## 5 Steps to a 5: AP Physics 1: Algebra-Based 2017 (2016)

### STEP __4__

### Review the Knowledge You Need to Score High

### CHAPTER 15

### Gravitation

**IN THIS CHAPTER**

**Summary:** The force on an object due to gravity is *mg* , where *m* is mass and *g* is the gravitational field. The gravitational field produced by an object of mass *m* is , where *d* is the distance from the object”s center. Want to be sure you know the difference between a gravitational field, a gravitational force, and the universal gravitation constant? This chapter explains these concepts.

**Definitions**

The **gravitational field** *g* near a planet tells how much 1 kg of mass weighs at a location. Near Earth”s surface, the gravitational field is 10 N/kg.

The **gravitational force** of a planet on any other object in the planet”s gravitational field is *mg* , where *m* is the mass of the object experiencing the force.

**Newton”s gravitation constant** is the universal constant *G* = 6 × 10^{−11} N·m^{2} /kg^{2} .

The **free-fall acceleration** (sometimes imprecisely called the **acceleration due to gravity** ) near a planet is, by an amazing coincidence of the universe, equal to the gravitational field near that planet. Near Earth, then, the free-fall acceleration is 10 m/s per second because the gravitational field is 10 N/kg.

The gravitational force is the weakest of the fundamental forces in nature. However, when enormous amounts of mass congregate—as in a star or a planet—the gravitational force becomes dominant. The picture as follows gives a hint of the scales involved in studying stars and planets—it”s the Earth and the Moon, but drawn to approximately the proper scale. Sure, the Earth seems humongous, especially when you”re caught in traffic. But *all* planets and stars seem small when the distances between them are considered.

The word *gravity* by itself is an ambiguous term. Begin this chapter by carefully reading the differences between all the various things that could be referred to by the word *gravity* . See the four preceding definitions, each of which relate to the word.

**Determining the Gravitational Field**

The gravitational field is a vector quantity—this means it has an amount and a direction. The direction is always toward the center of the Earth (or whatever is creating the gravitational field).

**FACT:** The amount^{ }^{1}^{ }of gravitational field depends on two things: the mass of the planet creating the field (*M* ) and the distance you are from that planet”s center (*d* ). The relevant equation for the gravitation field *g* produced by a planet is

Some books, and probably even the AP Exam, will use the variable *r* for the distance from the planet”s center. That”s fine, but know that this *r* does *not* necessarily stand for the radius of the planet—it means the distance from the planet”s center.

**Example 1:** A 20-kg rover sits on newly discovered Planet *Z* , which has twice the mass of Earth and twice the diameter of Earth.

You do not get a table of astronomical information on the AP Physics 1, Algebra-Based Exam. Nevertheless, you might well be asked to calculate the gravitational field near the surface of Planet *Z* . How can that be done without knowing the mass of Planet *Z* ? You”re expected to be fluent in semiquantitative reasoning.

Even though you don”t know the value of the mass of the Earth, you know that the mass of Planet *Z* is twice Earth”s mass. Whatever the exact mass, the numerator in the gravitational field equation will double for Planet *Z* .

The surface of a planet is one planet-radius away from the planet”s center; so here, *d* means the radius of the planet. Planet *Z* ”s diameter is twice Earth”s, which also means *Z* ”s radius is twice Earth”s. The *d* term in the denominator is doubled; because *d* is squared, the entire denominator is multiplied by 2^{2} , which is 4.

Combining these effects of mass and radius, the numerator is multiplied by 2, the denominator multiplied by 4, so the entire gravitational field of the Earth is multiplied by one-half. We know Earth”s gravitational field—that”s 10 N/kg. Planet *Z* produces a gravitational field of 5 N/kg at the surface. No calculator is necessary.

The formula shows that the gravitational field produced by a planet drops off rapidly as you get far from the planet”s center: if you double your distance from the planet”s center, then you cut in one-fourth the value of the gravitational field.

