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a / Gravity is the only really important force on the cosmic scale. This false-color representation of saturn's rings was made from an image sent back by the Voyager 2 space probe. The rings are composed of innumerable tiny ice particles orbiting in circles under the influence of saturn's gravity.

# Chapter 10. Gravity

a / Johannes Kepler found a mathematical description of the motion of the planets, which led to Newton's theory of gravity.

Cruise your radio dial today and try to find any popular song that would have been imaginable without Louis Armstrong. By introducing solo improvisation into jazz, Armstrong took apart the jigsaw puzzle of popular music and fit the pieces back together in a different way. In the same way, Newton reassembled our view of the universe. Consider the titles of some recent physics books written for the general reader: The God Particle, Dreams of a Final Theory. Without Newton, such attempts at universal understanding would not merely have seemed a little pretentious, they simply would not have occurred to anyone.

This chapter is about Newton's theory of gravity, which he used to explain the motion of the planets as they orbited the sun. Whereas this book has concentrated on Newton's laws of motion, leaving gravity as a dessert, Newton tosses off the laws of motion in the first 20 pages of the Principia Mathematica and then spends the next 130 discussing the motion of the planets. Clearly he saw this as the crucial scientific focus of his work. Why? Because in it he showed that the same laws of motion applied to the heavens as to the earth, and that the gravitational force that made an apple fall was the same as the force that kept the earth's motion from carrying it away from the sun. What was radical about Newton was not his laws of motion but his concept of a universal science of physics.

## 10.1 Kepler's laws

b / Tycho Brahe made his name as an astronomer by showing that the bright new star, today called a supernova, that appeared in the skies in 1572 was far beyond the Earth's atmosphere. This, along with Galileo's discovery of sunspots, showed that contrary to Aristotle, the heavens were not perfect and unchanging. Brahe's fame as an astronomer brought him patronage from King Frederick II, allowing him to carry out his historic high-precision measurements of the planets' motions. A contradictory character, Brahe enjoyed lecturing other nobles about the evils of dueling, but had lost his own nose in a youthful duel and had it replaced with a prosthesis made of an alloy of gold and silver. Willing to endure scandal in order to marry a peasant, he nevertheless used the feudal powers given to him by the king to impose harsh forced labor on the inhabitants of his parishes. The result of their work, an Italian-style palace with an observatory on top, surely ranks as one of the most luxurious science labs ever built. Kepler described Brahe as dying of a ruptured bladder after falling from a wagon on the way home from a party, but other contemporary accounts and modern medical analysis suggest mercury poisoning, possibly as a result of court intrigue.

Newton wouldn't have been able to figure out why the planets move the way they do if it hadn't been for the astronomer Tycho Brahe (1546-1601) and his protege Johannes Kepler (1571-1630), who together came up with the first simple and accurate description of how the planets actually do move. The difficulty of their task is suggested by figure c, which shows how the relatively simple orbital motions of the earth and Mars combine so that as seen from earth Mars appears to be staggering in loops like a drunken sailor.

c / As the Earth and Mars revolve around the sun at different rates, the combined effect of their motions makes Mars appear to trace a strange, looped path across the background of the distant stars.

Brahe, the last of the great naked-eye astronomers, collected extensive data on the motions of the planets over a period of many years, taking the giant step from the previous observations' accuracy of about 10 minutes of arc (10/60 of a degree) to an unprecedented 1 minute. The quality of his work is all the more remarkable considering that his observatory consisted of four giant brass protractors mounted upright in his castle in Denmark. Four different observers would simultaneously measure the position of a planet in order to check for mistakes and reduce random errors.

With Brahe's death, it fell to his former assistant Kepler to try to make some sense out of the volumes of data. Kepler, in contradiction to his late boss, had formed a prejudice, a correct one as it turned out, in favor of the theory that the earth and planets revolved around the sun, rather than the earth staying fixed and everything rotating about it. Although motion is relative, it is not just a matter of opinion what circles what. The earth's rotation and revolution about the sun make it a noninertial reference frame, which causes detectable violations of Newton's laws when one attempts to describe sufficiently precise experiments in the earth-fixed frame. Although such direct experiments were not carried out until the 19th century, what convinced everyone of the sun-centered system in the 17th century was that Kepler was able to come up with a surprisingly simple set of mathematical and geometrical rules for describing the planets' motion using the sun-centered assumption. After 900 pages of calculations and many false starts and dead-end ideas, Kepler finally synthesized the data into the following three laws:

##### Kepler's elliptical orbit law

The planets orbit the sun in elliptical orbits with the sun at one focus.

##### Kepler's equal-area law

The line connecting a planet to the sun sweeps out equal areas in equal amounts of time.

##### Kepler's law of periods

The time required for a planet to orbit the sun, called its period, is proportional to the long axis of the ellipse raised to the 3/2 power. The constant of proportionality is the same for all the planets.

Although the planets' orbits are ellipses rather than circles, most are very close to being circular. The earth's orbit, for instance, is only flattened by 1.7% relative to a circle. In the special case of a planet in a circular orbit, the two foci (plural of “focus”) coincide at the center of the circle, and Kepler's elliptical orbit law thus says that the circle is centered on the sun. The equal-area law implies that a planet in a circular orbit moves around the sun with constant speed. For a circular orbit, the law of periods then amounts to a statement that the time for one orbit is proportional to $$r^{3/2}$$, where $$r$$ is the radius. If all the planets were moving in their orbits at the same speed, then the time for one orbit would simply depend on the circumference of the circle, so it would only be proportional to $$r$$ to the first power. The more drastic dependence on $$r^{3/2}$$ means that the outer planets must be moving more slowly than the inner planets.

## 10.2 Newton's law of gravity

d / An ellipse is a circle that has been distorted by shrinking and stretching along perpendicular axes.

e / An ellipse can be constructed by tying a string to two pins and drawing like this with the pencil stretching the string taut. Each pin constitutes one focus of the ellipse.

f / If the time interval taken by the planet to move from P to Q is equal to the time interval from R to S, then according to Kepler's equal-area law, the two shaded areas are equal. The planet is moving faster during interval RS than it did during PQ, which Newton later determined was due to the sun's gravitational force accelerating it. The equal-area law predicts exactly how much it will speed up.

g / The moon's acceleration is $$60^2=3600$$ times smaller than the apple's.

h / Students often have a hard time understanding the physical meaning of $$G$$. It's just a proportionality constant that tells you how strong gravitational forces are. If you could change it, all the gravitational forces all over the universe would get stronger or weaker. Numerically, the gravitational attraction between two 1-kg masses separated by a distance of 1 m is $$6.67\times10^{-11}\ \text{N}$$, and this is what $$G$$ is in SI units.

i / Example 3. Computer-enhanced images of Pluto and Charon, taken by the Hubble Space Telescope.

j / The conic sections are the curves made by cutting the surface of an infinite cone with a plane.

k / An imaginary cannon able to shoot cannonballs at very high speeds is placed on top of an imaginary, very tall mountain that reaches up above the atmosphere. Depending on the speed at which the ball is fired, it may end up in a tightly curved elliptical orbit, 1, a circular orbit, 2, a bigger elliptical orbit, 3, or a nearly straight hyperbolic orbit, 4.

