- Ptolemy
- Kepler's First Law
- Conservation of Momentum
- One Body Problem
- Two Body Problem
- Kepler's Second Law
- A Path in Polar Coordinates
- Kepler's Second Law (Proof)
- Kepler's First Law (Proof)
- Function of Time
- Kepler's Third Law
- Hooke's Law

In the second century, Ptolemy misdirected most of his genius constructing elaborate formulas for the motions of the planets, based on two false premises that he considered self-evident. The sun, moon, and planets revolved around the earth, and, the circle was the perfect shape, the shape that the Gods would naturally use to construct the heavens. He placed circles within circles within circles, as though Mars were attached to a small wheel, spinning in a larger wheel, spinning on a track that went around the earth. His trig tables, a true leap forward in mathematics, enabled him to calculate the positions of the planets as they moved about in their "epicycles" - and his predictions were fairly accurate. Yet there were always discrepancies. The planet would be a bit to the left in winter, and a bit to the right in summer, and so on. The method of circles was only a first approximation.

By todays standards, we would say this theory was "ugly". Dozens of arbitrary and unrelated parameters are required to explain each orbit. There was no underlying mechanism that could explain these circles within circles. It looks contrived - a theory developed to fit the data - and it is.

At the beginning of the 16th century, Copernicus suggested that the planets might revolve around the sun. The center of the universe had shifted, yet the method of circles, with its revised, heliocentric parameters, still failed to predict planetary positions accurately.

In 1609, Kepler jettisoned the divine circles altogether, and discovered that each orbit was a perfect ellipse with the sun at one focus. This is his first law, and it changed astronomy forever.

Kepler also realized that the moon orbited the earth in an ellipse. The shape was not specific to solar orbits, Rather, it was a consequence of orbital mechanics everywhere. Any object in orbit around another object traces out an ellipse.

Today, we have modified the law slightly. Two objects are in orbit around their common center of mass c, and each object traces an Ellipse with c at one focus. If the objects have equal mass, they spin around each other like the ends of a dumbbell. Indeed, there are binary star systems where both stars are comparable in size. They swing around each other like two partners in a celestial dance. More often, one object is much larger than the other - like the sun and any or all of its planets. In this case c, the center of mass, lies just above the sun's surface, or deep within its interior, depending on which planet you are talking about. Jupiter travels around the sun every twelve years, yet the sun merely wabbles in response. Still, it does move, as jupiter tugs this way and that. We can detect these wabbles in other stars, and infer the presence of planets around them. How can we measure these tiny changes in the star's position from 100 lightyears away? Actually, we measure the change in the stars velocity as it moves towards and away from us. This is only a couple meters per second, but it alters the wave length of starlight, and our ultra sensitive spectrometers can detect these tiny frequency shifts. Thanks to this "wabble method", extra-solar planets are being discovered and catalogued every day.

Still other planets are betrayed by the tiny drop in starlight as they pass in front of the star. This is called the transit method. Still other planets will be seen directly, once the next generation of telescopes blocks the light from the parent star. We will literally see the planet. But these exciting developments in astronomy are beyond the scope of this book; so let's get back to Kepler's three laws.

Let n objects, sometimes referred to as "bodies", as in heavenly bodies, act upon each other through a force described by f(r), where f is a function of radius. Gravity is one such example, an inverse square law. This was not known in Kepler's day, but we know it now, so let's proceed.

Let n objects follow n separate paths in space. For convenience I will work in 3 dimensions, but this proof works in all dimensions. The paths are pi(t), as i runs from 1 to n. These have components xi(t), yi(t), and zi(t), indicating the x y and z coordinates of the paths as a function of time. There are thus 3n separate functions of time.

Let the ith object bi have a mass of mi, and let it be endowed with si stuff that generates the force. If the force is gravity then mi = si, but force could be electrostatic, whence si is the electric charge on each body.

Each function in time has its own differential equation. The acceleration of b1, in the x direction, is x1′′(t). This is force divided by mass, according to Newton's second law, which was still a century away. Thus m1 appears in the denominator. At the same time, the force between the first object and every other object is proportional to s1.

