Every physics textbook used for the introductory survey has a section on gravity. It is invariably about Newton's law of universal gravitation, and not much else. The Newtonian theory is what you're expected to know, if you claim to know anything about gravity. It is also the common standard of comparison for other theories of gravity, as in, the Aristotelian theory is unlike the Newtonian in that…, or, the general theory of relativity describes gravity differently than the Newtonian theory by saying that…For all these reasons, it's a good idea to start the science of gravity with Newton. For most of us, this is starting with the familiar, and making sure the things we have seen before are clear and correct. Familiarity can accommodate complacency and sloppiness, that is, imprecision, and that won't do for a science of gravity.

The presentation in this chapter will be in the straightforward style of a textbook, with little philosophical or methodological reflection. There will be no historical context, nothing about the process that led to or followed Newton's theory of universal gravitation, and no challenge about its being true or false or about things that are real or metaphorical. It will just be the nuts and bolts of the theory. This is pretty much what textbooks do. The goal is clarity in providing the stable, reliable background knowledge that a scientist needs to participate in the current paradigm. It's the important beginning of the expertise and authority it takes to be a peer, qualified to do peer review.

The central concept of Newtonian gravity is the description of the force. This is pure dynamics. The force of gravity between two objects is proportional to their masses m₁ and m₂, and inversely proportional to the square of the distance r between them. That is,

${\mathbf{F}}_{\mathrm{gravity}}\propto m_1{m}_2/{\mathbf{r}}^2$ (3.1)

The coupling property in the case of gravity is mass. This is the analog to electric charge; it is the property of an object that makes it both cause and response to the force of gravity. The constant of proportionality, the coupling constant, is called the gravitational constant G. Its value is discovered by empirical measurement rather than by derivation from other properties. It is simply an unexplained feature of nature.

$G=6.67\times {10}^{-11}{\mathrm{N}\mathrm{m}}^2/\mathrm{k}{\mathrm{g}}^2$

The N stands for Newtons, a unit of force. Putting all the pieces together we get the exact formula for the gravitational force between the two objects.

${\mathbf{F}}_{\mathrm{gravity}}=G{m}_1{m}_2/{\mathbf{r}}^2$ (3.2)

There are some important details implicit in the formula that should be made explicit.

First of all, the force of gravity is always attractive, unlike the electric force that can be either attractive or repulsive. Mass m has only one sign, always positive. Like masses attract, and all masses are alike in sign, so all masses attract.

Second, the coupling property m, the analog to q in the electric force, is the same property m in Newton's second law of motion, F = ma. So, with m in the formula for F_gravity, there is an m on both sides of the F = ma equation. It's the same m in the cause F_gravity as in the effect, the acceleration a. It's worth asking whether this is just a coincidence, another basic fact of nature that has no explanation. It is what it is. Or is there a deeper, more fundamental reason for this, and hence an explanation? The double-dipping of mass explains why everything on the Earth, heavy and light things alike, all fall to the ground at the same rate. Well it explains insofar as a coincidence itself can be cited as an explanation. We need to keep an eye out for a reason why what is called the gravitational mass, the m in the F_gravity, is identical to the inertial mass, the m in the F = ma formula. But for now, just take it as an empirically well-established fact.

A third noteworthy feature of the formula for the force of gravity is the 1∕r². This is exactly like the electric force, an inverse-square force. The force gets weaker with distance, and quickly, given the squared r. It gets weaker but it never goes to zero. This shows gravity to be a ubiquitous glue. It is an attraction between any two objects located at any two points in the universe. The attraction quickly becomes, as the physicists say, negligible, but it is always there. And since the magnitude of the force is a continuous function of position, position of one of the objects with respect to the other, we are on the way to describing it as a field.