### The sun's force on the planets obeys an inverse square law.

Kepler's laws were a beautifully simple explanation of what the planets did, but they didn't address why they moved as they did. Did the sun exert a force that pulled a planet toward the center of its orbit, or, as suggested by Descartes, were the planets circulating in a whirlpool of some unknown liquid? Kepler, working in the Aristotelian tradition, hypothesized not just an inward force exerted by the sun on the planet, but also a second force in the direction of motion to keep the planet from slowing down. Some speculated that the sun attracted the planets magnetically.

Once Newton had formulated his laws of motion and taught them to some of his friends, they began trying to connect them to Kepler's laws. It was clear now that an inward force would be needed to bend the planets' paths. This force was presumably an attraction between the sun and each planet. (Although the sun does accelerate in response to the attractions of the planets, its mass is so great that the effect had never been detected by the prenewtonian astronomers.) Since the outer planets were moving slowly along more gently curving paths than the inner planets, their accelerations were apparently less. This could be explained if the sun's force was determined by distance, becoming weaker for the farther planets. Physicists were also familiar with the noncontact forces of electricity and magnetism, and knew that they fell off rapidly with distance, so this made sense.

In the approximation of a circular orbit, the magnitude of the sun's force on the planet would have to be

$\begin{equation*} F= ma= mv^2/r . \end{equation*}$

Now although this equation has the magnitude, $$v$$, of the velocity vector in it, what Newton expected was that there would be a more fundamental underlying equation for the force of the sun on a planet, and that that equation would involve the distance, $$r$$, from the sun to the object, but not the object's speed, $$v$$ --- motion doesn't make objects lighter or heavier.

self-check:

If eq. [1] really was generally applicable, what would happen to an object released at rest in some empty region of the solar system?

Equation [1] was thus a useful piece of information which could be related to the data on the planets simply because the planets happened to be going in nearly circular orbits, but Newton wanted to combine it with other equations and eliminate $$v$$ algebraically in order to find a deeper truth.

To eliminate $$v$$, Newton used the equation

$\begin{equation*} v = \frac{\text{circumference}}{T} = \frac{2\pi r}{T} . \end{equation*}$

Of course this equation would also only be valid for planets in nearly circular orbits. Plugging this into eq. [1] to eliminate $$v$$ gives

$\begin{equation*} F = \frac{4\pi^2mr}{T^2} . \end{equation*}$

This unfortunately has the side-effect of bringing in the period, $$T$$, which we expect on similar physical grounds will not occur in the final answer. That's where the circular-orbit case, $$T \propto r^{3/2}$$, of Kepler's law of periods comes in. Using it to eliminate $$T$$ gives a result that depends only on the mass of the planet and its distance from the sun:

$\begin{multline*} F\propto m/r^2 . \shoveright{\text{[force of the sun on a planet of mass}}\\ \shoveright{\text{m at a distance r from the sun; same}}\\ \text{proportionality constant for all the planets]} \end{multline*}$

(Since Kepler's law of periods is only a proportionality, the final result is a proportionality rather than an equation, so there is no point in hanging on to the factor of $$4\pi ^2$$.)

As an example, the “twin planets” Uranus and Neptune have nearly the same mass, but Neptune is about twice as far from the sun as Uranus, so the sun's gravitational force on Neptune is about four times smaller.

self-check:

Fill in the steps leading from equation [3] to $$F\propto m/r^2$$.

### The forces between heavenly bodies are the same type of force as terrestrial gravity.

OK, but what kind of force was it? It probably wasn't magnetic, since magnetic forces have nothing to do with mass. Then came Newton's great insight. Lying under an apple tree and looking up at the moon in the sky, he saw an apple fall. Might not the earth also attract the moon with the same kind of gravitational force? The moon orbits the earth in the same way that the planets orbit the sun, so maybe the earth's force on the falling apple, the earth's force on the moon, and the sun's force on a planet were all the same type of force.

There was an easy way to test this hypothesis numerically. If it was true, then we would expect the gravitational forces exerted by the earth to follow the same $$F\propto m/r^2$$ rule as the forces exerted by the sun, but with a different constant of proportionality appropriate to the earth's gravitational strength. The issue arises now of how to define the distance, $$r$$, between the earth and the apple. An apple in England is closer to some parts of the earth than to others, but suppose we take $$r$$ to be the distance from the center of the earth to the apple, i.e., the radius of the earth. (The issue of how to measure $$r$$ did not arise in the analysis of the planets' motions because the sun and planets are so small compared to the distances separating them.) Calling the proportionality constant $$k$$, we have

\begin{align*} F_\text{earth on apple} &= k \: m_\text{apple} / r_\text{earth}^2 \\ F_\text{earth on moon} &= k \: m_\text{moon} / d_\text{earth-moon}^2 . \end{align*}

Newton's second law says $$a=F/m$$, so

\begin{align*} a_\text{apple} &= k \: / \: r_{earth}^2 \\ a_\text{moon} &= k \: / \: d_\text{earth-moon}^2 . \end{align*}

The Greek astronomer Hipparchus had already found 2000 years before that the distance from the earth to the moon was about 60 times the radius of the earth, so if Newton's hypothesis was right, the acceleration of the moon would have to be $$60^2=3600$$ times less than the acceleration of the falling apple.

Applying $$a=v^2/r$$ to the acceleration of the moon yielded an acceleration that was indeed 3600 times smaller than $$9.8\ \text{m}/\text{s}^2$$, and Newton was convinced he had unlocked the secret of the mysterious force that kept the moon and planets in their orbits.

### Newton's law of gravity

The proportionality $$F\propto m/r^2$$ for the gravitational force on an object of mass $$m$$ only has a consistent proportionality constant for various objects if they are being acted on by the gravity of the same object. Clearly the sun's gravitational strength is far greater than the earth's, since the planets all orbit the sun and do not exhibit any very large accelerations caused by the earth (or by one another). What property of the sun gives it its great gravitational strength? Its great volume? Its great mass? Its great temperature? Newton reasoned that if the force was proportional to the mass of the object being acted on, then it would also make sense if the determining factor in the gravitational strength of the object exerting the force was its own mass. Assuming there were no other factors affecting the gravitational force, then the only other thing needed to make quantitative predictions of gravitational forces would be a proportionality constant. Newton called that proportionality constant $$G$$, so here is the complete form of the law of gravity he hypothesized.