This seems easy enough, but think about reality for a moment. Just how often do you double your distance from the center of the Earth? The radius of the Earth is about 4,000 miles. Even when you fly in an airplane, you”re no more than about seven miles above the surface; so your distance from the center of the Earth is *still* about 4,000 miles.

The point is, unless you”re an astronaut, the gravitational field near the surface of a planet is a constant value. Don”t be tricked by the *d* ^{2} in the denominator—that only matters when you”re considering objects in space.

**Determining Gravitational Force**

**FACT:** The weight of an object—that is, the gravitational force of a planet on that object—is given by *mg* .

That 20-kg rover would weigh 200 N on Earth (20 kg times 10 N/kg). But on Planet *Z* , the rover weighs only 100 N (that”s 20 kg times 5 N/kg).

A weight of 100 N means that Planet *Z* pulls the rover downward with 100 N of force. What about the force of the rover on Planet *Z* ? That”s got to be so small it”s negligible, right?

Wrong. Newton”s Third Law says that the force of Planet *Z* on the rover is equal to the force of the rover on Planet *Z* . The rover pulls up on Planet *Z* with a force of 100 N.

Now, as you might suspect, you”d never notice or measure any effect from the rover”s 100-N force on Planet *Z* . Planet *Z* is enormously massive—in the neighborhood of 10^{24} kg. By *F* _{net} = *ma* , you can calculate that the *acceleration* provided to the planet is immeasurably small.^{ }^{2}^{ }It”s the force that”s the same, and the acceleration that”s different.

**Force of Two Planets on One Another—Order of Magnitude Estimates**

**Example 2:** The Earth has a mass of 6.0 × 10^{24} kg. The Sun has a mass of 2.0 × 10^{30} kg. The Earth orbits the sun in a circle of radius 1.5 × 10^{11} m.

**FACT:** The gravitational force of one object on another is given by

So it seems straightforward, if calculator intensive, to calculate the force the Sun exerts on the Earth (or vice versa). Just plug in the numbers. But that”s *not* a likely AP Physics 1 exercise! No one cares whether you can use the buttons on your calculator correctly.

Instead, you might be asked, “Which of the following is closest to the force of the Sun on the Earth?”

(A) 10^{12} N

(B) 10^{22} N

(C) 10^{32} N

(D) 10^{42} N

Look how far apart these answer choices are. Don”t use a calculator—instead make an *order of magnitude estimate* . Plug in just the powers of 10, and the answer will leap off the page. Leaving out the units of each individual term for simplicity, in the equation we have

That”s easy to simplify without a calculator—add exponents in the numerator, and then subtract the exponents in the denominator.

But that”s not one of the choices.

The only reasonable choice, though, is (B) 10^{22} N. The others are at least a factor of a billion too big or too small. Sure, you could have spent five minutes plugging in the more precise values into your calculator,^{ }^{3}^{ }getting 3.3 × 10^{22} N as the answer. The order of magnitude estimate is as precise as would ever be necessary on the AP Physics 1 exam, and it”s a lot easier, too.

**Gravitational Potential Energy**

Near the surface of the Earth, the potential energy provided by the gravitational force is

That”s plenty good enough for calculations with everyday objects.

However, if you”re talking about objects way out in space, the gravitational potential energy possessed by two objects near one another is given by

This equation is usually written with a negative sign; that negative sign simply indicates that this potential energy is *less* than zero; and in outer-space situations, zero potential energy means the objects are infinitely far away from each other.

This gravitational potential energy can be converted into kinetic energy—if two planets move toward one another due to their mutual gravitational attraction, you might be able to figure out how fast they move by calculating the gravitational potential energy possessed before and after they move. Any lost potential energy was converted into kinetic energy.

**Gravitational and Inertial Mass**

**Example 3:** Neil Armstrong had a mass of 77 kg when he went to the moon. The gravitational field on the Moon is one-sixth that on Earth.