It's worth noting a few numerical values of the magnitude of the gravitational force, to put things into quantitative perspective. It's easy to calculate the gravitational force between the Sun and the Earth. Look-up and put in the numbers for the mass of the Sun, the mass of the Earth, and the distance r between the two bodies, and do the math. The result is 3.6 × 10²² N. For comparison, that's about 8 × 10²¹ pounds. This is the force of the Sun on the Earth. The Earth weighs 8 × 10²¹ pounds in the gravitational pull of the Sun. What is the force of the Earth on the Sun? Exactly the same value. The formula for the force of gravity tells us the force of one object on another, without distinguishing which is the one and which is the other. This is exactly in keeping with Newton's third law of motion, that forces come in equal and opposite pairs. The Earth attracts the Sun exactly as forcefully as the Sun attracts the Earth. The Sun is bigger and more massive, but no more forceful than the Earth. The gravitational glue is a mutually shared property.

The formula reports the magnitude of the gravitational force. There is also the direction of the vector to keep track of, but this is easy. It is always attractive and exactly along the line between the two objects. It is a radial force. There is no tangential component of the vector, that is, no component perpendicular to the line connecting the two objects.

This description of the direction of the force vector would be ambiguous for an extended object like the Sun or the Earth or even an apple, since there is not just one point to which the vector could point. Physics deals with this by simplifying the sizable object as if all the mass was located at just one point. The terminology is to regard the Sun and the Earth as point particles. This is not fiction or idealization of mere convenience, since the force of gravity in fact does act as if all the mass is located at just one point. For a sphere of uniform density, the point is, not surprisingly, the geometric center of the object. If the object is oddly shaped or of asymmetric density, there is nonetheless a real point, the center of mass, that is the focus of gravitational force.

With this simplification, the gravitational situation of Sun and Earth can be represented by the simple vector diagram as shown in Figure 3.1. There is an equal and opposite force on each object, directed exactly toward the other.

Figure 3.1. The equal and opposite gravitational forces between the Sun and the Earth. The force on each body is directed toward the other. The magnitude of the force is exactly the same on both the Sun and Earth.

This two-body situation can be generalized to determine the gravitational field. As in the case of electricity, it will be the field generated by and surrounding just one of the objects, this time one object with mass. And again, the field will be in terms of what gravitational force the object would cause at every point in space. The field is determined at every point, even at empty points in space. The units of the field can be standardized by expressing it as force per mass, F∕m. This is simply the acceleration of a test mass, determined at every point. In fact, it's the acceleration of any mass at that point. This generalization follows from the formula for the gravitational force. To find the gravitational field produced by an object of mass M, now using the capital M to represent not a variable mass but a particular mass, first find the force on a test mass m: F = GMm∕r². The force per mass F∕m divides out the m, leaving GM∕r², the acceleration of gravity as a function of distance from the object M. This is the field, with the acceleration vectors all pointing directly to M, getting shorter as they get further from the source. A few of the field vectors are shown in Figure 3.2.

Figure 3.2. Gravitational-field vectors near a massive object M. The field always points directly toward M, and the strength decreases further away from M.

If the source of the gravitational field is the Sun, then M is the mass of the Sun M_Sun. Put the Earth in its place a distance r (not bold, since it is just the magnitude of the distance we are working with here) from the Sun, and the Earth will accelerate toward the Sun at a rate of GM_Sun∕r². When you put in the numbers, the result is 6 × 10^–3 m/s, a modest number, but acceleration nonetheless.

This raises two questions. If the Earth is accelerating directly toward the Sun, why doesn't it get closer to the Sun? And if the force of the Sun on the Earth has an equal and opposite force of the Earth on the Sun, why isn't the Sun accelerating?

We should answer the second question immediately, because the question itself is mistaken. The Sun does accelerate toward the Earth. The force on the Sun is equal to the force on the Earth, but the resulting acceleration is much less because of the Sun's much greater mass. Acceleration is inversely proportional to mass, and the Sun is 3.3 × 10⁵ times more massive than the Earth, so its acceleration is 3 × 10^–6 that of the Earth. There is also the complicating factor of the other planets that create gravitational forces on the Sun. Jupiter is particularly influential, because it is so massive. The Sun is pulled in many changing directions, and its resulting motion, its acceleration, is in response to the composite of all these planetary forces.

