you and the Earth attract each other just because you both have mass!—but at the level tested on the AP exam, it”s also a pretty easy concept. In fact, there are only a couple equations you need to know. The first is for gravitational force:

This equation describes the gravitational force that one object exerts on another object. *m* _{1} is the mass of one of the objects, *m* _{2} is the mass of the other object, *r* is the distance between the center of mass of each object, and *G* is called the “Universal Gravitational Constant” and is equal to 6.67 × 10^{−11} (*G* does have units—they are N·m^{2} /kg^{2} —but most problems won”t require your knowing them). The negative sign indicates that the force is attractive. We can leave it off unless we are doing calculus with the equation.

The mass of the Earth, *M _{E} *, is 5.97 × 10

^{24}kg. The mass of the sun,

*M*, is 1.99 × 10

_{S}^{30}kg. The two objects are about 154,000,000 km away from each other. How much gravitational force does Earth exert on the sun?

This is simple plug-and-chug (remember to convert km to m).

Notice that the amount of force that the Earth exerts on the sun is exactly the same as the amount of force the sun exerts on the Earth.

We can combine our knowledge of circular motion and of gravity to solve the following type of problem.

What is the speed of the Earth as it revolves in orbit around the sun?

The force of gravity exerted by the sun on the Earth is what keeps the Earth in motion—it is the centripetal force acting on the Earth.

*v* = 29,000 m/s. (Wow, fast … that converts to about 14 miles every second—much faster than, say, a school bus.)

Along with the equation for gravitational force, you need to know the equation for gravitational potential energy.

Why negative? Objects tend get pushed toward the lowest available potential energy. A long way away from the sun, the *r* term gets big, so the potential energy gets close to zero. But, since a mass is *attracted* to the sun by gravity, the potential energy of the mass must get lower and lower as *r* gets smaller.

We bet you”re thinking something like, “Now hold on a minute! You said a while back that an object”s gravitational potential energy equals *mgh* . What”s going on?”

Good point. An object”s gravitational PE equals *mgh* when that object is near the surface of the Earth. But it equals no matter where that object is.

Similarly, the force of gravity acting on an object equals *mg* (the object”s weight) only when that object is near the surface of the Earth.

The force of gravity on an object, however, *always* equals regardless of location.

**Kepler”s Laws**

Johannes Kepler, the late 1500s theorist, developed three laws of planetary motion based on the detailed observations of Tycho Brahe. You need to understand each law and its consequences.

*Planetary orbits are ellipses, with the sun at one focus*. Of course, we can apply this law to a satellite orbiting Earth, in which case the orbit is an ellipse, with Earth at one focus. (We mean the center of the Earth—for the sake of Kepler”s laws, we consider the orbiting bodies to be point particles.) In the simple case of a circular orbit, this law still applies because a circle is just an ellipse with both foci at the center.*An orbit sweeps out equal areas in equal times*. If you draw a line from a planet to the sun, this line crosses an equal amount of area every minute (or hour, or month, or whatever)—see__Figure 15.2__. The consequence here is that when a planet is close to the sun, it must speed up, and when a planet is far from the sun, it must slow down. This applies to the Earth as well. In the northern hemisphere winter, when the Earth is slightly closer to the sun,^{ }^{1}^{ }the Earth moves faster in its orbit. (You may have noticed that the earliest sunset in wintertime occurs about two weeks before the solstice—this is a direct consequence of Earth”s faster orbit.)

**Figure 15.2 Kepler”s second law. The area “swept out” by a planet in its orbit is shaded. In equal time intervals Δ t _{1} and Δt _{2} , these swept areas A _{1} and A _{2} are the same.**

*A planet”s orbital period squared is proportional to its orbital radius cubed*. In mathematics, we write this as*T*^{2}=*cR*^{3}. Okay, how do we define the “radius” of a non-circular orbit? Well, that would be average distance from the sun. And what is this constant*c*? It”s a different value for every system of satellites orbiting a single central body. Not worth worrying about, except that you can easily derive it for the solar system by solving the equation above for*c*and plugging in data from Earth”s orbit:*c*= 1 year^{2}/AU^{3}, where an “AU” is the distance from Earth to the sun. If you*really*need to, you can convert this into more standard units, but we wouldn”t bother with this right now.

**Energy of Closed Orbits**

When an object of mass *m* is in orbit around the sun, its potential energy is , where *M* is the mass of the sun, and *r* is the distance between the centers of the two masses.