The term “mass” is often colloquially defined as the amount of “stuff” in an object. You”re likely to see an AP question of the form, “What was Neil Armstrong”s mass on the Moon?” The answer is, still 77 kg. Sure, Neil”s weight on the Moon was smaller than his weight on Earth, because the gravitational field on the Moon is smaller, but since he didn”t cut off his leg or go on a starvation diet, his mass didn”t change.

**FACT:** Gravitational mass indicates how an object responds to a gravitational field.

**FACT:** Inertial mass indicates how an object accelerates in response to a net force.

**FACT:** In every experiment ever conducted, an object”s gravitational mass is equal to its inertial mass.

The AP Exam requires you to distinguish between the two meanings of mass. Simply put, if there”s acceleration involved, you”re talking about inertial mass. You”ll be asked to design experiments to measure each type of mass.

So let”s say you want to figure out who has more *gravitational* mass, you or Neil Armstrong. Just put each of you on a balance scale—the one with the bigger scale reading has more gravitational mass. You could use a spring scale, too—whoever compresses the springs more experiences more gravitational force in the same gravitational field, and so has more gravitational mass.

But to figure out experimentally who has more *inertial* mass, you”d put each of you in an identical buggy. Speed up each buggy using the same net force for the same amount of time. Whichever of you has sped up by more—that is, whichever experienced the greater acceleration under the same net force—has the smaller inertial mass.

And if your experiment gives contradictory results—say, that Neil has more inertial mass but that you have more gravitational mass—then you should reject that result as ridiculous. An exam question might pose just this type of situation in which you have to reject impossible results.^{ }^{4}

**Fundamental Forces: Gravity Versus Electricity**

The gravitational force is one of several “fundamental” forces in nature. The only other fundamental force on the AP Exam is the electrical force. In this context, “fundamental” means that all forces on the AP Exam are manifestations of one of these two forces.

Oh yeah? What about, say, friction?

All of the everyday forces that result from “contact” between two objects are fundamentally electrical. The electron shells in the outermost atomic layers of the objects interact, causing friction, normal forces, tensions, punches, etc. Really. But AP Physics 1 doesn”t care at all about the details of these interactions; just know that they”re electrical.

As discussed in __Chapter 16__ , the electrical force between two particles is generally much, much stronger than the gravitational force between them. The AP Exam is likely to ask you to explain why gravity is such a dominant everyday force, then.

The answer is twofold. First of all, consider the overwhelmingly massive scale of the Earth. It”s tens of thousands of miles in circumference, and a trillion trillion kilograms in mass. And even with that much stuff in it, the gravitational force it exerts on you is equal to the (electrical) contact force of the ground pushing up on you.

The bigger issue is that masses always cause a gravitational field that attracts other masses—there”s no such thing as a repulsive gravitational force. Electrical forces, though, can be both repulsive and attractive, depending on the sign of the charges interacting. Since in general the Earth is electrically neutral, all that unimaginably immense stuff in the Earth has very little net electrical effect, leaving gravity to keep you firmly attached to the ground.

** Practice Problems**

** 1 .** 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*.

(a) In terms of given variables and fundamental constants, what is the force each star exerts on the other?

(b) In terms of given variables and fundamental constants, what is the magnitude of the gravitational field at the surface of one of the stars due only to its own mass?

(c) In terms of given variables and fundamental constants, what is the magnitude of the gravitational field at the midpoint between the stars?

(d) Explain why the stars don”t crash into each other due to the gravitational force between them.

** 2 .** A space shuttle orbits Earth 300 km above the equator.

(a) Explain why it would be impractical for the shuttle to orbit 10 km above the Earth”s surface (about 1 km higher than the top of Mount Everest).

(b) A “geosynchronous orbit” means that the shuttle will always remain over the same spot on Earth. Explain and describe the calculations you would perform in order to determine whether this orbit is geosynchronous. You should not actually carry out the numerical calculations, just describe them in words and show them in symbols.

