We know where this is going. The astronomy and physics are about to come together in a law of universal gravitation, universal in the sense that a single cause will bring about both terrestrial and celestial effects. From the data and from the theoretical suggestions so far, Isaac Newton (1642–1727) discovered a force that is always exactly along the line between two objects and that decreases as the inverse of the square of the distance between them. These are the key features of the force vector, its direction and its magnitude. The goal of this chapter is to see how Newton got there.

Here is a summary of what Newton had to work with. From Galileo, both the Principle of Equivalence and the so-called time-squared law were key components in the derivation of the law of universal gravitation. Heavy things and light things fall at exactly the same speed, and that speed increases at a constant rate such that the distance fallen is proportional to the square of the time of falling. Both of these describe facts of terrestrial kinematics. From Kepler, the three laws of planetary motion were the most explicitly essential guides to universal gravitation. All three describe facts of celestial kinematics. It was Newton's accomplishment to pull all these together with a unifying force.

Newton himself provided some of the pieces of the puzzle. He, and independently Christiaan Huygens (1629–1695), determined from geometrical considerations that the magnitude of the acceleration of an object moving in a circle, the centripetal acceleration, depends on the square of the speed and inversely on the radius of the circle. This is a very general result that applies to circular motion on the Earth or in the sky. In this sense it is universal kinematics.

$a=v^2/r$ (7.1)

Newton also made the essential link between kinematics and dynamics with the second law of motion.

$F=ma$ (7.2)

In both of these equations we are describing only the magnitudes of the force and acceleration. So, even though they are vectors, we write the math without putting the variables in bold.

Equally important was Newton's third law of motion, that forces always come in pairs. If one object, the Sun, for example, exerts a force on another, a planet, then the planet exerts a force of equal magnitude on the Sun and exactly along the line between the two objects. This will bring some subtle but important changes to Kepler's model of planetary motion.

These were the ingredients. The challenge was to put them together into a theoretically coherent law that accurately matched the empirical data. As a general note on the logic of reasoning in this kind of situation, it is relatively easy to figure out how things will move if you already know the details of the cause of the motion and the initial conditions. This is the inference from dynamics to kinematics, from force to resulting acceleration. On the presumption that nature is deterministic rather than random, a cause determines the effect. But even in a deterministic world, a particular effect could have more than one possible cause. There may be more than one way to bring it about. This means that any inference from kinematics to dynamics is difficult. It is underdetermined in the sense that all the descriptive details on how things are moving do not have enough information on exactly what is causing the motion. Kinematics underdetermines dynamics, but this is exactly what Newton needs to do, “from the phenomena of motions to investigate the forces of nature.” And he is not alone. Again from the Preface to the Principia, “All the difficulty of [natural] philosophy seems to consist in this.” Figuring out the cause from the nature of the effect is a fundamental challenge, and accomplishment, of science. When it is explicitly presented in these terms of causation and inference it is called an inverse problem. This is to distinguish it from a direct problem, predicting the effect of a particular cause.

Newton's inverse problem was solved by first ignoring some of the details in the data. Just as Galileo was able to apply mathematics to experiments by judiciously discounting certain small imperfections in the phenomena, Newton had to idealize on the astronomical facts. The initial derivation of the law of universal gravitation assumed that the orbits of planets are perfect circles. It ignored, in other words, Kepler's first law. Remarkably, by the end of Newton's development of the theory of gravitation, the assumption of circularity disappeared, and the mathematics resulted in elliptical orbits. In this way the system of ideas that tied all the kinematic and dynamic pieces together was self-correcting.

Newton also started out neglecting any movement of the Sun. This too changed as the theory was fully developed, but initially the Sun's overwhelming mass allowed Newton to consider it unmoved by any of the planets, individually or collectively.

And finally, Newton regarded the Sun and each of the planets to be an amount of mass concentrated at a single point. He neglected the physical extension of the astronomical objects. He returned to this idealization after the fact, and, using the calculus, was able to explicitly prove that the gravitational effect of a spherical body is identical to that of a point particle. So, this idealized detail in the original derivation was not self-correcting, but it was harmless.

