In 1687 Isaac Newton published "Philosophiae Naturalis Principia Mathematica," arguably the most influential science book ever written. In addition to providing the framework still used for all of mechanics, the Principia introduces Newton's theory of universal gravitation, which posits an attractive force between all masses varying inversely as the square of their distance. A centerpiece of Principia is Newton's demonstration that Kepler's three laws of planetary motion are mathematically equivalent to his radically simpler inverse square central force law. This is an exceptionally difficult math problem, even for students who can bring all the developments of calculus in the intervening 350 years to bear on it. But every math problem has many solutions, and along one narrow path through the logic of this problem, each individual step makes no demands beyond the most elementary plane geometry and algebra. Here we go.
Kepler's three laws are: First, each planet orbits the sun in an ellipse with the sun at one focus. Second, the planet moves around its ellipse so that the radius drawn from the sun to the planet sweeps out equal areas in equal times. Third, the ratio of the cube of the semi-major axis of the orbital ellipse to the square of the time for a complete orbit is the same for every planet. We aim to show that if a planet begins from any point in the solar system and moving with any velocity, and thereafter is attracted to the sun by a force varying inversely as the square of its distance from the sun, it will orbit forever according to Kepler's laws. (We assume our planet is subject to no other forces.)
Our elementary path leads through three waypoints, which you may visit in any order: The first is to master the geometry of the focus points and tangent lines of an ellipse, in order to comprehend what Kepler is talking about. The second preliminary topic is circular motion, introducing the concept of velocity space. We must extend this study to the velocity space analog of circular motion, which is motion at a constant speed. The third preliminary is to demonstrate that all central force laws, whether an inverse square law or any other, cause motions obeying Kepler's equal area law. Note that our third preliminary step actually achieves a part of our final goal. With these three preliminaries fresh in our minds, the fact that an inverse square law orbit will be a Keplerian ellipse follows fairly easily. Kepler's third law, involving cubes and squares, requires several final algebraic steps to complete the demonstration.
Kepler's first law says that a planet P moves around the sun S in an ellipse with the sun at one focus; this figure explains what that means. A circle is the set of all points a given distance from a center. An ellipse is the set of all points P with a given sum of distances from two focus points S and O. You can imagine connecting pins at S and O by a string of length SP+PO, which you stretch taught by a pencil at P. By keeping the string taught, you can move the pencil along the ellipse. Instead of a single center, an ellipse has two focus points, S and O.
If we extend the sun-planet line SP to point Q such that the blue segments OP and PQ in the figure are equal, something interesting happens: Since SP+OP is the same for any point P on the ellipse, SP+PQ is also fixed, independent of P. Therefore, the point Q traces a circle with center S as P moves around the ellipse; the radius of this circle is the length of the string SP+PO.
Next, consider the midpoint M of the segment OQ. Since by our construction of Q, P is also equidistant from O and Q, the line PM is the perpendicular bisector of OQ, the set of all points equidistant from O and Q. The line PM appears to be tangent to the ellipse at P, that is PM just touches the ellipse at P without crossing inside. To prove this tangency, note that the sum of the distances from S and O is less than SP+PO inside the ellipse, and greater than SP+PO outside the ellipse. For every point N on PM, the sum SN+NO is equal to the sum SN+NQ, which is always greater than SP+PQ for any N except P, because SPQ is a straight line - the shortest distance between S and Q.
The figure is interactive. You can drag the point O to see how the shape of the ellipse depends on the distance from O to S (at fixed string length SQ). Notice that when O coincides with S, the ellipse becomes a circle. You can drag the point Q to watch what happens as P moves around the ellipse. Finally, you can drag the point M along PM to create a point N, to help you visualize the proof that SN+NQ is always greater than SPQ, so that line PM is entirely outside the ellipse, touching it only at point P.
Consider an object moving at a constant speed around a circle, as depicted in the next figure. We introduce the concept of vectors - quantities with both magnitude and direction which we represent as arrows in diagrams. The most basic vector is position measured from a given point. The black vector on the left side of the figure goes from the center of the circle to the orbiting object; its magnitude $r$ is the radius of the circular orbit. You can press and hold the button at the bottom of the figure to animate this motion.
At any instant of time, the object is moving tangent to the circle, at right angles to the radius vector $r$, indicated by the blue vector $v\tau$. Strictly speaking, any vector we draw in the position plane is a displacement vector, with its tail at the starting point of the displacement and its head at the final point. Hence, we label the blue vector by the product $v\tau$, the product of the speed $v$ and some short time interval $\tau$, so that $v\tau$ is the displacement the object would have made if it had continued in a straight line at constant speed for that short time.