(c) The radius of Earth is 6,400 km. At the altitude of the space shuttle, what fraction of the surface gravitational field *g* does the shuttle experience?

(d) When the shuttle was on Earth before launch, the shuttle”s mass (not including any fuel) was 2 × 10^{6} kg. At the orbiting altitude, what is the shuttle”s mass, not including fuel?

** 3 .** A satellite is in circular orbit around an unknown planet. A second, different satellite also travels in a circular orbit around this same planet, but with an orbital radius four times larger than the first satellite.

(a) Explain what information must be known in order to calculate the speed the first satellite travels in its orbit.

(b) Compared to the first satellite, how many times faster or slower is the second satellite”s speed?

(c) Bob the bad physics student says:

“The gravitational force of the planet on a satellite in circular orbit depends inversely on the orbital radius squared. Since the second satellite”s orbital radius is four times that of the first satellite, the second satellite experiences one-sixteenth the gravitational force that is exerted on the first satellite.”

Explain what is wrong with Bob”s explanation.

** 4 .** A spacecraft is positioned between the Earth and the Moon such that the gravitational forces on the spacecraft exerted by the Earth and the Moon cancel.

(a) Is this position closer to the Moon, closer to the Earth, or halfway in between?

(b) Are the gravitational forces on the spaceship (the force exerted by the Moon, and the force exerted by the Earth) a Newton”s third law force pair?

** Solutions to Practice Problems**

** 1 .** (a) The term in the denominator of Newton”s law of gravitation refers to the distance between the centers of the two stars. That distance is given as 2

*R*. So the answer is or . Notice you must use the notation given in the problem—this means a capital

*R*here. (The

*M*is squared because the equation for gravitational force multiplies the masses of the stars applying and experiencing the force. Since the masses of both stars are the same, you get

*M*·

*M*=

*M*

^{2}.)

(b) When calculating gravitational field at the surface of a star, the term in the denominator is the star”s radius. That”s *r* here. So . (The *M* is *not* squared here because the field is produced by a single star, not an interaction between two stars.)

(c) The left-hand star provides a field of at the location of the midpoint, pointing toward the left-hand star. The right-hand star also provides a field of , pointing toward the right-hand star. These fields with the same amount but opposite direction cancel out when they add as vectors, producing a net gravitational field of zero at the midpoint.

(d) The direction of a force is not the same as the direction of an object”s motion. Here, at any time a star is moving along an orbit, tangent to the radius of the orbit. The force applied by the other star will always be toward the center of the orbit, perpendicular to the direction of the star”s motion. When a force is applied perpendicular to an object”s velocity, the result is circular motion. The centripetal force changes the direction (not the amount) of the star”s speed, but the force itself always changes direction so that it is pulling toward the center of the circular motion.

** 2 .** (a) The shuttle must be above the atmosphere in order to maintain a circular orbit without continually burning fuel. If the Earth had no atmosphere, then a satellite could orbit at any distance from the surface that”s greater than the tallest mountain. But air resistance in the atmosphere does work on the shuttle, reducing its mechanical energy. At 300 km above the surface, though, the shuttle is above the atmosphere and experiences no forces in or against the direction of travel.

(b) This question requires a calculation: the force on the shuttle is , where *d* is the distance from the center of the Earth to the location of the shuttle. Since this force is a centripetal force, we can set it equal to , where *v* is the speed of the shuttle in orbit. If the orbit is geosynchronous, that doesn”t mean that the shuttle goes the same speed as a position on Earth, it means that the *period* of the orbit is 24 hours—it goes around the Earth in the same time as a location on the surface does.