The kinetic energy of the orbiting mass, of course, is *K* = ½*mv* ^{2} . The total mechanical energy of the mass in orbit is defined as *U* + *K* . When the mass is in a stable orbit, the total mechanical energy must be less than zero. A mass with positive total mechanical energy can escape the “gravitational well” of the sun; a mass with negative total mechanical energy is “bound” to orbit the sun.^{ }^{2}

All of the above applies to the planets orbiting in the solar system. It also applies to moons or satellites orbiting planets, when (obviously) we replace the “sun” by the central planet. A useful calculation using the fact that total mechanical energy of an object in orbit is the potential energy plus the kinetic energy is to find the “escape speed” from the surface of a planet … at *r* equal to the radius of the planet, set kinetic plus potential energy equal to zero, and solve for *v* . This is the speed that, if it is attained at the surface of the planet (neglecting air resistance), will cause an object to attain orbit.

** Practice Problems**

**Multiple Choice:**

__Questions 1 and 2:__

Two stars, each of mass *M* , form a binary system. The stars orbit about a point a distance *R* from the center of each star, as shown in the diagram above. The stars themselves each have radius *r* .

** 1 .** What is the force each star exerts on the other?

(A)

(B)

(C)

(D)

(E)

** 2 .** In terms of each star”s tangential speed

*v*, what is the centripetal acceleration of each star?

(A)

(B)

(C)

(D)

(E)

__Questions 3 and 4:__ In the movie *Return of the Jedi* , the Ewoks throw rocks using a circular-motion device. A rock is attached to a string. An Ewok whirls the rock in a horizontal circle above his head, then lets go, sending the rock careening into the head of an unsuspecting stormtrooper.

** 3 .** What force provides the rock”s centripetal acceleration?

(A) The vertical component of the string”s tension

(B) The horizontal component of the string”s tension

(C) The entire tension of the string

(D) The gravitational force on the rock

(E) The horizontal component of the gravitational force on the rock

** 4 .** The Ewok whirls the rock and releases it from a point above his head and to his right. The rock initially goes straight forward. Which of the following describes the subsequent motion of the rock?

(A) It will continue in a straight line forward, while falling due to gravity.

(B) It will continue forward but curve to the right, while falling due to gravity.

(C) It will continue forward but curve to the left, while falling due to gravity.

(D) It will fall straight down to the ground.

(E) It will curve back toward the Ewok and hit him in the head.

** 5 .** A Space Shuttle orbits Earth 300 km above the surface. Why can”t the Shuttle orbit 10 km above Earth?

(A) The Space Shuttle cannot go fast enough to maintain such an orbit.

(B) Kepler”s laws forbid an orbit so close to the surface of the Earth.

(C) Because *r* appears in the denominator of Newton”s law of gravitation, the force of gravity is much larger closer to the Earth; this force is too strong to allow such an orbit.

(D) The closer orbit would likely crash into a large mountain such as Everest because of its elliptical nature.

(E) Much of the Shuttle”s kinetic energy would be dissipated as heat in the atmosphere, degrading the orbit.

**Free Response:**

** 6 .** Consider two points on a rotating turntable: Point

*A*is very close to the center of rotation, while point

*B*is on the outer rim of the turntable. Both points are shown above. A penny could be placed on the turntable at either point

*A*or point

*B*.

(a) In which case would the speed of the penny be greater, if it were placed at point *A* , or if it were placed at point *B* ? Explain.

(b) At which point would the penny require the larger centripetal force to remain in place? Justify your answer.

(c) Point *B* is 0.25 m from the center of rotation. If the coefficient of friction between the penny and the turntable is *μ* = 0.30, calculate the maximum linear speed the penny can have there and still remain in circular motion.

** Solutions to Practice Problems**

** 1 . D—** In Newton”s law of gravitation,

the distance used is the distance between the centers of the planets; here that distance is 2*R* . But the denominator is squared, so (2*R* )^{2} = 4*R* ^{2} in the denominator here.

** 2 . E—** In the centripetal acceleration equation

the distance used is the radius of the circular motion. Here, because the planets orbit around a point right in between them, this distance is simply *R* .

** 3 . B—** Consider the vertical forces acting on the rock. The rock has weight, so

*mg*acts down. However, because the rock isn”t falling down, something must counteract the weight. That something is the vertical component of the rope”s tension. The rope must not be perfectly horizontal, then. Because the circle is horizontal, the centripetal force must be horizontal as well. The only horizontal force here is the horizontal component of the tension. (Gravity acts

*down*, last we checked, and so cannot have a horizontal component.)

** 4 . A—** Once the Ewok lets go, no forces (other than gravity) act on the rock. So, by Newton”s first law, the rock continues in a straight line. Of course, the rock still must fall because of gravity. (The Ewok in the movie who got hit in the head forgot to let go of the string.)

** 5 . E—** A circular orbit is allowed at any distance from a planet, as long as the satellite moves fast enough. At 300 km above the surface Earth”s atmosphere is practically nonexistent. At 10 km, though, the atmospheric friction would quickly cause the Shuttle to slow down.