We can treat the velocity, which is the rate of displacement, as a vector in its own right, as opposed to a displacement dependent on an arbitrary short time interval $\tau$. Therefore, we introduce a second plane, the velocity plane indicated on the right side of the figure. In this plane, we draw the tail of the velocity vector at a fixed point representing zero velocity rather than as a displacement from the moving point at the tip of the object. Furthermore, the length of vectors in the velocity plane represents the speed an object is moving (rather than its position). Notice that tip of the velocity vector orbits a circle in the velocity plane, staying exactly ninety degrees ahead of the orbit in the position plane.
To a large degree, Newtonian mechanics takes place in this velocity plane, not in the ordinary position plane. This is because $F=ma$, Newton's second law, associates force with acceleration, which is the rate of change of velocity. Just as velocity times a short time interval $v\tau$ is a displacement vector in position space, acceleration times time $g\tau$ is a displacement in the velocity plane - the change in velocity in some short time $\tau$, drawn orange in the figure. Because the acceleration vector stays ninety degrees ahead of the velocity vector, which stays ninety degrees ahead of the position vector, the acceleration always points directly toward the center of the orbital circle. Here the acceleration will be due to a gravitational attraction of the orbiting object toward the center of the circle in the position plane, so we use the symbol $g$ (rather than $a$) for acceleration.
Let $T$ be the time it takes to complete one orbit. Since the perimeter of the circle is $2\pi r,$ the speed $v$ is $2\pi r/T.$ Similarly, the perimeter of the velocity orbit in the velocity plane is $2\pi v,$ so the acceleration $g$ is $2\pi v/T.$ Now the angle between any fixed direction and the radius or velocity vector changes at a constant rate, which we call $\omega.$ If we use radian units to measure angle, then $\omega = 2\pi/T,$ because there are $2\pi$ radians in one complete circle. Thus, our formulas relating orbital radius, speed, and acceleration are simply $v=\omega r$ and $g=\omega v.$ In fact, the reason people use radians to measure angle instead of degrees is to make these important formulas as simple as possible: In radians, velocity is just radius times the rate of change of direction (heading angle), and acceleration is velocity times the rate of change of direction, at least for uniform circular motion.
Since planets do not move in circular orbits, we need to broaden our study to more complicated kinds of motion. The true shape of planetary orbits had been sought by ancient Greek and medieval Islamic astronomers before culminating in Kepler's eillipses, but it turns out that the position plane is the wrong place to look for orbital shapes. The genius of Newton was to recognize that when you look at planetary motion in the velocity plane, it is circular after all. Therefore, we continue this preliminary exercise by studying motion with a constant speed, that is, motion constrained to a circle in velocity space, rather than constrained to a circle in position space.
As a first step away from uniform circular motion, we turn to an object moving at constant speed but with a time varying sideways acceleration $g.$ The following figure demonstrates this general case of constant speed motion.
Press the button at the bottom to watch the point move at constant speed. Slide the button left or right to deflect its course to the left or right - more positive or more negative rate of change of direction $\omega.$ Think of the slider as the tiller of a boat or the steering wheel of a car. (In the figure, the point will jump to the opposite side if you drive off the visible part of the position plane.) The sideways acceleration $g$ varies in proportion to $\omega,$ with $g=\omega v$ as for uniform circular motion. The orange acceleration vector always remains perpendicular to the blue velocity vector, which is why the speed of the point remains constant while you steer it around. Hold the button at a fixed position and watch the point move in a circle - the uniform circular motion we studied before. Unlike that uniform case, a varying angular speed $\omega$ only refers to the angular rate of change of the velocity vector, because there is no longer a center in the position plane.
Our second step is to study motion constrained to a velocity space circle centered on a non-zero velocity. In the previous figure, you can drag the point at the center of the velocity plane away from the center of the circle to any other point you please. That new point becomes the origin where velocity is zero. The displaced blue arrow remains the velocity of the point in position space, which is now the sum of the two gray velocity vectors. The first is the now non-zero center velocity, running from the origin (tail of the blue arrow) to the center of the velocity space circle. The second is a radius vector, with its tail at the center and its head at the head of the blue arrow. The length $v$ of the gray radius vector remains fixed, and both the length and direction of the center vector remain fixed, as the point moves. Again, you can animate the motion and control the magnitude of the acceleration by holding and sliding the button at the bottom. The center velocity acts like a steady current, making the point harder to steer.