The answer to the first question, the one about the acceleration of the Earth, opens the discussion of planetary motion, the kinematics of orbit. Keeping something in orbit requires an inward, central force. If you whirl a ball on a string around over your head, you can feel the pull. More to the point, you have to supply the pull to keep the ball from flying off, out of orbit. Any turning, as the ball is constantly turning, requires a force toward the center of the curve. When you drive around a bend, the force comes from friction between the road and the tires. If there's ice on the road this friction is lost and you can't make the turn. This inward force is called a centripetal force, regardless of what causes it, whether it's tension in the sting, friction on the road, or gravity. It's always a central force, pointing to the center of the curve, and gravity has exactly that feature. Gravity is the centripetal force that holds the planets in orbit around the Sun. It also holds man-made satellites in orbit around the Earth.

We know the direction of the gravitational force vector is right for the job, always pointing to the center of the orbit, but what about the magnitude of the force? It is possible to mathematically prove that the magnitude of the gravitational force is exactly the centripetal force required for orbit, but it will be better to get the result by a conceptual argument. This way it will make sense.

The faster the orbiting object is going, the more force it takes to make it turn. This is why you slow down to make a turn on ice, as there is less friction, less centripetal force, to pull the car inward and around the bend. This is also why, if you spin the ball on the string too fast, the string breaks. The string just can't handle the tension needed to supply the centripetal force. So expect the centripetal force to be proportional to the speed of the object in orbit. In fact, it's proportional to the square of the speed, indicating that a little more speed requires a lot more force. So, F ∝ v²$F\propto v^2$. (Again, these variables are not bold, since they represent only the magnitudes of each property.)

The more massive the orbiting object the more force required to hold it in orbit. If you spin a bowling ball on the string, rather than a tennis ball, it will take more force on your part, and more tension in the string, to hold it in orbit. So, F ∝ m.

Making a tighter turn requires more force. You're more likely to skid on a sharp turn than a gentle turn. In orbit, the smaller the radius r of the circle, the sharper the turning, so small r requires big F. In other words, F ∝ 1∕r.

Put these pieces together and we have the formula for centripetal force, the force needed to keep a mass m in orbit of radius r, if the speed of the object is v$v$.

$F_{\mathrm{centripetal}}=m{v}^2/r$ (3.3)

This results in an acceleration, since any force causes an acceleration a = F∕m. This is centripetal acceleration, along the line of the force, directed toward the center of the orbit.

$a_{\mathrm{centripetal}}=v^2/r$ (3.4)

An object in orbit is constantly accelerating, changing velocity in direction but not magnitude. Its speed remains constant, since its speed is always along the tangent of the circle, as in Figure 3.3. The acceleration of the Earth toward the Sun is a matter of changing direction, always being steered toward the Sun to stay in orbit. The Earth doesn't speed up or get closer to the Sun, just as the whirling tennis ball is neither speeding up nor getting closer to your hand. This is the nature of circular motion.

Figure 3.3. Two positions in the orbit of an object. The velocity changes from v₁ to v₂. The speed remains constant, so the lengths of the two vectors v₁ and v₂ are the same, but the direction changes. Δv represents the change in the direction of the velocity. This change in velocity is the acceleration, the centripetal acceleration. It points toward the orbital center.

Now put the gravity of a central body M as the source of the centripetal force on an orbiting body m. That is,

$F_{\mathrm{gravity}}=F_{\mathrm{centripetal}}$ (3.5)

$GMm/r^2=m{v}^2/r$ (3.6)

Note the m on both sides of the equation, and reduce the r² to get

$GM/r=v^2$ (3.7)

This describes the requirements for any orbit of anything, planet, Moon, satellite. It will be so useful throughout the modern science of gravity that we'll give it a name. Call it the orbit equation.