The next few pages include a few mathematical steps in Newton's derivation of universal gravitation. You can skip the math and still get the basic logic by skimming ahead to Equation (7.9) and the paragraph that starts, “To summarize the logic…”

With these initial simplifications, Newton was first able to link Kepler's third law to the inverse-square decrease in the force of attraction between two astronomical bodies. If the ratio of the orbital radius cubed to the period squared is the same for every planet, it follows that the magnitude of the force between the Sun and a planet decreases by the inverse-square law. In a succinct logical form, if r³∕P² = a constant, then F ∝ 1∕r².

Here is a sketch of the proof.

Considering the planetary orbit to be circular, the planet's acceleration is, by Equation (7.1), v²∕r$v^2∕r$. By Newton's second law, Equation (7.2), this requires a force that is the mass times the acceleration.

$F_{\mathrm{gravity}}=m{v}^2/r$ (7.3)

The speed v is the distance the planet travels per time, namely the circumference of its orbit divided by the time it takes for one complete orbit, the period P. The circumference is 2πr, so the required force is as follows.

$F_{\mathrm{gravity}}=m{(2\pi r/P)}^2/r$ (7.4)

Multiply the right-hand side by r∕r, that is, by 1, and regroup.

$F_{\mathrm{gravity}}=\left(1/r^2\right)\times [\left(4{\pi }^{2}m\right)\times \left(r^3/P^2\right)]$ (7.5)

Everything in the square bracket is a constant, including the (r³∕P²), since that is the constant specified by Kepler's third law. The result is that the force on the planet is some constant times 1∕r². In other words, we have the following.

$F_{\mathrm{gravity}}\propto 1/r^2$ (7.6)

The next step was to apply this idea to the orbit of the Moon, and compare the force holding the Moon in orbit to the force that accelerates a dropped stone or a falling apple. For this comparison, assume all of the mass of the Earth is located at a single point at the center. The distance to the Moon, the radius of its (approximately) circular orbit, is measured from the center of the Earth to the center of the Moon. Using our current value and units, this distance is R_M = 3.84 × 10⁸m. We'll use the capital R to denote a specific value of distance between two specific things, in this case the Earth and the Moon. Lower case r will be used for general, variable distance between unspecified objects. The dropped stone on the surface of the Earth is at a distance from the point source of gravity that is just equal to the radius of the Earth itself, R_E. That is, the stone is just like the Moon, left to its natural motion, but at a distance R_E = 6.38 × 10⁸m. The ratio of the two distances, R_E∕R_M is 1/60.

Since it moves in a circular orbit, the Moon accelerates toward the center at a rate of v²/R_M$v^2/R_{\mathrm{M}}$. The period of its orbit is roughly 28 days, or more precisely, 2.3 × 10⁶ s. When you do the math, the Moon's acceleration is a_M = 2.7 × 10^–3 m/s². The stone, or any other object dropped on the surface of the Earth, accelerates toward the center at a rate of 9.8 m/s², the acceleration of gravity g. The ratio of acceleration on the Earth to acceleration of the Moon is g/a_M = 3600$g/a_{\mathrm{M}}=3600$. Note, as Newton did, that this is the square of the inverse of the ratio of distances. And, since there is already reason to believe that the force holding the Moon in orbit depends on the square of the inverse of the distance, and that acceleration is proportional to the force, it follows that it is the same force holding the Moon in orbit as is the cause of the stone's falling. Thus gravity is universal, and astronomy and physics are united.

The acceleration of gravity, the rate of falling on the Earth, does not depend on the mass of the falling object. The only way this is possible is if the mass appears on both sides of the equation F = ma. It has to be on the left, in the formula for F, in order to cancel with its appearance on the right. So we now have two components of the force of gravity, mass m$m$ and 1/r²$1/r^2$.