In the following, j runs from 2 to n, and d is shorthand for the distance between b1 and bj. The first factor on the right extracts the x component of the vector from location 1 to location j, the second is the force generating stuff on b1 and bj, and the third is the force / distance relationship. This is added up over all the objects in the system, save the first object, which exerts no force on itself.

x1′′ = ∑ (xj-x1)/d × s1×sj × f(d) / m1

Convention is important here. For gravity, f is a positive constant over d2. Put b2 to the right of b1 in the x direction, both having mass, and b2 pulls b1 to the right, which is a positive acceleration in the x direction, consistent with the sign of f. But if force is electric charge then f is a negative constant over d2. If b1 and b2 both have positive charge then force comes out negative, accelerating b1 to the left, away from b2.

There is of course one such equation for each coordinate and each object, giving a system of 3n differential equations in 3n unknowns, where each "unknown" is a function of t.

Assume the paths of the objects satisfy these differential equations, and add a constant c to each xi(t). Notice that the equations are still satisfied. Therefore the position of the origin is arbitrary. Let's put it at the center of mass at time 0, which is the sum of pi(0)×mi, divided by w, where w is the total mass of the system. With the origin in position, the sum over pi(0)×mi is 0.

Next, notice that any constant velocity ct can be added to each xi, and the equations are still satisfied. The system evolves the same way, whether it is fixed in space, or whether it is sliding across the sky. Set the net velocity to 0 at time 0. By net velocity, I really mean momentum. Thus the sum of pi′(0)×mi = 0.

Now do something clever. Multiply the 3 equations for pi, with xi′′, yi′′, and zi′′ on the left, through by mi, and do this for each i, and add all 3n equations together. The left side is the sum over acceleration times mass. That is, the sum of pi′′×mi. For each i < j, the right side includes (xi-xj)/d × si×sj × f(d) and (xj-xi)/d × sj×si × f(d). These cancel, and the right side is 0. This is the mathematical embodiment of Newton's third law: "For every action there is an equal and opposite reaction." Forces pull things together, or push things apart, pro rata.

The weighted sum of acceleration is 0. Integrate, and the weighted sum of velocity is constant. Thus momentum is conserved. It is neither created nor destroyed. In our frame of reference, momentum is 0 at time 0, thus it remains 0 forever.

Integrate again, and the weighted sum of position is constant. Again, this is 0 at time 0, hence it is 0 forever. The center of mass is conserved. The entire system does not suddenly jump to the right. If some objects fly off to the right, other objects fly off to the left to compensate. At all times, the origin is the center of mass of all the objects, and the swarm of objects has no net momentum.

If there is only one object, it has to sit at the origin forever, else the center of mass leaves the origin. In other words, it just sits there. If we had included a constant velocity to our equations, as described above, then our isolated heavenly body would continue on forever at that velocity. This is Newton's first law: "An object at rest remains at rest, and an object in motion remains in motion, unless acted upon by an outside force." Of course Newton did not discover these differential equations under a tree, and derive his first law from there. Instead, he deduced his first law by observing nature, then derived his second law, "force equals mass times acceleration", as a necessary condition for the first law; then he invented calculus to compute the trajectories of objects as they are acted upon by various forces. Newton validated his own three laws, then went on to proved Kepler's three laws, thereby confirming his theories of motion and gravity. Calculus, newtonian mechanics, orbital mechanics, gravity, optics, and the first reflecting telescope - it is difficult to overstate Newton's accomplishments.

The following proofs are fundamentally the same as those crafted by Newton, but they use modern notation, which makes them more concise and easier to understand.

Assume two bodies are in play, such as the earth and the sun. At time 0, let them have vectors v1 and v2, relative to the origin. As above, the origin is set to the center of mass, or m1v1 + m2v2 over w. The origin lies along a line joining v1 and v2, and closer to the body with the most mass. The center of mass does not change, hence this relationship persist throughout time. The bodies may swing around each other, or fly apart, or crash together, but they are always on opposite sides of the origin, and their distances to the origin are inversely related to their masses.

At any time t, the force is m1m2f(d), times x1-x2, or x2-x1. The forces are equal and opposite. This is Newton's third law, "For every action there is an equal and opposite reaction." The two bodies pull mutually upon each other, or push each other away (e.g. two protons repelling each other). Of course the actions are "equal" only when adjusted by mass. Push yourself away from a tree, and you do indeed push the tree, and the earth, away from you, but that reaction is imperceptible, given the enormous mass of the earth.