For a particular M, for example the Sun or the Earth as the center of an orbit, there are only two variables in the orbit equation, v and r. This shows that there is a fixed relation between the speed and radius of an orbit. Orbit at a particular distance r requires a specific speed v. The distance to the Moon, for example, is 3.8 × 10⁸ m. This is the distance from the center of the Earth to the center of the Moon, and that's the r to put in the orbit equation. There is only one possible orbital speed at this distance. From the orbit equation, that speed is 1 × 10³ m/s. Notice there was no need to know the mass of the Moon, since the orbit equation works for anything, independent of its mass. This is one of the wonders of gravity. Put a marble in the same orbit as the Moon and it will have to go exactly as fast as the Moon. The Earth's force on the two objects, marble and Moon, is very different, but the acceleration is the same. The dynamics depend on the mass of the orbiting object, but the kinematics do not. This, we will discover, poses a challenge to figuring out the mass of a distant astronomical object, using observations of its orbit.

The challenge can be met in some circumstances. For example, if M is very much greater than the mass m of the orbiting object, as the mass of the Sun is much larger than the mass of the Earth, we can determine the mass of M from the orbit of m. This is from the orbit equation. Do it for the Sun–Earth orbital system and calculate the mass of the Sun: GM∕r = v². Solve for M. Recall that r is the Earth–Sun distance (center to center), and v is the distance the Earth travels per time, that is, the circumference of its orbital circle divided by one year, the period of orbit. The result is M_Sun = 2 × 10³⁰ kg. In this way you could also find the mass of the Earth, using the orbital details of the Moon.

There are two noteworthy limitations to this technique using gravity and orbits to determine mass. The simple formula only works under the assumption that the central object sits still. That never actually happens, since, by Newton's third law, there is always a force on both objects, M and m, so M must be accelerating. We have neglected the acceleration of M by assuming it is so small as to make little difference, M being so much larger than m. The spirit of this chapter is basic Newtonian theory, and this shows just how basic the analysis is at this point.

A second limitation is more profound and more to the concerns of scientific method. Any theory of gravity will have to be tested, and that will require knowing the masses of distant objects – it's a theory of universal gravitation, after all – to see if the formula for F_gravity matches the observations. But we've used the theory of gravity to determine the masses. It would be circular reasoning to use a theory to supply key parameters used in its own testing. We'll have to deal with this when the Newtonian theory, and the general theory of relativity, are put to tests. For now note that it's one example of a more general challenge in testing theories about dynamics. Dynamics are hard to verify; kinematics are easy.

For objects in hand, things on the Earth like apples and stones, we can measure the mass directly, independent of any theory of gravity. So, testing the theory of gravity on the Earth avoids the problematic circularity. As this is a theory of universal gravitation, the formula must be the same on the surface of the Earth as in the empty space of the Solar system: F_gravity = GMm∕r², where M is the mass of the Earth, m is the mass of the apple or stone or what have you, and r is the radius of the Earth. The gravitational field of the Earth is as if all of the Earth's mass is at the one point, the center of the Earth. The distance between the apple and the source of gravity is the distance between the apple and the center of the Earth. This is r.

The force of the Earth's gravity on an object is its weight. From the formula, the weight is proportional to mass m. And from Newton's second law, the force will result in acceleration a = F∕m, that is, a = GM∕r². This is the same acceleration as for an object in orbit, as it should be, since this is universal gravitation, but for an object dropped on the surface of the Earth there is no tangential velocity. In this case the object accelerates, picks up speed, heading for the source, the center of the Earth. Regardless of the mass of the object, this acceleration is 9.8 m/s² on the surface of the Earth, the so-called acceleration of gravity, abbreviated with the letter g. Drop an apple or a stone and it speeds up at a rate of g = 9.8 m/s² as it falls. Toss the apple straight up and it slows down at a rate of 9.8 m/s² until the speed is zero at the top of its flight, and then it speeds up at 9.8 m/s² as it falls. Since this is independent of the mass of the object, heavy things and light things fall at the same rate. As with the orbit equation, the dynamics, in this case the weights, are different but the kinematics, the acceleration and trajectories, are the same. All of this is assuming that gravity is the only force in play. It ignores, for example, the force of air resistance. Since there generally is air in the room when things are dropped, this equal acceleration for all will be difficult to demonstrate.