$F_{\mathrm{gravity}}\propto m/r^2$ (7.7)

The force holding the Moon in orbit, the force of gravity, is proportional to the mass of the moon M_M (again using the capitalized notation to signify a specific value of a specific object) and the inverse of the distance squared 1∕R_M². Now use Newton's third law to point out that there must also be an equal and opposite force on the Earth caused by the Moon. By this symmetry we could have done the derivation from the perspective of the Moon, asking about the acceleration of the Earth. That puts the mass of the Earth, the accelerating object, on the right-hand side of the F = ma equation. And again, to make the acceleration independent of the mass, that factor has to be on the left-hand side of the equation as well, that is, in the formula for the force. Both masses, those of the Earth and the Moon, have to be in the formula for the force of gravity. This is required by the Principle of Equivalence. In general, for any two masses m₁ and m₂ separated by a distance r, we have the following.

$F_{\mathrm{gravity}}\propto m_1{m}_2/r^2$ (7.8)

All that's left is to determine the constant of proportionality, the so-called gravitational constant G. There is no principle or law that can be used to derive this constant. It is an empirical matter. It turns out to be G = 6.7 × 10^–11m³/kg s². All together, Newton's law of universal gravitation is in the following formula.

$F_{\mathrm{gravity}}=G{m}_1{m}_2/r^2$ (7.9)

There were more steps in Newton's derivation, but they can be summarized without the mathematical detail. He demonstrated that a central force, that is, a force directed exactly along the line between planet and Sun, results in Kepler's second law, that the planet sweeps out equal areas in equal times. (For this proof, see Cushing Reference Cushing1998, pp. 115–117.) This works for any central force, not just one with a 1∕r² dependence. When you add the 1∕r², the resulting orbit turns out to be an ellipse. (Again, Cushing Reference Cushing1998 gives a clear account of this, on pp. 119–122.) A central, inverse-square force, in other words, entails Kepler's first law. And though Newton started with the approximation of circular orbits, his analysis ended up with ellipses.

To summarize the logic of Newton's solution of the inverse problem, his derivation of the law of universal gravitation, start with Kepler's third law and Newton's own F = ma. From these the inverse-square relation, the 1∕r², follows. And, when combined with the Principle of Equivalence, that inertial mass is the same thing as gravitational-source mass, and Newton's third law of motion, it follows that the force on the Moon, and other orbiting objects, is the same as the force on a falling apple or stone, and it depends on the masses of the two bodies, Earth and Moon, or Earth and apple. The force of gravity is universal. Finally, combine the inverse-square relation with the central direction of the force, and both Kepler's second and first laws follow. The orbits of planets are elliptical.

The orbits of the planets are elliptical, but there is a subtle difference between Newton's model and Kepler's. It comes from Newton's third law, the one about equal and opposite forces. The Sun exerts a force on a planet to hold it in orbit and cause it to accelerate, so the planet must exert exactly the same force on the Sun. The Sun must be accelerating and in orbit. The rate of the Sun's acceleration is much less than that of the planet, since the Sun is more massive. We could use the orbit equation (Equation (3.7)) to show that this means the radius of the Sun's orbit is much smaller than that of the planet. The Sun and planet orbit the same point, the center of mass between the two. This point is at the focus of the ellipse. In other words, the planetary orbit is, as Kepler reported, elliptical, but the Sun is not at the hearth. The planet orbits an empty point in space, a void point, as shown in Figure 7.1.

Figure 7.1. The Newtonian model of planetary motion. This is Kepler's model, revised as necessary following Newton's laws of motion. Both the planet P and Sun S are in motion. They follow elliptical orbits around a common focus. The planet's orbit is much larger than the Sun's, but the orbital period is the same. The focus of the orbits is at the center of mass between S and P. It's a void point, but its location is determined by the masses of the real physical objects S and P.