To make the problem simpler, I am going to temporarily suspend Newton's third law. Instead of two bodies, there will be one body flying through a force field that magically emanates from the origin. The gravity source is at the origin, and it does not move. It is anchored in space. When one body is much heavier than the other, this is a good approximation. The sun barely moves in response to the earth. It is, for all practical purposes, fixed at the origin, as the earth revolves around it. Newton assumed this was the case. His proof was later extended to two bodies orbiting around their common center of mass. Either approach is equivalent to the other.

If b1 (the first body) is twice as far from the origin as b2, this ratio persists for all time. Any other distance ratio would pull the center of mass away from the origin. When b1 moves 6 meters in, b2 moves 3 meters in; and when b1 moves 10 meters out, b2 moves 5 meters out.

In the above example, the distance from b1 to b2 is always 3/2 times the distance from b1 to the origin. Assume f is well behaved, like an inverse square or an inverse cube. If d is multiplied by 3/2, then f(d) is multiplied by some other constant, for every value of d. In the case of gravity, multiplying d by 3/2 multiplies force by 4/9, and that works for any d. Now, move b2 to the origin, and multiply f by 4/9. Lock b2 in place, and the force experienced by b1 is the same as that produced by the earlier two body problem. b1 responds to this force just as it did before. It is sufficient to chart the path of a single object moving through a fixed force field in space.

Draw an imaginary line from the sun to the earth, and it sweeps out area at a constant rate as the earth moves around the sun. This is Kepler's second law, which he phrased as "Equal areas in equal time." It isn't very interesting when the orbit is a perfect circle. The planet moves around at a fixed speed, and sweeps out area at a constant rate. But the time = area principle makes a big difference when the orbit is a long ellipse. The planet has to move very fast at its closest approach, so that the shorter line sweeps out the same area. At its farthest distance, the planet moves slowly.

This has practical applications. Most satellites run in circular orbits at or near the equator, and many are synchronized with the rotation of the earth. In other words, the orbital period is 24 hours. They seem to hang there in space, high above the equator, broadcasting mind-numbing television down to our sedentary population below. However, this does not serve the needs of countries near the north pole. A geosynchronous satellite is practically on the horizon, and communication can be blocked by something as simple as a tree or a nearby building. Therefore, Russia has placed its molniya communication satellites in highly elliptical orbits. They swing by the southern hemisphere quickly, just a few hundred kilometers above the earth, then they drift slowly across the northern hemisphere, some 40,000 km overhead. During most of their orbit, they can serve the needs of the Russian people; and with a substantial fleet in service, there are always a few satellites overhead.

Assume the acceleration on an object is always directed towards or away from a fixed point, say the origin. The force need not be inverse square; any radial force will do. The path lies in a plane, and sweeps out area at a constant rate.

The orbital plane is defined by the position and velocity vectors at time 0. If these are colinear, then we have an orbital line. The object is falling directly towards the origin. Either way, the object does not suddenly jump out of line, or out of plane. This is entirely intuitive. The acceleration is always within the plane, so the velocity vector remains in the plane, and the object remains in the plane. Differential geometry provides a formal proof, but that is beyond the scope of this book.

Since the orbit lies in a plane, switch to polar coordinates r(t) and θ(t), where all the force, and all the acceleration, is radial. But first, some formulas for velocity and acceleration for an arbitrary path in polar coordinates.

If a path is contained in the xy plane, you have the option of using polar coordinates r(t) and θ(t), rather than x(t) and y(t). Conversion is straightforward: x = r×cos(θ) and y = r×sin(θ). Use the product and chain rules to find the velocity.

x′ = r′×cos(θ) - r×sin(θ)×θ′

y′ = r′×sin(θ) + r×cos(θ)×θ′

That's fine, but we usually want the velocity expressed in the directions of r and θ, rather than the x and y coordinates. To switch to the new orthonormal basis, take the dot product of the above velocity vector with the unit radial vector and the unit tangent vector. These vectors are cos(θ),sin(θ) and -sin(θ),cos(θ) respectively. Dot each with x′,y′, and the velocity, measured along r and θ, is r′ and r×θ′. The radial speed is the change in r, and the tangential speed is the change in θ times r. When the particle is far from the origin, a small change in θ translates to a high tangential velocity.