The kinematics of free-fall are the same for any mass m. They are also the same at any place on the surface of the Earth, assuming the Earth to be a perfect sphere of uniform density. The kinematics are the same wherever you set up your coordinate system. You could even set it up in a train car. Drop something in the train car and it falls, accelerates straight down at 9.8 m/s². We didn't specify the mass of the object. We didn't even say whether the train was moving or not. As long as it's not turning, slowing down or speeding up, the law of free-fall, and the phenomenon of free-fall, will be the same on the train, moving or not. And if the window shades were down, you couldn't tell if the train was moving, at least not by doing experiments with gravity, experiments like dropping objects and measuring their acceleration.

There is an important generalization to be made from this. The results of any experiments in mechanics, and hence the laws that govern these experiments, are the same in all reference frames, at any place, moving or not, as long as the motion of the reference frame is steady, that is, at a constant velocity. Galileo pointed this out, and used it to argue that no experiment on the Earth could reveal whether the Earth was moving or not, since the outcome of the experiment would be the same either way. This general principle plays a fundamental role in both mechanics and the science of gravity. Einstein gave it a name, the Principle of Relativity. Note how misleading that name is. The principle says that the laws of physics are not relative to the reference frame; they are invariant, the same in all reference frames. Principle of Invariance would be a better name, but we are stuck; it's the Principle of Relativity.

Put the train in motion at some constant speed v, have someone on the train drop a stone, and consider what the trajectory of the falling stone looks like to someone standing on the ground. Assume they can see into the train car, and they can see the stone for the entire duration of the fall. The stone accelerates down at 9.8 m/s². It also has a constant horizontal speed v, as it stays with the train. The Principle of Relativity requires that the two components of motion, vertical and horizontal, are independent of each other. There is no horizontal force on the stone. It got its initial horizontal velocity from the train, but once let go, that force is gone. There is now only the force of gravity, straight down. The result of acceleration down plus constant speed horizontally is a curved trajectory. More precisely, it's a parabola. The arc gets steeper as the stone falls, since the stone is not just moving down, it's speeding up as it moves down.

Now do exactly the same experiment but without the train. A stone with some horizontal velocity is let go. That's the same as just throwing the stone horizontally. It is, in the language of mechanics, a projectile. The horizontal component of velocity continues unchanged, while the vertical component of velocity increases at 9.8 m/s². The trajectory is a parabola, as it was when the train was in the picture. The horizontal distance between the release of the stone, the throw, and its hitting the ground is called the range. This is shown in Figure 3.4.

Figure 3.4. Several moments at equal time intervals during the flight of a horizontally thrown projectile. The horizontal component of velocity is the same at each moment, while the vertical component of velocity steadily increases. The resulting trajectory is a parabola.

In a reference frame moving horizontally with the projectile we would record the object's motion as being simply straight down, as if simply dropped. This is back on the train, dropping the stone. If instead of a tossed stone the projectile is a bullet shot horizontally from a gun, the kinematics are exactly the same. The trajectory is a parabola, and the greater horizontal velocity results in a longer range. If you could speed along-side the bullet, you would see it simply fall straight down to the ground, at 9.8 m/s².

The parabolic trajectory of a projectile curves down and eventually meets the ground. If the horizontal velocity is really fast, the horizontal distance, the range, is really long. We've been assuming the ground is flat, but in fact the Earth is round, and the surface curves gently down and away from the flight of the projectile. If the horizontal velocity is great enough, making the range long enough, the Earth will curve away just as fast as the trajectory curves down. The projectile continues to fall (at 9.8 m/s²) but continues to miss the Earth. It will do this all the way around the Earth and the projectile is now a satellite; it's in orbit. Figure 3.5 shows some projectiles with successively higher velocities, until the one that finally makes it into orbit.

Figure 3.5. Projectiles thrown horizontally with progressively increasing velocity. With higher initial velocity, the range increases. At some point the initial horizontal velocity will be sufficient to put the object in orbit.