There is an important difference between the void point in the Newtonian model and void points in previous models. The location of the void point for Newton is determined by the locations and properties, specifically masses, of real objects. The void point is at the center of mass of the objects involved. This is unlike the Aristotelian situation, in which the possible void point is simply at the center of the universe, independent of what the universe contains. And it is unlike the Ptolemaic void points that are strictly kinematic, with no dynamic role or consequences, and whose position and motion are freely adjustable to match the appearances of the planets, that is, to save the phenomena. In the Newtonian model, it's not the point in space that is having the dynamic effect; it's the mass of Sun and planet balanced at the unoccupied point in space. So, it's a void point, and a dynamic void point, but it is allowable by Machian principles that require real things to be the physical causes of motion.

Newton solved the first part of the basic problem of natural philosophy, “from the phenomena of motions to investigate the forces of nature…” He, and his successors, went on to great success with the second part as well, “from these forces to demonstrate the other phenomena.” The law of universal gravitation proved to be useful in explaining a variety of happenings in nature, and in getting a variety of things done. Its applications were genuinely universal.

The Newtonian theory explained the ocean tides. The mathematics are a bit daunting, but the basic mechanism is not. We can understand what's going on by using the modern concept of a field. This was not available to Newton, but with a few basic field vectors, directly from his formula for the force of gravity, the tidal phenomenon is apparent. Figure 7.2 shows a few of the vectors of gravitational force on a test mass at various places near the gravitational source M. The important features are the decrease in length (magnitude) of the force at greater distance from M, and the radial orientation of the vectors; they all point exactly toward M. These properties follow from the facts that the gravitational force decreases as 1∕r² and is a central force. In the language of field theory, this is an inhomogeneous field, since the direction and magnitude of the vectors are different at different points in space.

Figure 7.2. The gravitational field produced by a mass source M. The field is radial, meaning that all of the field vectors point directly toward M. The strength of the field decreases at greater distance from M. These two features make this an inhomogeneous field; the magnitude and orientation of the vectors are variables of position.

Consider four small test particles, each of mass m, at each of the points A, B, C, and D in Figure 7.2. A falls faster than B, since it is closer to the source M, so their tendency is to separate, to get further apart. C and D fall at the same rate, but they each have a component of force toward the other. Their tendency is to move closer together, squeezed by the converging, inhomogeneous field. In free-fall, this quartet would bulge out along the AB fall-line, and squeeze in along the CD line. If it were a water balloon in free-fall in this inhomogeneous field, it would in fact elongate, as in Figure 7.3.

Figure 7.3. The distorting effect on a fluid in the inhomogeneous field of mass M. This is a tidal effect, and the oceans on the Earth will bulge out at points A and B, in at points C and D, in the gravitational field of the Moon. There will be high tides at A and B, low tides at C and D.

The Earth is in free-fall toward the Moon. The Earth has a gravitational force on the Moon, and by Newton's third law the Moon has the equal and opposite force on the Earth. Like the Sun and planets, the Earth and Moon each orbits the center of mass between them. So, the Earth is in orbit, and that is simply free-fall – the only force acting on the Earth in this system is the gravity of the Moon – with sufficient tangential orbital velocity to avoid hitting the Moon. Fluids, like water, on the Earth will react just as the fluids in the freely falling water balloon. They will bulge, rise, at points closest to and furthest from the gravitational source M (now the Moon), and they will sink inward, lower, at points in between, C and D. High tides at A and B, low tides at C and D. With the Earth rotating as well as orbiting in the Earth–Moon system, those points are not stationary on the ground. Just as the position of the Moon changes in the sky, so do the positions of high and low tides.

Newton's law of universal gravitation seems to be on to something fundamental in nature, since it can, with a single simple formula, account for such diverse phenomena as the orbits of planets, falling apples and stones, the ocean tides, and our own feeling of weight. It was a remarkable accomplishment. But it came at considerable metaphysical cost, requiring a dynamic connection between two things with no contact and even no intermediate physical stuff between them. This is the action at a distance that Galileo and then Mach found mystical and (hence) unscientific. But it works, and that may be enough to make it acceptably scientific. It works especially well when it is conceptually upgraded to being a thing itself, a field.