Differentiate again, using the chain rule, to get acceleration. Dot this with the radial and tangential unit vectors, as we did with velocity. I'll spare you the algebra. The acceleration away from the origin is r′′ - r×θ′2. The first term is the radial acceleration, and the second is the centripetal acceleration. If the particle is tracing a perfect circle, the second term gives the acceleration needed to keep the particle in its orbit.

The acceleration in the tangential direction is rθ′′ + 2r′θ′. The first term is angular acceleration magnified by the radial distance, and the second term is the coriolis force. It takes extra force to keep a skater spinning at the same rate, as she extends her arms. Her hands, moving outward, represent r′×θ′.

Recall that the two body problem has been reduced to an object in the plane, responding to a force field that emanates from the origin. Switch to polar coordinates, so that the path is r(t) and θ(t). The acceleration is always radial, and never tangential. Review the formula for tangential acceleration in the previous section. This can be set to 0.

rθ′′ + 2r′θ′ = 0

(r2θ′)′ / r = 0 { expand by the product rule and get the above }

(r2θ′)′ = 0 { r is assumed to be nonzero throughout time }

r2θ′ = k { k is some constant }

½r2θ′ = ½k

If r is a function of θ, measuring the distance from the sun to the planet as a function of angle, the area of a tiny slice of pie of angle ε is ½r2ε. This is the area of a circle of radius r, cut down to a sliver with arc length ε. The area covered by the orbit, from time a to time b, is the integral of ½r2θ′. This is ½k×(b-a). As the angle moves from its position at time a to its position at time b, an imaginary line from the origin to the orbiting body sweeps out an area that is proportional to time. This proves Kepler's second law.

It's time to prove Kepler's first law: planets orbit the sun in ellipses. More generally, (and Kepler had no way of knowing this), a planet traces out a conic section as it responds to the gravity of the sun. If a comet is in orbit it travels in an ellipse, sometimes a long elliptical ellipse, perhaps returning every 76 years. But if a comet is simply passing through our solar system, it's path is bent by the sun into a hyperbola. It will not return, and if you are able to graph its conic section, and see that it is a hyperbola, you know it will never return; it is on its way to other stars.

Again, the sun is the origin, the planet follows a path around the sun, and this path is r(t) θ(t) in polar coordinates. Let u and v be the unit radial and tangential vectors of the path. Remember that u and v are functions of time. At any time t, they point away from the origin and around the circle respectively. Actually, u is cos(θ),sin(θ), where θ is a function of t, and v is -sin(θ),cos(θ). Meanwhile, r(t) is a scalar function, the distance from the origin. That fixes the position at p(t) = r(t)×u(t). The perpendicular unit vector v is a notational convenience.

The acceleration a(t) is g/r2 times u, plus 0 times v, where g is a suitable constant that describes the force field generated by the origin. If g is negative, the force is attractive; if g is positive, the force is repulsive. In orbital mechanics, i.e. gravity, g is negative. Remember that a is merely shorthand for p′′.

The tangential velocity is rθ′v. Take the cross product of position and tangential velocity. I'm taking advantage of a third dimension to build the cross product, but I don't think there's any harm; it's just a mathematical convenience. Picture the orbit as counterclockwise, hence v runs 90 degrees ahead of u, and z = u cross v stands up from the orbital plane, like the z axis. I will use × between two vectors to indicate the cross product. Thus u×v is the unit vector z. Radial distance × tangential speed is r2θ′, which is the constant k described above. The cross product of position and tangential velocity is kz.

What is the cross product of position and true velocity? Velocity is tangential velocity plus radial velocity, and cross product distributes over addition; hence we can cross position with radial velocity, and with tangential velocity, and add them up. Since position and radial velocity are parallel, this term drops out, leaving position cross tangential velocity, which was derived in the previous paragraph. Therefore p×p′ = kz.

Think about this geometrically. As the planet moves around in its orbit, a vector stands up out of plane. This vector always has height k, which is the area of the parallelogram determined by position and velocity. This holds even for a hyperbolic trajectory. The position vector grows and grows, as the object flies away from the origin, but velocity and position are almost parallel, hence the area of the associated parallelogram is still k.