** 6 .** (a) Both positions take the same time to make a full revolution. But point

*B*must go farther in that same time, so the penny must have bigger speed at point

*B*.

(b) The coin needs more centripetal force at point *B* . The centripetal force is equal to *mv* ^{2} /*r* . However, the speed itself depends on the radius of the motion, as shown in *part (a)* . The speed of a point at radius *r* is the circumference divided by the time for one rotation *T, v* = 2*πr* /*T* . So the net force equation becomes, after algebraic simplification, *F _{net} *= 4mπ

^{2}

*r*/

*T*

^{2}. Because both positions take equal times to make a rotation, the coin with the larger distance from the center needs more centripetal force.

(c) The force of friction provides the centripetal force here, and is equal to *μ* times the normal force. Because the only forces acting vertically are *F* _{N}_{ }and *mg, F* _{N}_{ }= *mg* . The net force is equal to *mv* ^{2} /*r* , and also to the friction force *mmg* . Setting these equal and solving for *v* ,

Plug in the values given (*r* = 0.25 m, *μ* = 0.30) to get *v* = 0.87 m/s. If the speed is faster than this, then the centripetal force necessary to keep the penny rotating increases, and friction can no longer provide that force.

** Rapid Review**

- When an object travels in a circle, its velocity vector is directed tangent to the circle, and its acceleration vector points toward the center of the circle.
- A centripetal force keeps an object traveling in a circle. The centripetal force is simply whatever net force is directed toward the center of the circle in which the object is traveling.
- Newton”s law of gravitation states that the gravitational force between two objects is proportional to the mass of the first object multiplied by the mass of the second divided by the square of the distance between them. This also means that the gravitational force felt by one object is the same as the force felt by the second object.
- Predictions of Kepler”s laws: Planets undergo elliptical orbits with the sun at one focus; at points in an orbit closer to the sun, a planet moves faster; the smaller a planet”s mean distance from the sun, the shorter its orbital period, by
*T*^{2}∝*R*^{3}. (The symbol “∝” means “is proportional to.”)

^{1}^{ }*Please* don”t say you thought the Earth must be farther away from the sun in winter because it”s cold. When it”s winter in the United States, it”s summer in Australia, and we”re all the same distance from the sun!

^{2}^{ }It can be shown that for a planet in a stable, *circular* orbit, the kinetic energy is half the absolute value of the potential energy. This isn”t something important enough to memorize, but it might help you out sometime if you happen to remember.

**CHAPTER 15**

**Gravitation and Circular Motion**

** 1 .** A binary star system consists of two identical stars each with a mass of

*M*that orbit each other about the center of mass of the system as shown. The orbital velocity of the stars is

(A)

(B)

(C)

(D)

(E)

** 2 .** A sprinter runs around an oval track at a constant speed, starting at point A, proceeding around the track counterclockwise, and returning to point A. Which graph best represents the magnitude of the sprinter”s acceleration as she runs one complete lap?

(A)

(B)

(C)

(D)

(E)

** 3 .** A pendulum of mass (

*m*) is swinging in a horizontal circle of radius (

*r*) at a constant rate, as shown in the figure. The angle that the pendulum string makes with the vertical is θ, and the tension of the string is

*T*. Which of the following statements concerning this arrangement is correct?

(A) The gravity force and the tension cancel each other out.

(B) The velocity of the mass is constant.

(C) The mass accelerates in the direction of the tension force.

(D) The velocity of the mass is:

(E) The tension force is:

** 4 .** A satellite in geosynchronous orbit around Earth has an orbital time period of

*T*, as shown in the figure. If the mass of the satellite is doubled while keeping the orbital radius (

*R*) the same, the new time period of the satellite”s orbit will be

(A)

(B)

(C) *T*

(D)

(E) 2*T*

** Answers**

__1__ .**E** —Gravity supplies the centripetal force that allows the stars to orbit, which we find by using the equation:

__2__ .**A** —The sprinter only accelerates when changing direction around the curved portions of the tracks from points A to C and from points D to F.

__3__ .**D** —The mass has a centripetal acceleration toward the center of the horizontal circle. Thus, choices A, B, and C are not correct. The vertical portion of tension must cancel out the gravity force: *T* cos θ = *mg* . Thus, choice E is incorrect. The horizontal portion of tension supplies the centripetal acceleration toward the center of the circular path:

__4__ .**C** —Use the following equation to calculate the orbital time period:

The velocity of circular orbit is

Substituting the velocity into the force equation, we get

Note that the mass of the satellite is not in the time period equation. This means the satellite mass does not influence the orbital time period at all!