Another application of the law of universal gravitation highlights just how effective and useful it is to think in terms of fields. As a consequence of the 1∕r² in the force of gravity, it is possible to escape the gravitational bonds of the source such as the Sun or the Earth. Since the force decreases quickly with distance from the source – it's the inverse of the distance squared – it's possible to start out going fast enough that you never slow down to zero and fall. What goes up, doesn't come down, as long as it starts going up fast enough. The particular speed required is the escape velocity. It depends on the mass of the object you are leaving, the Sun or a planet, say, and where you start. The closer you are to the source, and the more massive the source, the more speed it will take to get away.

It's not difficult to calculate exactly what the escape velocity is, as a function of the mass and distance of the source. Then we can put in the numbers, for example, the mass and radius of the Earth, to see how fast you would have to go to escape the Earth's gravitational pull. It's particularly easy using fields, and the concept of energy.

The kinetic energy K of an object is defined as $K=1/2m{v}^2$. It is the energy of movement, and something that is not moving, v = 0$v=0$, has no kinetic energy. Kinetic energy is a scalar, and this is what makes it easy to work with. A related property is work W. It's related in that, as defined in physics, work also requires that something moves. It is also a scalar. The work done by a force F is defined as W = Fx$W=Fx$, where x is the distance the object moves under the influence of the force. This is a simplified definition; it works only if the force is constant and the motion is exactly along the direction of the force. It can be generalized to variable force and any orientation of motion, but this basic case will serve.

If you hold a heavy stone at arm's length, no work is done. You are indeed applying a force to oppose the gravitational force on the stone; you have to push up with a force equal to the stone's weight, F = mg. But since there is no change of position, no x, there is no work done. If you lift the stone through a vertical distance x, then you do work by an amount Fx, that is, mgx. If you apply a force horizontally to an object, that will cause it to accelerate. This work changes the velocity of the object, and consequently changes its kinetic energy.

Focusing on this last feature of work, the defining formula can be rewritten to make it explicit that the work done on an object is equal to the change in its kinetic energy ΔK$\Delta K$. That is, W = ΔK$W=\Delta K$. But what happened in the case of lifting the stone? You did work to move the stone some vertical distance x, but its speed, and hence its kinetic energy, did not change during the lifting. To cover this, we expand the idea of energy. There is kinetic energy, the energy of motion, and there is potential energy V, a kind of stored energy that can be released to do work. In the lifting case, the work increases the potential energy by positioning the stone higher in the gravitational field. If released from that height, the stone would pick up speed, and the work done pays off in the form of kinetic energy. Potential energy is energy in terms of position in the field. Higher up in the gravitational field means greater potential energy. And it doesn't matter how the object got higher up. The gravitational field near the surface of the Earth is approximately homogeneous, as shown in Figure 7.4. The change in potential energy ΔV between points A and B depends only on the vertical distance x between the two points: ΔV = mgx$\Delta V=mgx$. The total work done to get from A to B is path-independent. This is one of the simplifying virtues of working with scalars. The result depends only on the end-points.

Figure 7.4. The potential energy difference between points at different heights in a homogeneous gravitational field. The potential energy at B is greater than the potential energy at A by an amount that depends only on the vertical distance x (and the mass of an object at point B and A). It is independent of the path taken from A to B.

The most general relation between work and energy is that the work done by a force is the change in total energy E, kinetic plus potential. If there is no external force applied and hence no work done, then the total energy is constant. This is the law of conservation of energy.