##### Newton's law of gravity

$\begin{multline*} F = \frac{Gm_1m_2}{r^2} \shoveright{\text{[gravitational force between objects of mass}}\\ \shoveright{\text{ m_1 and m_2, separated by a distance r; r is not}}\\ \text{the radius of anything ]} \end{multline*}$

Newton conceived of gravity as an attraction between any two masses in the universe. The constant $$G$$ tells us how many newtons the attractive force is for two 1-kg masses separated by a distance of 1 m. The experimental determination of $$G$$ in ordinary units (as opposed to the special, nonmetric, units used in astronomy) is described in section 10.5. This difficult measurement was not accomplished until long after Newton's death.

##### Example 1: The units of $$G$$

$$\triangleright$$ What are the units of $$G$$?

$$\triangleright$$ Solving for $$G$$ in Newton's law of gravity gives

$\begin{equation*} G = \frac{Fr^2}{m_1m_2} , \end{equation*}$

so the units of $$G$$ must be $$\text{N}\!\cdot\!\text{m}^2/\text{kg}^2$$. Fully adorned with units, the value of $$G$$ is $$6.67\times10^{-11}\ \text{N}\!\cdot\!\text{m}^2/\text{kg}^2$$.

##### Example 2: Newton's third law

$$\triangleright$$ Is Newton's law of gravity consistent with Newton's third law?

$$\triangleright$$ The third law requires two things. First, $$m_1$$'s force on $$m_2$$ should be the same as $$m_2$$'s force on $$m_1$$. This works out, because the product $$m_1m_2$$ gives the same result if we interchange the labels 1 and 2. Second, the forces should be in opposite directions. This condition is also satisfied, because Newton's law of gravity refers to an attraction: each mass pulls the other toward itself.

##### Example 3: Pluto and Charon
$$\triangleright$$ Pluto's moon Charon is unusually large considering Pluto's size, giving them the character of a double planet. Their masses are $$1.25\times10^{22}$$ and $$1.9x10^{21}$$ kg, and their average distance from one another is $$1.96\times10^4$$ km. What is the gravitational force between them?

$$\triangleright$$ If we want to use the value of $$G$$ expressed in SI (meter-kilogram-second) units, we first have to convert the distance to $$1.96\times10^7\ \text{m}$$. The force is

$\begin{multline*} \frac{ \left(6.67\times10^{-11}\ \text{N}\!\cdot\!\text{m}^2/\text{kg}^2\right) \left(1.25\times10^{22}\ \text{kg}\right) \left(1.9 \times10^{21}\ \text{kg}\right) }{\left(1.96\times10^7\ \text{m}\right)^2} \\ = 4.1\times10^{18}\ \text{N} \end{multline*}$

The proportionality to $$1/r^2$$ in Newton's law of gravity was not entirely unexpected. Proportionalities to $$1/r^2$$ are found in many other phenomena in which some effect spreads out from a point. For instance, the intensity of the light from a candle is proportional to $$1/r^2$$, because at a distance $$r$$ from the candle, the light has to be spread out over the surface of an imaginary sphere of area $$4\pi r^2$$. The same is true for the intensity of sound from a firecracker, or the intensity of gamma radiation emitted by the Chernobyl reactor. It's important, however, to realize that this is only an analogy. Force does not travel through space as sound or light does, and force is not a substance that can be spread thicker or thinner like butter on toast.

Although several of Newton's contemporaries had speculated that the force of gravity might be proportional to $$1/r^2$$, none of them, even the ones who had learned Newton's laws of motion, had had any luck proving that the resulting orbits would be ellipses, as Kepler had found empirically. Newton did succeed in proving that elliptical orbits would result from a $$1/r^2$$ force, but we postpone the proof until the chapter 15 because it can be accomplished much more easily using the concepts of energy and angular momentum.

Newton also predicted that orbits in the shape of hyperbolas should be possible, and he was right. Some comets, for instance, orbit the sun in very elongated ellipses, but others pass through the solar system on hyperbolic paths, never to return. Just as the trajectory of a faster baseball pitch is flatter than that of a more slowly thrown ball, so the curvature of a planet's orbit depends on its speed. A spacecraft can be launched at relatively low speed, resulting in a circular orbit about the earth, or it can be launched at a higher speed, giving a more gently curved ellipse that reaches farther from the earth, or it can be launched at a very high speed which puts it in an even less curved hyperbolic orbit. As you go very far out on a hyperbola, it approaches a straight line, i.e., its curvature eventually becomes nearly zero.

Newton also was able to prove that Kepler's second law (sweeping out equal areas in equal time intervals) was a logical consequence of his law of gravity. Newton's version of the proof is moderately complicated, but the proof becomes trivial once you understand the concept of angular momentum, which will be covered later in the course. The proof will therefore be deferred until section 15.7.

self-check:

Which of Kepler's laws would it make sense to apply to hyperbolic orbits?

◊ Solved problem: Visiting Ceres — problem 10 ◊ Solved problem: Geosynchronous orbit — problem 16 ◊ Solved problem: Why $$a$$ equals $$g$$ — problem 11 ◊ Solved problem: Ida and Dactyl — problem 12 ◊ Solved problem: Another solar system — problem 15 ◊ Solved problem: Weight loss — problem 19 ◊ Solved problem: The receding moon — problem 17

##### Discussion Questions

How could Newton find the speed of the moon to plug in to $$a=v^2/r?$$

Two projectiles of different mass shot out of guns on the surface of the earth at the same speed and angle will follow the same trajectories, assuming that air friction is negligible. (You can verify this by throwing two objects together from your hand and seeing if they separate or stay side by side.) What corresponding fact would be true for satellites of the earth having different masses?

What is wrong with the following statement? “A comet in an elliptical orbit speeds up as it approaches the sun, because the sun's force on it is increasing.”

Why would it not make sense to expect the earth's gravitational force on a bowling ball to be inversely proportional to the square of the distance between their surfaces rather than their centers?

Does the earth accelerate as a result of the moon's gravitational force on it? Suppose two planets were bound to each other gravitationally the way the earth and moon are, but the two planets had equal masses. What would their motion be like?

Spacecraft normally operate by firing their engines only for a few minutes at a time, and an interplanetary probe will spend months or years on its way to its destination without thrust. Suppose a spacecraft is in a circular orbit around Mars, and it then briefly fires its engines in reverse, causing a sudden decrease in speed. What will this do to its orbit? What about a forward thrust?

## 10.3 Apparent weightlessness

If you ask somebody at the bus stop why astronauts are weightless, you'll probably get one of the following two incorrect answers:

(1) They're weightless because they're so far from the earth.