The orange acceleration vector must remain normal to the gray radius vector - not to the blue velocity vector - in order to keep the blue vector on its circular path. The magnitude of the acceleration is still $g=\omega v$, but now $v$ is the radius of the velocity circle, the second gray vector. The speed of the point in position space, which is the length of the blue vector, changes depending on which way it is moving relative to the center velocity.
Gravity is an attraction between any two masses, according to Newton. The strength of attraction is proportional to mass, and in the case of the sun and the planets, the mass of the sun is so much larger than the mass of any planet, that to a good approximation you can ignore the attractions of the planets to one another, and consider only their attraction toward the sun. (It turns out that this is also the accuracy of Kepler's Laws, which themselves are only approximate.) A force which is always directed toward a fixed point - in this case the sun - is called a central force. For any central force, Newton's first two laws of motion turn out to produce Kepler's equal area law, as the next figure indicates.
Here S is the sun, the center of attraction, and P is the planet at distance $r$ from the sun, moving with velocity $v.$ Again, we can draw the blue velocity vector in this position plane by plotting the point V the planet would reach in a time $\tau$ if it continued in a straight line at its current speed. The shaded blue triangle is the area which the radius vector would sweep out were it to follow that course. This area is half the parallelogram SPVW; we have added point W such that SW is parallel to PV and the same length.
Now according to Newton's first law of motion, if there were no force acting on the planet P, it would continue along PV at a constant speed $v.$ You can drag the point P along this no-force trajectory. Notice that no matter where the point lies on the line PV, the area of the parallelogram (hence of triangle SPV) remains unchanged, since the parallelogram always has base SW and lies between the parallel lines SW and PV.
On the other hand, according to Newton's second law of motion, any force on the planet P directed towards (or away from) S causes its velocity to change in the direction PS, which is parallel to VW. You can drag the point V along the line VW to see all the different velocities a central force toward (or away from) S could produce. Once again, the area of SPVW (hence of SPV) remains the same, no matter how strong the deflection in the direction of S.
The shaded area is what would be swept over by the radius if P were to continue along its trajectory for a time $\tau.$ In other words, the rate of change of area swept out by the radius SP is the shaded area divided by $\tau.$ In order to calculate how the point P actually moves, we can break the motion into a sequence of very short time steps (much smaller than the $\tau$ in the figure). In each time step, the velocity changes only slightly, so we can use the current velocity to compute a new position. Similarly, the position changes only slightly, so we can use the acceleration due to the central force at the current position to compute the change in velocity. We take the next step from the new position with the new velocity. What the figure shows is that both parts of each of these time steps - the calculation of the change in position and the calculation of the change in velocity - leave the rate of change of swept area unchanged. That is, the rate of change of swept area remains constant as the planet P moves along its path as long as its acceleration is centrally directed, no matter how that force changes along its trajectory.
Instead of the area SPV swept out by the radius vector, we will write a formula for twice that area, SPVW. The rate of change of parallelogram area is called the angular momentum of the planet P relative to the center S. (Actually, it is the angular momnetum per unit planetary mass.) What we have shown is that for a central force directed toward S, the angular momentum remains constant throughout the orbit; we will denote this rate of change of parallelogram area by $L.$ And this conservation of angular momentum is equivalent to Kepler's law that the planet's radius vector sweeps out equal areas in equal times.
The area of SPVW is the length of its base SP, which is the sun-planet distance $r,$ times the component of the velocity normal to SP times $\tau.$ This normal component of the velocity is $\omega r,$ where $\omega$ is the angular speed of P around S, measured in radians per unit time. That is, the area SPVW is $(r)(\omega r)(\tau).$ The rate of change of this area is the angular momentum $L=\omega r^2.$ This formula relates the unchanging angular momentum to the changing distance $r$ and angular speed $\omega.$
Our three preliminary studies - the construction of an ellipse and its tangent lines from its two focus points, motion with velocity constrained to a circle, and the fact that central force laws produce orbits which conserve angular momentum - fit together like puzzle pieces. We now assemble the puzzle, showing that an inverse square law for gravity is equivalent to elliptical planetary orbits with the sun at one focus.