If velocity and position really are parallel, k = 0, and the following equations drop to 0 = 0. This proof assumes the orbit defines a plane. The object is not falling directly towards, or flying away from, the origin.

Consider the cross product a×kz. Substitute for a, and kz, and get this.

gu/r2 × r2θ′z

Here is where we use the fact that f is inverse square. Cancel r2 and get gu × θ′z, or -gθ′v.

That's kz cross acceleration, how about kz cross velocity? The derivative of a dot product, or a cross product, can be computed using the traditional product rule. Use this to expand the derivative of velocity cross kz. The derivative of velocity is acceleration, and we already know a×kz is -gθ′v. Since kz is a constant vector, its derivative is 0. Therefore (p′×kz)′ is -gθ′v, just like a×kz.

Remember that u is a function of θ, is a function of t, and verify, by the chain rule, that the derivative of -gu(θ(t)) is -gθ′ times v, which is also the derivative of p′×kz. Integrate, and p′×kz is -g×(u+e), for some constant vector e.

Take the dot product with p to get a triple scalar product.

p . p′ × kz = p . -g(u+e)

p × p′ . kz = p . -g(u+e)

p×p′ = kz { this was demonstrated earlier }

kz . kz = p . -g(u+e)

k2 = p . -g(u+e)

p . -g(u+e) = k2

p.u + p.e = -k2/g

m = k2/g { notational convenience }

p.u + p.e = -m

r + p.e = -m

r + r×u.e = -m

We are working in 3 dimensions, so e might have a component in the z direction, but the projection of e into the orbital plane has the same dot product with u as e itself. So assume e lives in the orbital plane.

If e = 0 then r is constant, giving a perfect circle with radius -k2/g. Clearly g must be negative. This makes sense - you can't be in a circular orbit unless the force is attractive. Set this case aside and assume e is nonzero.

Let φ be the angle between u and e. Although e is constant, u varies with time, hence φ is actually a function of t.

#f'(t) = u(t) - e

r + r×|e|×cos(φ) = -m

At t = 0, p(0) is a point on the curve. Let p(0) lie on the positive x axis. Thus θ(0) = 0, and u(0) points along the positive x axis. Let j be φ(0). This is the angle between u(0) and e, a value between 0 and 2π. As t advances, θ advances, and u turns in lockstep. At all times, φ = j+θ.

r + r×|e|×cos(j+θ) = -m

Expand the cosine using the angle addition formula, replace r×cos(θ) with x, replace r×sin(θ) with y, replace r with sqrt(x2+y2), move the linear terms to the right, and square both sides. The result is a quadratic equation in x and y, which is a conic section.

Now that we know it's a conic section, select a point on the curve that is closes to the origin, and let that be your starting point. In other words, p(0) is closer to 0 than p(±ε), for small ε. As before, let the positive x axis contain p(0). Thus θ(0) = 0, and u(0) points along the positive x axis. The velocity runs straight up, parallel to the positive y axis. (This has to be so, since p(0) is closest to the origin.) Cross this with kz to find a vector that points along the positive x axis. It is also -g×(u+e). This means e runs along the x axis. It is either parallel or antiparallel to u. In the earlier equation, the offset j is either 0 or π.

r + r×|e|×cos(j+θ) = -m

j = 0 yields |e|×cos(θ), and j = π yields -|e|×cos(θ). Treat e as a real number, positive or negative, and cover both cases simultaneously.

r + r×e×cos(θ) = -m

r + ex = -m { moving back to rectangular coordinates }

sqrt(x2+y2) = -m - ex

x2 + y2 = m2 + 2emx + e2x2

(1-e2)x2 - 2emx + y2 = m2

fx2 - 2emx + y2 = m2 { notational convenience }

f×(x-em/f)2 - e2m2/f + y2 = m2 { assuming e is not 1 }

f×(x-em/f)2 + y2 = m2/f

This is a conic section with center at em/f. Unless you have protons repelling each other, m, like g, is negative. The semimajor axis, the distance from the center to the left and right edge of the ellipse, is m/f. e = 0 and f = 1 gives a circle. Set this aside and let e be somewhere between -1 and 1, so that f is still positive. If e is negative then the center is positive, and the origin is closer to the left edge of the ellipse. If e is positive then the center is negative, and the origin is closer to the right edge of the ellipse. Negating e keeps f the same, and slides the ellipse over, moving the center to the other side of the origin. The shape of the ellipse, how round or how flat, depends on the magnitude of e.