Apply the law of conservation of energy to an example of tossing a stone straight up. What goes up, must come down. How fast will it be going when it comes down? If it starts with speed v, say 10 m/s, then it will return with exactly the same speed, 10 m/s. As it goes up, the kinetic energy decreases while the potential energy increases. The total energy stays the same. At the top of its flight, the stone's vertical speed is zero in the instant it must stop and then head down. At the top, all the energy is potential energy. As it goes down, the potential energy decreases while the kinetic energy increases. When it reaches the same vertical position from which it was tossed at 10 m/s, all the potential energy will be returned as kinetic. In other words, the kinetic energy will be the same returning as when it was launched, so the speed will be the same. What goes up at 10 m/s will come down at 10 m/s (at the altitude of the toss). And since the calculation is done in terms of scalars, work and energy, this result does not depend on the direction of the throw. Throw the stone straight up at 10 m/s, or at a 45° angle at 10 m/s, and when it returns to the ground it will be going 10 m/s.

Gravity is a conservative force. All the energy put in to move around in the field gets returned when you get back to where you started. No energy is lost, dissipated, as it would be to the force of friction. Friction is a non-conservative force. There is no potential energy associated with the force of friction, and there is no field in which to determine the energy of position. But with gravity and other conservative forces, there is a determinate property of position and path through the field, and so there is potential energy.

The analysis of potential energy has so far been assuming the gravitational field to be homogeneous, as shown in Figure 7.4. This is approximately true for the field near the surface of the Earth or the Sun, but we know that in fact the field is not homogeneous. The force gets weaker with distance as 1∕r², and the field lines converge toward the center of the source. The variable magnitude of force means that calculating the work to move an object from one altitude to another is not easy. It requires calculus. The procedure is to figure the work for a tiny distance, over which the force is essentially constant, and then add up all these increments of work from A to B. The general procedure, which we won't do, shows that if the force varies as 1∕r², then the potential energy varies as −1∕r. The minus sign makes sense, since the potential energy increases as r increases. Further from the ground means more energy stored. Without the minus sign v would be decreasing as 1∕r. The potential energy of a mass m at a distance r from a planet or the Sun with mass M is as follows.

$V_{\mathrm{gravity}}=-GMm/r$ (7.10)

A coordinate-system choice has been made in this formula, with the convention that V = 0 an infinite distance away from the source. This way, the source is understood to create a potential well, as shown in Figure 7.5, where the potential energy V is plotted as a function of distance r from the source.

Figure 7.5. A graph of gravitational potential energy V_gravity as a function of distance r from the source. The source, a star or planet, has physical radius R, so the graph and the formula V = –GMm∕r are only for values of r greater than R. At infinite distance, r → ∞, the potential energy is zero. The graph can be used to determine how much kinetic energy, and hence how much speed, is needed to get from one distance r₁ to another r₂. If r₂ → ∞, the calculated speed is exactly the escape velocity at point r₁.

If this is the potential well specifically for the Earth, then objects on the ground are at distance R_E, the radius of the Earth. An object dropped from an altitude r₁ will have potential energy V₁, as shown on Figure 7.5. Since it starts with zero velocity (it's dropped, not thrown) it starts with zero kinetic energy. Its total energy E is therefore all potential, that is, V₁. When it reaches the surface of the Earth, distance R_E, it will have acquired kinetic energy exactly equal to the change in potential energy between the position r₁ and R_E. Call this value of kinetic energy K₁. And knowing the kinetic energy, and the mass of the object, we could calculate the object's speed. K₁ is also the kinetic energy, and associated speed, needed to launch the object from the surface of the Earth to reach an altitude r₁. As it rises above the Earth, it loses speed, and hence kinetic energy, and gains potential energy. The point where the kinetic energy goes to zero, and where the vertical flight stops, is where all the energy has become potential, that is, r₁. To launch a rocket to an altitude r₁, the rocket needs initial kinetic energy such that, as it loses speed, v = 0$v=0$ when r = r₁.

Greater altitude requires more energy, greater launch-speed. To get to r₂ from the surface of the Earth requires greater kinetic energy. The rocket has to climb higher up the well. But if had started from r₁, to get to r₂ would require only the difference between V₁ and V_2.