(2) They're weightless because they're moving so fast.

The first answer is wrong, because the vast majority of astronauts never get more than a thousand miles from the earth's surface. The reduction in gravity caused by their altitude is significant, but not 100%. The second answer is wrong because Newton's law of gravity only depends on distance, not speed.

The correct answer is that astronauts in orbit around the earth are not really weightless at all. Their weightlessness is only apparent. If there was no gravitational force on the spaceship, it would obey Newton's first law and move off on a straight line, rather than orbiting the earth. Likewise, the astronauts inside the spaceship are in orbit just like the spaceship itself, with the earth's gravitational force continually twisting their velocity vectors around. The reason they appear to be weightless is that they are in the same orbit as the spaceship, so although the earth's gravity curves their trajectory down toward the deck, the deck drops out from under them at the same rate.

Apparent weightlessness can also be experienced on earth. Any time you jump up in the air, you experience the same kind of apparent weightlessness that the astronauts do. While in the air, you can lift your arms more easily than normal, because gravity does not make them fall any faster than the rest of your body, which is falling out from under them. The Russian air force now takes rich foreign tourists up in a big cargo plane and gives them the feeling of weightlessness for a short period of time while the plane is nose-down and dropping like a rock.

## 10.4 Vector addition of gravitational forces

l / Gravity only appears to pull straight down because the near perfect symmetry of the earth makes the sideways components of the total force on an object cancel almost exactly. If the symmetry is broken, e.g., by a dense mineral deposit, the total force is a little off to the side.

m / A, who is outside a spherical shell of mass, feels gravitational forces from every part of the shell --- stronger forces from the closer parts, and weaker ones from the parts farther away. The shell theorem states that the vector sum of all the forces is the same as if all the mass had been concentrated at the center of the shell. B, at the center, is clearly weightless, because the shell's gravitational forces cancel out. Surprisingly, C also feels exactly zero gravitational force.

n / The asteroid Toutatis, imaged by the space probe Chang'e-2 in 2012, is shaped like a bowling pin.

Pick a flower on earth and you move the farthest star. -- Paul Dirac

When you stand on the ground, which part of the earth is pulling down on you with its gravitational force? Most people are tempted to say that the effect only comes from the part directly under you, since gravity always pulls straight down. Here are three observations that might help to change your mind:

• If you jump up in the air, gravity does not stop affecting you just because you are not touching the earth: gravity is a noncontact force. That means you are not immune from the gravity of distant parts of our planet just because you are not touching them.
• Gravitational effects are not blocked by intervening matter. For instance, in an eclipse of the moon, the earth is lined up directly between the sun and the moon, but only the sun's light is blocked from reaching the moon, not its gravitational force --- if the sun's gravitational force on the moon was blocked in this situation, astronomers would be able to tell because the moon's acceleration would change suddenly. A more subtle but more easily observable example is that the tides are caused by the moon's gravity, and tidal effects can occur on the side of the earth facing away from the moon. Thus, far-off parts of the earth are not prevented from attracting you with their gravity just because there is other stuff between you and them.
• Prospectors sometimes search for underground deposits of dense minerals by measuring the direction of the local gravitational forces, i.e., the direction things fall or the direction a plumb bob hangs. For instance, the gravitational forces in the region to the west of such a deposit would point along a line slightly to the east of the earth's center. Just because the total gravitational force on you points down, that doesn't mean that only the parts of the earth directly below you are attracting you. It's just that the sideways components of all the force vectors acting on you come very close to canceling out.

A cubic centimeter of lava in the earth's mantle, a grain of silica inside Mt. Kilimanjaro, and a flea on a cat in Paris are all attracting you with their gravity. What you feel is the vector sum of all the gravitational forces exerted by all the atoms of our planet, and for that matter by all the atoms in the universe.

When Newton tested his theory of gravity by comparing the orbital acceleration of the moon to the acceleration of a falling apple on earth, he assumed he could compute the earth's force on the apple using the distance from the apple to the earth's center. Was he wrong? After all, it isn't just the earth's center attracting the apple, it's the whole earth. A kilogram of dirt a few feet under his backyard in England would have a much greater force on the apple than a kilogram of molten rock deep under Australia, thousands of miles away. There's really no obvious reason why the force should come out right if you just pretend that the earth's whole mass is concentrated at its center. Also, we know that the earth has some parts that are more dense, and some parts that are less dense. The solid crust, on which we live, is considerably less dense than the molten rock on which it floats. By all rights, the computation of the vector sum of all the forces exerted by all the earth's parts should be a horrendous mess.

Actually, Newton had sound reasons for treating the earth's mass as if it was concentrated at its center. First, although Newton no doubt suspected the earth's density was nonuniform, he knew that the direction of its total gravitational force was very nearly toward the earth's center. That was strong evidence that the distribution of mass was very symmetric, so that we can think of the earth as being made of layers, like an onion, with each layer having constant density throughout. (Today there is further evidence for symmetry based on measurements of how the vibrations from earthquakes and nuclear explosions travel through the earth.) He then considered the gravitational forces exerted by a single such thin shell, and proved the following theorem, known as the shell theorem:

If an object lies outside a thin, spherical shell of mass, then the vector sum of all the gravitational forces exerted by all the parts of the shell is the same as if the shell's mass had been concentrated at its center. If the object lies inside the shell, then all the gravitational forces cancel out exactly.

For terrestrial gravity, each shell acts as though its mass was at the center, so the result is the same as if the whole mass was there.

The second part of the shell theorem, about the gravitational forces canceling inside the shell, is a little surprising. Obviously the forces would all cancel out if you were at the exact center of a shell, but it's not at all obvious that they should still cancel out perfectly if you are inside the shell but off-center. The whole idea might seem academic, since we don't know of any hollow planets in our solar system that astronauts could hope to visit, but actually it's a useful result for understanding gravity within the earth, which is an important issue in geology. It doesn't matter that the earth is not actually hollow. In a mine shaft at a depth of, say, 2 km, we can use the shell theorem to tell us that the outermost 2 km of the earth has no net gravitational effect, and the gravitational force is the same as what would be produced if the remaining, deeper, parts of the earth were all concentrated at its center.

The shell theorem doesn't apply to things that aren't spherical. At the point marked with a dot in figure n, we might imagine that gravity was in the direction shown by the dashed arrow, pointing toward the asteroid's center of mass, so that the surface would be a vertical cliff almost a kilometer tall. In reality, calculations based on the assumption of uniform density show that the direction of the gravitational field is approximately as shown by the solid arrow, making the slope only about $$60°$$.1 This happens because gravity at this location is more strongly affected by the nearby “neck” than by the more distant “belly.” This slope is still believed to be too steep to keep dirt and rocks from sliding off (see problem 11, p. 229).

self-check:

Suppose you're at the bottom of a deep mineshaft, which means you're still quite far from the center of the earth. The shell theorem says that the shell of mass you've gone inside exerts zero total force on you. Discuss which parts of the shell are attracting you in which directions, and how strong these forces are. Explain why it's at least plausible that they cancel.