The semiminor axis has length m/sqrt(f). In an ellipse, the focus squared is the semimajor axis squared minus the semiminor axis squared. Evaluate m2/f2 - m2/f, remembering that f = 1-e2, and get e2m2/f2. This gives a focus of em/f, which is also the distance from the center to the origin. The origin is the focus, confirming part B of Kepler's first law: the planets orbit the sun in an ellipse, with the sun at one focus.

Bear in mind, you don't get to select e arbitrarily. Rather, e is determined by the initial conditions of position, velocity, and g. Usually p(0) and g are fixed, while velocity and e vary in lockstep.

Picture a rocket traveling around the earth in a perfect circle. At the bottom of the circle, fire the engine parallel to the direction of travel, thus increasing the tangential velocity. This puts more energy into the orbit. With new initial conditions, e is nonzero. The top of the orbit raises in response. Using a playground analogy, a push from a friend at the bottom of your swing allows you to go higher. The circle extends into an ellipse. Another burn at the bottom of the orbit makes the ellipse even longer. Finally, e could become 1, whence x2 drops out altogether, leaving a parabola. (In our equations the parabola runs sideways, not up; just turn your head.) The rocket is no longer in orbit. It has just enough energy to escape earth's gravity. This is escape velocity, about 25,000 miles per hour for the Earth. If we had fired the engine even longer, e would exceed 1, f would go negative, and the shape would be a hyperbola. In the extreme, e could be vary large, so large that earth's gravity is barely noticed. The rocket streaks away in a line that is just slightly bent by earth's gravity. This is a hyperbola that is nearly flat.

When an asteroid flies by the earth, it's path is a hyperbola. In general, an object cannot be captured by earth's gravity. If it is in orbit now, it was always in orbit; just retrace time. However, earth and its moon can conspire to capture an asteroid and place it in orbit. This is the three body problem, which is far more complicated.

We proved the orbit is a conic, but this does not predict the position of the object at any time t. In fact, there is no algebraic formula for this. However, one can approximate position as a function of time with great accuracy, using standard numerical analysis techniques. Nobody does this better than the Jet Propulsion Lab. A space craft takes off from Cape Canaveral, travels half a billion miles to reach Mars, and lands at its predetermined destination, with only a couple minor course corrections (delta v is 1 or 2 meters per second) along the way. Amazing!

The third law relates the orbital period (what we call a year) with the mean distance from the sun. But what is "mean distance" anyways? For a long ellipse, the average distance as a function of time is exaggerated, since the planet spends most of its time far from the sun. Instead, take the average across the entire path, as though the planet traveled at constant speed. It is the shape of the orbit that counts.

There are really two foci, one at the sun and one out in space. Consider the average distance to both foci. By symmetry, this is twice the average distance to either focus alone. Now - the distance to one focus, plus the distance to the other focus, is equal to the major axis. Thus the mean distance is the semimajor axis.

By Kepler's second law, area and time are proportional, according to the constant k. Thus the orbital period is the area of the ellipse divided by k.

The area is π times the semimajor axis times the semiminor axis. Continuing the notation used above, this is πm2/f3/2. Bring in 1/k, then square the result. Remember that k2 downstairs cancels an m upstairs. Ignoring the constants of proportionality, you get m3/f3. This is the cube of the semimajor axis. Therefore, the period of revolution squared is proportional to the mean distance cubed. This is Kepler's third law.

As you move outward in our solar system, the planets take longer to revolve around the sun. Mars is half again as far from the sun as Earth, approximately, hence a martian year is sqrt(1.53) or about 1.85 earth years.