The potential-energy well and the law of conservation of energy can be used to derive the exact formula for escape velocity. Think of it as escape kinetic energy. How much kinetic energy is required to escape the well? To just exactly make the escape, the launched object will slow to zero velocity at infinite distance, that is, v = 0 when r → ∞. At that point, K = 0. By the choice of coordinate system in Figure 7.5, V = 0, also at infinite distance. The total energy E is the sum of K and V, so at infinite distance, E = K + V = 0 + 0 = 0. Actually, since total energy is a constant, the total energy is zero at all points and all times in the flight of an object that has exactly what it takes to escape. So, K + V = 0, and K = –V. This is the energy requirement for escape. Putting in the specifics for K and V, ½ mv² = GMm∕r, and solving for v$v$, we find the escape velocity.

$v_{\mathrm{escape}}={\left(2GM/r\right)}^{1/2}$ (7.11)

The escape velocity depends on where you start, r, and on the mass of the gravitational source, M. It does not depend on the mass m of the escaping object, the rocket or the stone; m dropped out of the equation, as it usually does with gravity when the equation is kinematics. Gravitation is universal both in the sense of affecting everything everywhere, but also in the sense of being egalitarian in its effect. Whether it's a rocket or a stone, the necessary escape velocity from the Earth is the same. And when you put in the numbers for the mass and radius of the Earth, the radius because we are launching from the Earth's surface, the escape velocity from the surface of the Earth is 1.1 × 10⁴ m/s, that is, 11 km/s.

The escape velocity from any mass source M depends on where you are. The closer to M, the stronger the gravity and so the faster you have to go to escape. This has an interesting consequence for very dense bodies, that is, bodies with a lot of mass M compacted inside a small radius R. It means you can get very close to the mass, so the escape velocity must be very fast, perhaps even faster than the speed of light. You can see where this is going, or maybe you can't. The idea of a black hole, an object so dense that light cannot escape its gravitational field, was suggested not long after Newton derived the 1∕r² formula for the force of gravity. It's a pretty simple idea. If the Earth were to shrink in size but not in mass, and you stayed at the same distance r from the center, the escape velocity would remain at 11 km/s. That's because M and r don't change in the v_escape$v_{\text{escape}}$ formula. But if you rode in on the surface of the shrinking Earth, r would decrease so v_escape$v_{\text{escape}}$ would increase. If the shrinking continues you will get to a radius so small that v_escape$v_{\text{escape}}$ equals the speed of light c = 3 × 10⁸ m/s. The calculation is easy and, for the mass of the Earth, the radius at which the escape velocity is equal to the speed of light is about 10 mm. All of the mass of the Earth compressed to a sphere that small would create a black hole. This early understanding of black holes is a bit simplistic, in that it ignores some important properties of light, but it gets the basic idea across that gravity can affect even light.

There is more you can do with Newton's law of universal gravitation, the basic F_gravity = Gm₁m₂/r²$F_{\mathrm{gravity}}=G{m}_1{m}_2/r^2$ formula. The orbit equation (Equation (3.7)) that prescribes the orbital speed and radius for any planet, moon, or man-launched satellite is easily derivable. This, and the correlation between gravitational potential energy and kinetic energy – speed again – are the basis of contemporary space travel. That came later, centuries after Newton but, even at Newton's time, the applications of the theory were many and varied. It may be fair to summarize the accomplishment in terms of coherence and simplicity. What had appeared to be very different kinds of phenomena governed by different laws were brought together in a systematic connection between Earth and sky, cause and effect, force and motion. The pieces now fit together in the efficient package of Newton's second law and the law of universal gravitation. These are some of the conceptual virtues, simplicity and coherence, discussed in Chapter 4. Add to these the empirical virtues of explaining so many phenomena and being so useful in getting things done (it is rocket science!), and the theory seems well justified by scientific standards. All that remained was explicitly testing the theory in the pattern of precise prediction and follow-up observation. A scientific theory must be both testable and tested. Newton's theory was both, and two of the most informative tests, informative about gravity and about scientific method, will be described in the next chapter.