Just how fast would the projectile have to go to get into orbit? First figure out the range of a normal bullet, to see how close it comes. A typical muzzle velocity is 1000 m/s. If the bullet is fired horizontally, shoulder-high (approximately 2 meters off the ground), it will go about 600 meters before hitting the ground. There's not much curvature of the surface of the Earth in 600 meters, so the bullet won't even come close to making it into orbit. To calculate the speed it would take to achieve orbit may seem a complicated task, comparing the curvature of the trajectory and the curvature of the Earth. But there's an easy way to do it. Just use the orbit equation, Equation (3.7). What is the velocity required for an orbital radius equal to the radius of the Earth (plus the 2 meters shoulder-height)? This is a very idealized thought-experiment, assuming a perfectly spherical Earth with no obstructions over 2 meters high, but it gives us an idea of the speed required. When you do the math, the orbital speed is about 8000 m/s, eight times the speed of the bullet. Since the required orbital speed decreases for higher orbits – the r is in the denominator – real satellites have speeds somewhat less than this. They nonetheless need a significant boost, first the vertical to get to the orbital elevation, and then sideways to achieve the orbital speed.

As always, the mass of the object, the bullet or the satellite, never entered into the calculation. That's because it was pure kinematics, and with gravity, all masses respond the same. This fact is sufficiently important as to deserve a proper name. It's called the Principle of Equivalence, mentioned in Chapter 1. There are two versions of the principle at this point, and there will be one more as the science of gravity progresses. Later versions don't reject or revise the earlier; they add to them.

Once again it's Galileo who first clarified the principle. Heavy things and light things fall at the same rate. This is the first version of the Principle of Equivalence. The acceleration of gravity is independent of the mass, and even the weight, of the object. In modern terms, this means different dynamics – different weight, different force – do not cause different kinematics – acceleration. This is because the mass of the stone is on both sides of the equation, the gravitational force side, and the second law of motion side, the F = ma.

The mass of the stone will cancel, as long as the mass that is the coupling source of gravity is identical to the mass that is the inertial resistance to force. And this is the second way to express the Principle of Equivalence. This is Newton's formulation of the principle, that the gravitational mass m_g of an object, the property that causes the force of gravity, is equivalent to the inertial mass m_i, the property that resists the effect of any force, gravity included: m_g = m_i, so they cancel.

The equivalence of gravitational and inertial mass has the consequence of everything falling at the same acceleration, 9.8 m/s². It has the additional, related, consequence that the kinematic effects of gravity disappear in a reference frame that is falling along with the stone, falling at 9.8 m/s². If you do your physics in an elevator instead of a train, and allow the elevator to free-fall by cutting the cable, objects in this lab will not fall to the floor. The dropped stone will just float, stationary at the point of release. There will be no gravitational effects in the freely falling reference frame. The dynamics disappear with this kinematic change of frame.

The reverse is true as well, that is, an accelerating reference frame will result in gravitational effects, even if there are no massive objects around, no gravitational field. Do your physics in a spaceship, just drifting through space without the rockets burning, far from any gravitational sources. Nothing will fall in any direction, and the released stone will just float, exactly as it did in the falling elevator. But now fire the rockets so the spaceship is accelerating. All the loose objects, including the released stone, will fly toward the back of the ship, just as you are pressed back into the seat when a jet accelerates down the runway. In fact, all the loose objects will accelerate toward the back of the ship at the same rate, equal to the acceleration of the ship itself. The effect will exactly duplicate the effect of gravity, with the back of the ship as the new down. The dynamics appear with this kinematic change of reference frame.

This leads to a third version of the Principle of Equivalence. The effects of gravity are identical to the effects of an accelerating reference frame. This is the version that will eventually get us to the general theory of relativity.

We have identified two important principles in this chapter, the Principle of Equivalence and the Principle of Relativity. Both principles were recognized and used by Galileo. Both were named and strictly applied to the science of gravity by Einstein. Both play roles in the Newtonian theory of gravity, but it was Einstein's precision and enforcement of the principles that led to the downfall of the theory we have been working with in this chapter. We'll follow this principled undoing of the Newtonian theory, but first we should see what came before and what led to Newton's formula of the force of gravity.