##### Discussion Questions

If you hold an apple, does the apple exert a gravitational force on the earth? Is it much weaker than the earth's gravitational force on the apple? Why doesn't the earth seem to accelerate upward when you drop the apple?

When astronauts travel from the earth to the moon, how does the gravitational force on them change as they progress?

How would the gravity in the first-floor lobby of a massive skyscraper compare with the gravity in an open field outside of the city?

In a few billion years, the sun will start undergoing changes that will eventually result in its puffing up into a red giant star. (Near the beginning of this process, the earth's oceans will boil off, and by the end, the sun will probably swallow the earth completely.) As the sun's surface starts to get closer and close to the earth, how will the earth's orbit be affected?

## 10.5 Weighing the earth

p / A simplified version of Cavendish's apparatus.

Let's look more closely at the application of Newton's law of gravity to objects on the earth's surface. Since the earth's gravitational force is the same as if its mass was all concentrated at its center, the force on a falling object of mass $$m$$ is given by

$\begin{equation*} F = G \: M_\text{earth} \: m \: / \: r_\text{earth}^2 . \end{equation*}$

The object's acceleration equals $$F/m$$, so the object's mass cancels out and we get the same acceleration for all falling objects, as we knew we should:

$\begin{equation*} g = G \: M_\text{earth} \: / \: r_\text{earth}^2 . \end{equation*}$

o / Cavendish's apparatus. The two large balls are fixed in place, but the rod from which the two small balls hang is free to twist under the influence of the gravitational forces.

Newton knew neither the mass of the earth nor a numerical value for the constant $$G$$. But if someone could measure $$G$$, then it would be possible for the first time in history to determine the mass of the earth! The only way to measure $$G$$ is to measure the gravitational force between two objects of known mass, but that's an exceedingly difficult task, because the force between any two objects of ordinary size is extremely small. The English physicist Henry Cavendish was the first to succeed, using the apparatus shown in figures o and p. The two larger balls were lead spheres 8 inches in diameter, and each one attracted the small ball near it. The two small balls hung from the ends of a horizontal rod, which itself hung by a thin thread. The frame from which the larger balls hung could be rotated by hand about a vertical axis, so that for instance the large ball on the right would pull its neighboring small ball toward us and while the small ball on the left would be pulled away from us. The thread from which the small balls hung would thus be twisted through a small angle, and by calibrating the twist of the thread with known forces, the actual gravitational force could be determined. Cavendish set up the whole apparatus in a room of his house, nailing all the doors shut to keep air currents from disturbing the delicate apparatus. The results had to be observed through telescopes stuck through holes drilled in the walls. Cavendish's experiment provided the first numerical values for $$G$$ and for the mass of the earth. The presently accepted value of $$G$$ is $$6.67\times10^{-11}\ \text{N}\!\cdot\!\text{m}^2/\text{kg}^2$$.

Knowing $$G$$ not only allowed the determination of the earth's mass but also those of the sun and the other planets. For instance, by observing the acceleration of one of Jupiter's moons, we can infer the mass of Jupiter. The following table gives the distances of the planets from the sun and the masses of the sun and planets. (Other data are given in the back of the book.)

 average distance from the sun, in units of the earth’s average distance from the sun mass, in units of the earth’s mass sun — 330,000 mercury 0.38 0.056 venus 0.72 0.82 earth 1 1 mars 1.5 0.11 jupiter 5.2 320 saturn 9.5 95 uranus 19 14 neptune 30 17 pluto 39 0.002
##### Discussion Questions

It would have been difficult for Cavendish to start designing an experiment without at least some idea of the order of magnitude of $$G$$. How could he estimate it in advance to within a factor of 10?

Fill in the details of how one would determine Jupiter's mass by observing the acceleration of one of its moons. Why is it only necessary to know the acceleration of the moon, not the actual force acting on it? Why don't we need to know the mass of the moon? What about a planet that has no moons, such as Venus --- how could its mass be found?

## 10.6 Dark energy (optional)

Until recently, physicists thought they understood gravity fairly well. Einstein had modified Newton's theory, but certain characteristrics of gravitational forces were firmly established. For one thing, they were always attractive. If gravity always attracts, then it is logical to ask why the universe doesn't collapse. Newton had answered this question by saying that if the universe was infinite in all directions, then it would have no geometric center toward which it would collapse; the forces on any particular star or planet exerted by distant parts of the universe would tend to cancel out by symmetry. More careful calculations, however, show that Newton's universe would have a tendency to collapse on smaller scales: any part of the universe that happened to be slightly more dense than average would contract further, and this contraction would result in stronger gravitational forces, which would cause even more rapid contraction, and so on.

When Einstein overhauled gravity, the same problem reared its ugly head. Like Newton, Einstein was predisposed to believe in a universe that was static, so he added a special repulsive term to his equations, intended to prevent a collapse. This term was not associated with any interaction of mass with mass, but represented merely an overall tendency for space itself to expand unless restrained by the matter that inhabited it. It turns out that Einstein's solution, like Newton's, is unstable. Furthermore, it was soon discovered observationally that the universe was expanding, and this was interpreted by creating the Big Bang model, in which the universe's current expansion is the aftermath of a fantastically hot explosion.2 An expanding universe, unlike a static one, was capable of being explained with Einstein's equations, without any repulsion term. The universe's expansion would simply slow down over time due to the attractive gravitational forces. After these developments, Einstein said woefully that adding the repulsive term, known as the cosmological constant, had been the greatest blunder of his life.

q / The WMAP probe's map of the cosmic microwave background is like a “baby picture” of the universe.

This was the state of things until 1999, when evidence began to turn up that the universe's expansion has been speeding up rather than slowing down! The first evidence came from using a telescope as a sort of time machine: light from a distant galaxy may have taken billions of years to reach us, so we are seeing it as it was far in the past. Looking back in time, astronomers saw the universe expanding at speeds that were lower, rather than higher. At first they were mortified, since this was exactly the opposite of what had been expected. The statistical quality of the data was also not good enough to constitute ironclad proof, and there were worries about systematic errors. The case for an accelerating expansion has however been supported by high-precision mapping of the dim, sky-wide afterglow of the Big Bang, known as the cosmic microwave background. This is discussed in more detail in section 27.4.