There is another radial force that leads to an ellipse, this time with the sun at the center rather than one focus. Unlike gravity, the force increases with distance. The farther away from the origin, the stronger the pull. This does not appear in astronomy, but it does appear in nature. Take a spring, like the one in your car's suspension, and anchor it to the floor. Let its resting height be 0. Pull up on the spring, so that the top of the spring has a positive displacement d. The spring pulls back with a force proportional to d. Then push down on the spring until its height is -d. The spring pushes up with a force proportional to d. Fold the sign into the force, and the force exerted by the spring is proportional to its displacement d. The constant of proportionality is negative, always moving the spring back to the center. This is Hooke's law.

Put a ball on the spring and pull it up to a distance d, then let go. The spring pulls the ball back to its resting height, but by then the ball is moving, and has some inertia. The ball keeps moving, and why not - at 0 the spring exerts no force, so of course the ball keeps moving. The ball descends until the spring has enough force to push it back up. The ball rises, sails past the center, and reaches a height at which the spring will pull it back. In summary, the ball oscillates back and forth. If the spring were ideal, with no mass and no internal friction, and if there were no air friction either, the ball would oscillate back and forth forever.

What is the position of the ball as a function of time? The acceleration is force divided by mass (one more nod to Newton), hence acceleration is proportional to distance. Let y(t) be the displacement and write this differential equation.

y′′ = -v2y

y = c×sin(vt + θ)

The constants c and θ are constants of integration, but v is inherited from the differential equation. Start the ball higher and c is larger, a larger sine wave, from c to -c, but the frequency v is the same. Galileo observed and documented this phenomenon when he noticed that an oil lamp, suspended from the ceiling of a cathedral, had the same period of oscillation regardless of its arc. Whether a narrow or wide swing, the lamp's oscillating frequency is the same. This is because the swinging lamp is subject to a differential equation that approximates the one above, (though the approximation is only valid for modest angles of displacement). This led to the development of pendulum clocks, far more accurate than any time piece that came before.

Another application of the fixed frequency of a spring is the tuning fork. The stem of the fork acts like a stiff spring, while the end vibrates back and forth at a fixed frequency. Tap the fork gently and the tone is quiet, strike the fork firmly and the tone is louder, induced by wider sweeps of the fork, yet the frequency is the same.

Attach the Earth to the sun by an imaginary, ideal spring. Throughout time the spring pulls the Earth towards the sun, inducing an acceleration of gr, where g is a negative constant. This vector has an x and a y component. The x component is (x/r)gr, or xg, which is proportional to x, and similarly for y. Thus the spring can be replaced with two imaginary springs, one pulling in the x direction and one pulling in the y direction. These springs are independent of each other, except that they both come from the radial force, and they both have the same constant g. Both springs induce the same period of oscillation. There is a definite orbit, a definite path around the sun that repeats.

Scale x and y so that the parametric equations are sin(vt+a) and sin(vt+b). These linear transformations map conic sections to conic sections. If the path was an ellipse, it is still an ellipse, and conversely. Pull a and b back to values between 0 and 2π. This does not change the path. Finally, start time at the average of a and b, so that the equations become x = sin(t-j) and y = sin(t+j). expand by the angle addition formula, and x+y is a multiple of sin(t), and x-y is a multiple of cos(t). Some linear combination of (x+y)2 and (x-y)2 equals 1. This is a conic section. In fact there are no linear terms, so the center is at the origin.

There is no such thing as escape velocity, since the force increases with distance. The only possible conic section is an ellipse. However, unlike our solar system, The sun is at the center of the ellipse, and the planet's position, as a function of time, obeys a concise trigonometric formula.

Kepler's second law is valid for any radial force, so let's try it out here. Let the ellipse be a×cos(t),sin(t). consider the area at time t. Shrink the ellipse back to a circle and the angle is t, and the area is t/2. Stretch this back out and the area is at/2. Area is proportional to time, as it should be.

Intuitively, every radial force defines a periodic orbit around the sun, assuming the planet does not have escape velocity. Set t = 0 when the planet reaches a local maximum in its trajectory. The velocity is tangential, and the acceleration is radial. As time moves forward, the planet moves counterclockwise in its orbit, gaining speed and losing altitude. Potential energy is converted into kinetic energy. If time runs in reverse, the same thing happens running clockwise. The trajectories are mirror images of each other. Follow both paths around 180 degrees, and they must meet at the same point on the other side of the origin. The orbit is a continuous loop, though the path might not be an algebraic form in the plane, such as a circle or an ellipse.