We begin with the formula we wrote for angular momentum, which we rewrite as an equation for the angular speed $\omega$ of the planet around its center of attraction, the sun: $\omega=L/r^2,$ where the angular momentum $L$ (twice the rate of change of area swept out by the radius vector) remains constant as the planet orbits. In other words, for any central force law, the angular speed of the planet around the sun is inversely proportional to the square of its distance from the sun - the closer it gets, the faster it goes around. If the gravitational acceleration of the planet toward the sun also varies inversely as the square of the distance - $g=M/r^2,$ where the constant $M,$ Newton argues, is proportional to the mass of the sun - that means its acceleration will be proportional to its angular speed. In fact, $g=\omega M/L,$ where $M$ is the solar mass constant, and $L$ is the constant angular momentum of the planet as it orbits.
From our study of circular motion in velocity space, we found that acceleration toward the center in velocity space is proportional to angular speed around that center: $g=\omega v,$ where $v$ is the radius of the circle in velocity space. Thus, if we could arrange for the center of the circle in velocity space to somehow coincide with the sun in position space, we would know the planetary orbit is a circle in velocity space with radius $v=M/L.$
Our diagram of ellipse geometry shows us how to arrange for the velocity space circle to connect to a Keplerian ellipse in exactly this way: The three vectors OS, SQ, and OQ resemble the decompostion of the velocity into a drift component (OS) and a radial component (SQ) to make a total velocity (OQ), which is constrained to lie on a circle in velocity space. The only difference is that we must rotate the velocity diagram ninety degrees clockwise relative to the OSQ position space triangle. With that rotation, the blue velocity vector indeed becomes parallel to the ellipse tangent, since that tangent is perpendicular to OQ. Furthermore, the drift component velocity is now perpendicular to OS, in other words vertical, and the radial velocity component is perpendicular to the sun-planet direction SQ or SP. Since the radial velocity component is perpendicular to PS, the orange acceleration vector is parallel to PS. That is, the acceleration of the planet P is directly toward the sun S. The puzzle pieces indeed fit perfectly.
You can drag the points O and Q around as before to show that our construction works for any ellipse shape and for any position of the planet P on its orbital ellipse.
A planet which obeys $F=ma$ with the force directed toward the sun and varying inversely as the square of its distance will indeed orbit in an ellipse with the sun at one focus. We already found that for any central force law the planet's radius vector will sweep over equal areas in equal times, so we have proven the first two of Kepler's laws are equivalent to Newton's inverse square law. We have learned the additional fact, which Kepler did not notice, that the velocity vector of the planet orbits in a circle in the velocity plane.
Since Kepler's first two laws merely describe the motion of a single planet, he wanted a third law to relate the orbits of the different planets. More than a decade elapsed between his publication of his ellipse and equal area laws and his discovery of the third law, that the cubes of the semi-major axes are proportional to the squares of the orbital periods. In contrast, Newton explains Kepler's first two laws as consequences of an inverse square force law, $g=M/r^2,$ in which the constant of proportionality $M$ is a property of the sun - in fact, a measure of the mass of the sun. Therefore, in the Newtonian system, the same force law already applies to all the planets, so he does not need any additional laws relating the orbits of different planets.
Our proof that the inverse square law produces elliptical orbits completely sidestepped the question of when the planet reaches any point on its orbit. Since we know that the rate of change of area swept out by the radius is constant, $L/2$ in fact, we can use the area between any fixed radius, say the point of closest approach to the sun, and any point P on the ellipse to compute the time when the planet will reach P. We will not derive this equation of time in detail (it requires some trigonometry). However, we do need a formula for the orbital period $T$ - the length of the planet's year.
The area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is $\pi ab,$ a generalization of the celebrated formula $\pi r^2$ for a circle. Since the constant angular momentum $L$ is twice the rate of change of the area, the time for one complete orbit must be $T=2\pi ab/L.$
Now SQ, the string length for constructing the ellipse, is twice its semi-major axis, or $2a.$ We define $c$ to be the distance from the center of the ellipse to either focus, so that OS is $2c.$ In the velocity plane, we have already named the radius of the circle $v=M/L.$ We now give the name $u$ to the drift component of the velocity. Since the velocity plane triangle is similar to OSQ in the position plane, the ratio OS/SQ equals $c/a=u/v.$
To derive Kepler's cube-square law, consider the moment when the planet P is equidistant from the foci O and S. At that instant, the total velocity is horizontal with magnitude $\sqrt{v^2-u^2}$, and the point P is a distance $b$ above OS. Since $u/v=c/a,$ $\sqrt{v^2-u^2}=vb/a.$ The angular momentum is therefore $L=vb^2/a.$ Since $v=M/L,$ we have $M=v^2b^2/a.$ We can eliminate the velocity $v$ to find an expression relating $M,$ $L,$ $a,$ and $b,$ namely $M=L^2a/b^2.$ The final step is to square our expression for the period $T^2=4\pi^2 a^2b^2/L^2.$ Rearranging this, $4\pi^2 a^3/T^2 = L^2a/b^2,$ which is $a^3/T^2=M/(4\pi^2).$ Thus, we identify Kepler's third law constant as Newton's solar mass constant $M$ divided by $4\pi^2.$ Since the mass $M$ is a property of the sun, the constant is the same for every planetary orbit, completing our demonstration that the Newtonian law of gravity produces Keplerian orbits.