To determine the period of the orbit, set the speed equal to the orbit”s circumference (2π*d* ) divided by the period, which I”ll call *T* . Here”s how our equation looks now:

There is lots of algebra here, but notice that all values are things that can be looked up: *G* , the mass of the Earth, the mass of the shuttle (which cancels anyway), and *d* , which is the radius of the Earth plus 300 km. Solve this equation for *T* . If the value of *T* turns out to be 24 hours, then the orbit is geosynchronous; if not, the orbit is not geosynchronous.^{ }^{5}

(c) Your first instinct might be that you need the mass of the Earth to answer this question, because the equation for the gravitational field is . On one hand, we could calculate the mass of Earth knowing that the gravitational field near the surface is 10 N/kg—just plug in the given value of *R* and *G* from the table of information. That will work. It”s more elegant to solve in variables: The fraction we want is

The *G* and *M* cancel, whatever their value. Work the improper fraction to get that the gravitational field 300 km above the surface is of *g* at the surface.

(d) Mass is the amount of “stuff” in an object. It doesn”t matter where that object is in the universe, 1 kg of mass is 1 kg of mass. Unless the shuttle loses a wing, its mass is still 2 × 10^{6} kg.

** 3 .** (a) Set the gravitational force on the satellite equal to the formula for centripetal force:

We”ll want to be able to solve for *v* . The mass of the satellite cancels, and *G* is a universal constant. But we”ll need to know *d* , the distance of the satellite”s orbit from the center of the planet. And we”ll need to know the mass of the planet or the period of the orbit.

(b) When we solve for *v* , we get . The second satellite has four times the orbital radius, which is represented by *d* . Multiplying by four in the denominator under the square root multiplies the whole expression by one-half. The second satellite”s speed is one-half as large.

(c) The equation for the force of the planet on the satellite is . Sure, Bob is right about the inverse square dependence on *d* , but he”s assumed that the satellites have the same masses as each other. If they do, Bob is correct; if not, then, the mass of the satellite shows up in the numerator of the force equation.

** 4 .** (a) The relevant equation here comes from setting the forces on the spacecraft (of mass

*m*) equal:

The mass of the spacecraft cancels, as does the *G* . Since the mass of the Earth is bigger than the mass of the Moon, the equation shows that the distance of this location from Earth *d* _{earth} must be larger than the distance of the location from the Moon *d* _{moon} . The location is closer to the Moon.

(b) A Newton”s third law force pair cannot act on the same object. The force of the Moon on the spacecraft is paired with the force of the spacecraft on the Moon; the force of the Earth on the spacecraft is paired with the force of the spacecraft on the Earth.

** Rapid Review**

- The amount of gravitational field depends on two things: the mass of the planet creating the field (
*M*) and the distance you are from that planet”s center (*d*). The relevant equation for the gravitation field*g*produced by a planet is . - The weight of an object—that is, the gravitational force of a planet on that object—is given by
*mg*. - The gravitational force of one object on another is given by .
- Gravitational mass indicates how an object responds to a gravitational field.
- Inertial mass indicates how an object accelerates in response to a net force.
- In every experiment ever conducted, an object”s gravitational mass is equal to its inertial mass.

__ ^{1}__ The exam will refer not to the “amount,” but to the “magnitude” of a vector quantity. Just translate in your head.

__ ^{2}__ And the rover”s force on the planet is certainly not the

*net*force on the planet, so this

*F*

_{net}=

*ma*calculation is silly anyway.

__ ^{3}__ And then spend another 10 minutes swearing at the calculator because you left out a parenthesis, or you forgot to type a negative sign, or you forgot a decimal.

__ ^{4}__ Well, if you really have repeatable and unassailable results of this nature, you should publish. I guarantee you”ll win a Nobel Prize for your work, which is a bit more likely to earn college credit than a 5 on the AP Exam.

__ ^{5}__ If you do all the plugging and chugging—which are

*not*necessary—you”ll find that the orbit is not geosynchronous. Geosynchronous orbits are somewhere in the 35,000-km range above Earth”s surface.