So now Einstein's “greatest blunder” has been resurrected. Since we don't actually know whether or not this self-repulsion of space has a constant strength, the term “cosmological constant” has lost currency. Nowadays physicists usually refer to the phenomenon as “dark energy.” Picking an impressive-sounding name for it should not obscure the fact that we know absolutely nothing about the nature of the effect or why it exists.

Dark energy is discussed in more detail on p. 27.4.4.

## 10.7 A gravitational test of Newton's first law (optional)

This section describes a high-precision test of Newton's first law. The left panel of figure r shows a mirror on the moon. By reflecting laser pulses from the mirror, the distance from the earth to the moon has been measured to the phenomenal precision of a few centimeters, or about one part in $$10^{10}$$. This distance changes for a variety of known reasons. The biggest effect is that the moon's orbit is not a circle but an ellipse, with its long axis about 11% longer than its short one. A variety of other effects can also be accounted for, including such exotic phenomena as the slightly nonspherical shape of the earth, and the gravitational forces of bodies as small and distant as Pluto. Suppose for simplicity that all these effects had never existed, so that the moon was initially placed in a perfectly circular orbit around the earth, and the earth in a perfectly circular orbit around the sun.

r / Left: The Apollo 11 mission left behind a mirror, which in this photo shows the reflection of the black sky. Right: A highly exaggerated example of an observation that would disprove Newton's first law. The radius of the moon's orbit gets bigger and smaller over the course of a year.

If we then observed something like what is shown in the right panel of figure r, Newton's first law would be disproved. If space itself is symmetrical in all directions, then there is no reason for the moon's orbit to poof up near the top of the diagram and contract near the bottom. The only possible explanation would be that there was some preferred frame of reference of the type envisioned by Aristotle, and that our solar system was moving relative to it. Another test for a preferred frame was described in example 3 on p. 239.

One could then imagine that the gravitational force of the earth on the moon could be affected by the moon's motion relative to this frame. The lunar laser ranging data3 contain no measurable effect of the type shown in figure r, so that if the moon's orbit is distorted in this way (or in a variety of other ways), the distortion must be less than a few centimeters. This constitutes a very strict upper limit on violation of Newton's first law by gravitational forces. If the first law is violated, and the violation causes a fractional change in gravity that is proportional to the velocity relative to the hypothetical preferred frame, then the change is no more than about one part in $$10^7$$, even if the velocity is comparable to the speed of light.

## Vocabulary

ellipse — a flattened circle; one of the conic sections

conic section — a curve formed by the intersection of a plane and an infinite cone

hyperbola — another conic section; it does not close back on itself

period — the time required for a planet to complete one orbit; more generally, the time for one repetition of some repeating motion

focus — one of two special points inside an ellipse: the ellipse consists of all points such that the sum of the distances to the two foci equals a certain number; a hyperbola also has a focus

## Notation

$$G$$ — the constant of proportionality in Newton's law of gravity; the gravitational force of attraction between two 1-kg spheres at a center-to-center distance of 1 m

## Summary

{}

Kepler deduced three empirical laws from data on the motion of the planets:

• Kepler's elliptical orbit law: The planets orbit the sun in elliptical orbits with the sun at one focus.
• Kepler's equal-area law: The line connecting a planet to the sun sweeps out equal areas in equal amounts of time.
• Kepler's law of periods: The time required for a planet to orbit the sun is proportional to the long axis of the ellipse raised to the 3/2 power. The constant of proportionality is the same for all the planets.
Newton was able to find a more fundamental explanation for these laws. Newton's law of gravity states that the magnitude of the attractive force between any two objects in the universe is given by

$\begin{equation*} F=Gm_1m_2/r^2 . \end{equation*}$

Weightlessness of objects in orbit around the earth is only apparent. An astronaut inside a spaceship is simply falling along with the spaceship. Since the spaceship is falling out from under the astronaut, it appears as though there was no gravity accelerating the astronaut down toward the deck.

Gravitational forces, like all other forces, add like vectors. A gravitational force such as we ordinarily feel is the vector sum of all the forces exerted by all the parts of the earth. As a consequence of this, Newton proved the shell theorem for gravitational forces:

If an object lies outside a thin, uniform shell of mass, then the vector sum of all the gravitational forces exerted by all the parts of the shell is the same as if all the shell's mass was concentrated at its center. If the object lies inside the shell, then all the gravitational forces cancel out exactly.

## Homework Problems

t / Problem 8.

u / Problem 12.

v / Problem 21.

1. Roy has a mass of 60 kg. Laurie has a mass of 65 kg. They are 1.5 m apart.
(a) What is the magnitude of the gravitational force of the earth on Roy?
(b) What is the magnitude of Roy's gravitational force on the earth?
(c) What is the magnitude of the gravitational force between Roy and Laurie?
(d) What is the magnitude of the gravitational force between Laurie and the sun?(answer check available at lightandmatter.com)

2. During a solar eclipse, the moon, earth and sun all lie on the same line, with the moon between the earth and sun. Define your coordinates so that the earth and moon lie at greater $$x$$ values than the sun. For each force, give the correct sign as well as the magnitude. (a) What force is exerted on the moon by the sun? (b) On the moon by the earth? (c) On the earth by the sun? (d) What total force is exerted on the sun? (e) On the moon? (f) On the earth?(answer check available at lightandmatter.com)

3. Suppose that on a certain day there is a crescent moon, and you can tell by the shape of the crescent that the earth, sun and moon form a triangle with a $$135°$$ interior angle at the moon's corner. What is the magnitude of the total gravitational force of the earth and the sun on the moon? (If you haven't done problem 2 already, you might want to try it first, since it's easier, and some of its results can be recycled in this problem.) (answer check available at lightandmatter.com)

s / Problem 3.

4. How high above the Earth's surface must a rocket be in order to have 1/100 the weight it would have at the surface? Express your answer in units of the radius of the Earth.(answer check available at lightandmatter.com)

5. The star Lalande 21185 was found in 1996 to have two planets in roughly circular orbits, with periods of 6 and 30 years. What is the ratio of the two planets' orbital radii?(answer check available at lightandmatter.com)

6. In a Star Trek episode, the Enterprise is in a circular orbit around a planet when something happens to the engines. Spock then tells Kirk that the ship will spiral into the planet's surface unless they can fix the engines. Is this scientifically correct? Why?