Kepler's three laws describe planetary motion without giving any insight into what causes the orbital shapes or speeds. Newton's law of motion $F=ma,$ on the other hand, applies to the motion of everyday objects on earth. While there isn't any more cause for Newton's inverse square law of gravitational force $g=M/r^2$ than for Kepler's laws, it is far simpler and the fact that planets respond to this simple force law by moving in Keplerian orbits shows us that planets move according to the same rules as rocks on earth. A planet is just a large rock, moving as rocks move. Despite its apparently simpler form, Newtonian gravity is really dramatically more complicated than any prior description of planetary motion because of its universal character. Newtonian gravity acts between every pair of objects, between every atom of the sun or planet or rock. Before Principia, no one had any idea that the force holding the planets in their orbits around the sun is identical to the force that makes a rock fall to earth.
Newtonian gravity gives us a framework for comparing the acceleration of the moon in orbit around the earth to the downward acceleration of a rock: A sidereal month is 27.3 days or about 2.36 million seconds and the moon is about 60 earth radii away, so $\omega^2 r$ for the moon is $(2\pi/2.36\times 10^6)^2\times 60=4.3\times 10^{-10}$ earth radii per second per second. According to the inverse square law, the acceleration of gravity at the surface of the earth, 60 times closer, should be $60^2=3600$ times greater. Since the radius of the earth is about $6.4\times 10^6$ meters, the gravity holding the moon in its orbit would cause an acceleration of $4.3\times 10^{-10} \times 3600\times 6.4\times 10^6=9.9$ meters per second per second at the surface of the earth according to the inverse square law. When you drop a rock, it falls to earth with an acceleration of 9.8 meters per second per second, "which answers pretty nearly," as Newton put it; the inverse square law describes how both the moon and the rock move. No one before Newton connected the two.
Because of all of the gravitational attractions to other planets as well as to the sun, the planetary orbits should not quite be ellipses with the sun at one focus sweeping out equal areas in equal times - their mutual attractions as they move should cause them to wobble about in slightly irregular orbits. The moon is an extreme case, because its attraction to the sun is an appreciable fraction of its attraction to the earth, which causes the orbit of the moon to be particularly irregular, deviating obviously from a Keplerian ellipse. Newton himself spent years computing these orbital perturbations, particularly for the moon. In every case, the observed deviations from perfect Keplerian ellipses and equal area sweep rates match what you expect from all those additional inverse square attractions. Unlike any previous theory, Newtonian gravity predicts the complexity we observe in the real world, assigning to it the same cause as the approximate simple elliptical motion.
You can enjoy the simple, approximate planetary motion according to Kepler's laws by pressing and holding the button at the bottom of our final drawing. The figure shows both the elliptical orbit with the sun at one focus in the position plane, and the circular orbit with offset center in the velocity plane. You can drag the gray focal point of the ellipse in the position plane to change the shape of the ellipse. The eight planets of the solar system all describe nearly circular orbits, with the separation between their focal points only a few percent of the diameter of their orbit. Comets have highly eccentric orbits, spending a long time far from the sun, then dramatically plunging close to the sun.
Using no math beyond high school geometry and algebra, we have demonstrated that planetary motion obeying Newton's laws of motion and of universal gravitation will also obey Kepler's laws. However, the length and depth of our reasoning far exceeds what most people ever experience in high school, or even in college. After all, this is a famously hard math problem, so it has cost you some time to follow the reasoning all the way through. It was time well spent because, apart from the pleasure of solving any puzzle, this particular problem launched the science of physics. A lot has changed in the past three and a half centuries, but the force of Newton's style of reasoning remains irresistible.