7. (a) Suppose a rotating spherical body such as a planet has a radius $$r$$ and a uniform density $$\rho$$, and the time required for one rotation is $$T$$. At the surface of the planet, the apparent acceleration of a falling object is reduced by the acceleration of the ground out from under it. Derive an equation for the apparent acceleration of gravity, $$g$$, at the equator in terms of $$r$$, $$\rho$$, $$T$$, and $$G$$.(answer check available at lightandmatter.com)
(b) Applying your equation from a, by what fraction is your apparent weight reduced at the equator compared to the poles, due to the Earth's rotation?(answer check available at lightandmatter.com)
(c) Using your equation from a, derive an equation giving the value of $$T$$ for which the apparent acceleration of gravity becomes zero, i.e., objects can spontaneously drift off the surface of the planet. Show that $$T$$ only depends on $$\rho$$, and not on $$r$$.(answer check available at lightandmatter.com)
(d) Applying your equation from c, how long would a day have to be in order to reduce the apparent weight of objects at the equator of the Earth to zero? [Answer: 1.4 hours]
(e) Astronomers have discovered objects they called pulsars, which emit bursts of radiation at regular intervals of less than a second. If a pulsar is to be interpreted as a rotating sphere beaming out a natural “searchlight” that sweeps past the earth with each rotation, use your equation from c to show that its density would have to be much greater than that of ordinary matter.
(f) Astrophysicists predicted decades ago that certain stars that used up their sources of energy could collapse, forming a ball of neutrons with the fantastic density of $$\sim10^{17}\ \text{kg}/\text{m}^3$$. If this is what pulsars really are, use your equation from c to explain why no pulsar has ever been observed that flashes with a period of less than 1 ms or so.

8. You are considering going on a space voyage to Mars, in which your route would be half an ellipse, tangent to the Earth's orbit at one end and tangent to Mars' orbit at the other. Your spacecraft's engines will only be used at the beginning and end, not during the voyage. How long would the outward leg of your trip last? (Assume the orbits of Earth and Mars are circular.) (answer check available at lightandmatter.com)

9. (a) If the earth was of uniform density, would your weight be increased or decreased at the bottom of a mine shaft? Explain.
(b) In real life, objects weigh slightly more at the bottom of a mine shaft. What does that allow us to infer about the Earth?

10. (solution in the pdf version of the book) Ceres, the largest asteroid in our solar system, is a spherical body with a mass 6000 times less than the earth's, and a radius which is 13 times smaller. If an astronaut who weighs 400 N on earth is visiting the surface of Ceres, what is her weight?

11. (solution in the pdf version of the book) Prove, based on Newton's laws of motion and Newton's law of gravity, that all falling objects have the same acceleration if they are dropped at the same location on the earth and if other forces such as friction are unimportant. Do not just say, “$$g=9.8\ \text{m}/\text{s}^2$$ -- it's constant.” You are supposed to be proving that $$g$$ should be the same number for all objects.

12. (solution in the pdf version of the book) The figure shows an image from the Galileo space probe taken during its August 1993 flyby of the asteroid Ida. Astronomers were surprised when Galileo detected a smaller object orbiting Ida. This smaller object, the only known satellite of an asteroid in our solar system, was christened Dactyl, after the mythical creatures who lived on Mount Ida, and who protected the infant Zeus. For scale, Ida is about the size and shape of Orange County, and Dactyl the size of a college campus. Galileo was unfortunately unable to measure the time, $$T$$, required for Dactyl to orbit Ida. If it had, astronomers would have been able to make the first accurate determination of the mass and density of an asteroid. Find an equation for the density, $$\rho$$, of Ida in terms of Ida's known volume, $$V$$, the known radius, $$r$$, of Dactyl's orbit, and the lamentably unknown variable $$T$$. (This is the same technique that was used successfully for determining the masses and densities of the planets that have moons.)

13. If a bullet is shot straight up at a high enough velocity, it will never return to the earth. This is known as the escape velocity. We will discuss escape velocity using the concept of energy later in the course, but it can also be gotten at using straightforward calculus. In this problem, you will analyze the motion of an object of mass $$m$$ whose initial velocity is exactly equal to escape velocity. We assume that it is starting from the surface of a spherically symmetric planet of mass $$M$$ and radius $$b$$. The trick is to guess at the general form of the solution, and then determine the solution in more detail. Assume (as is true) that the solution is of the form $$r= kt^p$$, where $$r$$ is the object's distance from the center of the planet at time $$t$$, and $$k$$ and $$p$$ are constants.
(a) Find the acceleration, and use Newton's second law and Newton's law of gravity to determine $$k$$ and $$p$$. You should find that the result is independent of $$m$$.(answer check available at lightandmatter.com)
(b) What happens to the velocity as $$t$$ approaches infinity?
(c) Determine escape velocity from the Earth's surface.(answer check available at lightandmatter.com) ∫

14. Astronomers have recently observed stars orbiting at very high speeds around an unknown object near the center of our galaxy. For stars orbiting at distances of about $$10^{14}\ \text{m}$$ from the object, the orbital velocities are about $$10^6$$ m/s. Assuming the orbits are circular, estimate the mass of the object, in units of the mass of the sun, $$2\times10^{30}$$ kg. If the object was a tightly packed cluster of normal stars, it should be a very bright source of light. Since no visible light is detected coming from it, it is instead believed to be a supermassive black hole.(answer check available at lightandmatter.com)

15. (solution in the pdf version of the book) Astronomers have detected a solar system consisting of three planets orbiting the star Upsilon Andromedae. The planets have been named b, c, and d. Planet b's average distance from the star is 0.059 A.U., and planet c's average distance is 0.83 A.U., where an astronomical unit or A.U. is defined as the distance from the Earth to the sun. For technical reasons, it is possible to determine the ratios of the planets' masses, but their masses cannot presently be determined in absolute units. Planet c's mass is 3.0 times that of planet b. Compare the star's average gravitational force on planet c with its average force on planet b. [Based on a problem by Arnold Arons.]

16. (solution in the pdf version of the book) Some communications satellites are in orbits called geosynchronous: the satellite takes one day to orbit the earth from west to east, so that as the earth spins, the satellite remains above the same point on the equator. What is such a satellite's altitude above the surface of the earth?

17. As is discussed in more detail in example 3 on p. 386, tidal interactions with the earth are causing the moon's orbit to grow gradually larger. Laser beams bounced off of a mirror left on the moon by astronauts have allowed a measurement of the moon's rate of recession, which is about 1 cm per year. This means that the gravitational force acting between earth and moon is decreasing. By what fraction does the force decrease with each 27-day orbit? \hwhint{hwhint:receding-moon} (solution in the pdf version of the book) [Based on a problem by Arnold Arons.]

18. Suppose that we inhabited a universe in which, instead of Newton's law of gravity, we had $$F=k\sqrt{m_1m_2}/r^2$$, where $$k$$ is some constant with different units than $$G$$. (The force is still attractive.) However, we assume that $$a=F/m$$ and the rest of Newtonian physics remains true, and we use $$a=F/m$$ to define our mass scale, so that, e.g., a mass of 2 kg is one which exhibits half the acceleration when the same force is applied to it as to a 1 kg mass.
(a) Is this new law of gravity consistent with Newton's third law?
(b) Suppose you lived in such a universe, and you dropped two unequal masses side by side. What would happen?
(c) Numerically, suppose a 1.0-kg object falls with an acceleration of 10 $$\text{m}/\text{s}^2$$. What would be the acceleration of a rain drop with a mass of 0.1 g? Would you want to go out in the rain?
(d) If a falling object broke into two unequal pieces while it fell, what would happen?
(e) Invent a law of gravity that results in behavior that is the opposite of what you found in part b. [Based on a problem by Arnold Arons.]

19. (a) A certain vile alien gangster lives on the surface of an asteroid, where his weight is 0.20 N. He decides he needs to lose weight without reducing his consumption of princesses, so he's going to move to a different asteroid where his weight will be 0.10 N. The real estate agent's database has asteroids listed by mass, however, not by surface gravity. Assuming that all asteroids are spherical and have the same density, how should the mass of his new asteroid compare with that of his old one?
(b) Jupiter's mass is 318 times the Earth's, and its gravity is about twice Earth's. Is this consistent with the results of part a? If not, how do you explain the discrepancy?(solution in the pdf version of the book)

20. Where would an object have to be located so that it would experience zero total gravitational force from the earth and moon?(answer check available at lightandmatter.com)

21. The planet Uranus has a mass of $$8.68\times10^{25}$$ kg and a radius of $$2.56\times10^4$$ km. The figure shows the relative sizes of Uranus and Earth.
(a) Compute the ratio $$g_U/g_E$$, where $$g_U$$ is the strength of the gravitational field at the surface of Uranus and $$g_E$$ is the corresponding quantity at the surface of the Earth.(answer check available at lightandmatter.com)

22. The International Space Station orbits at an average altitude of about 370 km above sea level. Compute the value of $$g$$ at that altitude.(answer check available at lightandmatter.com)

w / Problem 23: New Horizons at its closest approach to Jupiter. (Jupiter's four largest moons are shown for illustrative purposes.) The masses are:
sun: $$1.9891\times10^{30}\ \text{kg}$$
Jupiter: $$1.8986\times10^{27}\ \text{kg}$$
New Horizons: 465.0 kg

23. On Feb. 28, 2007, the New Horizons space probe, on its way to a 2015 flyby of Pluto, passed by the planet Jupiter for a gravity-assisted maneuver that increased its speed and changed its course. The dashed line in the figure shows the spacecraft's trajectory, which is curved because of three forces: the force of the exhaust gases from the probe's own engines, the sun's gravitational force, and Jupiter's gravitational force. Find the magnitude of the total gravitational force acting on the probe. You will find that the sun's force is much smaller than Jupiter's, so that the magnitude of the total force is determined almost entirely by Jupiter's force. However, this is a high-precision problem, and you will find that the total force is slightly different from Jupiter's force.(answer check available at lightandmatter.com)

24. On an airless body such as the moon, there is no atmospheric friction, so it should be possible for a satellite to orbit at a very low altitude, just high enough to keep from hitting the mountains. (a) Suppose that such a body is a smooth sphere of uniform density $$\rho$$ and radius $$r$$. Find the velocity required for a ground-skimming orbit.(answer check available at lightandmatter.com)
(b) A typical asteroid has a density of about $$2\ \text{g}/\text{cm}^3$$, i.e., twice that of water. (This is a lot lower than the density of the earth's crust, probably indicating that the low gravity is not enough to compact the material very tightly, leaving lots of empty space inside.) Suppose that it is possible for an astronaut in a spacesuit to jump at $$2\ \text{m}/\text{s}$$. Find the radius of the largest asteroid on which it would be possible to jump into a ground-skimming orbit.(answer check available at lightandmatter.com)

25. The figure shows a region of outer space in which two stars have exploded, leaving behind two overlapping spherical shells of gas, which we assume to remain at rest. The figure is a cross-section in a plane containing the shells' centers. A space probe is released with a very small initial speed at the point indicated by the arrow, initially moving in the direction indicated by the dashed line. Without any further information, predict as much as possible about the path followed by the probe and its changes in speed along that path.

x / Problem 25.

26. Approximate the earth's density as being constant. (a) Find the gravitational field at a point P inside the earth and half-way between the center and the surface. Express your result as a ratio $$g_P/g_S$$ relative to the field we experience at the surface. (b) As a check on your answer, make sure that the same reasoning leads to a reasonable result when the fraction 1/2 is replaced by the value 0 (P being the earth's center) or the value 1 (P being a point on the surface).

27. The earth is divided into solid inner core, a liquid outer core, and a plastic mantle. Physical properties such as density change discontinuously at the boundaries between one layer and the next. Although the density is not completely constant within each region, we will approximate it as being so for the purposes of this problem. (We neglect the crust as well.) Let $$R$$ be the radius of the earth as a whole and $$M$$ its mass. The following table gives a model of some properties of the three layers, as determined by methods such as the observation of earthquake waves that have propagated from one side of the planet to the other.

 region outer radiusR massM mantle 1 0.69 outer core 0.55 0.29 inner core 0.19 0.017

The boundary between the mantle and the outer core is referred to as the Gutenberg discontinuity. Let $$g_s$$ be the strength of the earth's gravitational field at its surface and $$g_G$$ its value at the Gutenberg discontinuity. Find $$g_G/g_s$$.(answer check available at lightandmatter.com)

\begin{handson}{}{The shell theorem}{\onecolumn}

This exercise is an approximate numerical test of the shell theorem. There are seven masses A-G, each being one kilogram. Masses A-F, each one meter from the center, form a shape like two Egyptian pyramids joined at their bases; this is a rough approximation to a six-kilogram spherical shell of mass. Mass G is five meters from the center of the main group. The class will divide into six groups and split up the work required in order to calculate the vector sum of the six gravitational forces exerted on mass G. Depending on the size of the class, more than one group may be assigned to deal with the contribution of the same mass to the total force, and the redundant groups can check each other's results.

\includegraphics[width=78mm]{../share/mechanics/figs/ex-octahedron}

1. Discuss as a class what can be done to simplify the task of calculating the vector sum, and how to organize things so that each group can work in parallel with the others.

2. Each group should write its results on the board in units of piconewtons, retaining five significant figures of precision. Everyone will need to use the same value for the gravitational constant, $$G=6.6743\times10^{-11}\ \text{N}\!\cdot\!\text{m}^2/\text{kg}^2$$.

3. The class will determine the vector sum and compare with the result that would be obtained with the shell theorem.

\end{handson}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.

##### Footnotes
[1] Hudson et al., Icarus 161 (2003) 346
[2] Section section 19.5 presents some of the evidence for the Big Bang.
[3] Battat, Chandler, and Stubbs, http://arxiv.org/abs/0710.0702