Kepler's Second Law

Return to Contents

Recall that Kepler's laws assert:

1) The shape of each planet's orbit is an ellipse with the sun at one focus.

2) If an imaginary line is drawn from the sun to the planet, the line will sweep out equal areas in space in equal periods of time for all points in the orbit.

3) The ratio of the cube of the semimajor axis of the ellipse (i.e., the average distance of the planet from the sun) to the square of the planet's period (the time it needs to complete one revolution around the sun) is the same for all the planets.

In one beautiful calculation, using his Calculus, Newton showed that all of these would follow if the planets, in their motion around the Sun, obeyed his Universal Law of Gravitation. And, of course, he developed a theory of Lunar motion on the same principles. Thus, these laws about the motion of the planets, derived from empirical observation, could now be deduced.

Newton's Universal Law of Gravitation

Let us start with the Law of Universal Gravitation as Newton stated it. Given two objects separated by a distance d from each other, there is a tendency for each to accelerate towards the other, that is, each will feel its velocity change towards the other, with a rate of change of velocity that is, for each, proportional to the mass of the other, and inversely proportional to d squared, the square of the distance separating them. The precise formula is this. A body of mass is attracted towards a body of mass with a force which is equal to:

where is a unit vector ( ) directed from towards .

According to Newton, force is a vector that is equal to mass times acceleration

where the acceleration (the rate of change of velocity) is given by

where unit vector is directed along the line separating the "centers of mass" of the two bodies. Of course, its direction, as well as its strength, changes continually.

On this page we will see why (from a 3 dimensional viewpoint) that it follows from Newton's Gravitational law that each of the planets should move in a plane that contains the Sun. That allowed us to use a polar coordinate representation for the motion in the Polar Coordinates section of the Satellite Orbits Chapter. We will justify that assumption here and then we will revisit the conserved quantity that we discovered in Motion in a Gravitational Field.

If m is the mass of a planet, then in polar coordinates, its distance from the Sun, and the time rate of change of its angle measured in radians, we called the constant quantity

the angular momentum of the planet. And we saw that the angular momentum is always conserved. This is an example of a "conservation law" and we will use the 3-dimensional Cartesian representation to establish the first of those conservation laws on this page. This is called: Conservation of Angular Momentum. It is essentially Kepler's second law: if an imaginary line is drawn from a planet to the center of the Sun, that line sweeps out equal areas in equal times. In the Galilean language of the Introduction, "The area swept out increases uniformly with time."

Deduction of Conservation of Angular Momentum

Let us suppose then that there is a three-dimensional system of coordinates with the Sun at the Origin. Each point in space is then represented with three coordinates

An object (such as a planet) whose position changes with time may then be represented by a space curve (which is a

function of time)

Now according to what we saw in Satellite Orbits, the velocity of such a moving point is given by

where the "dot" indicates time derivative. And the acceleration is given by

In terms of these coordinates, Newton's Universal law may be stated (for M the mass of the Sun) as

where

is the "unit vector" in the direction of the point and the negative sign accounts for the direction of the acceleration.

Also, recall that when you "multiply" a vector by a number, it means that you multiply each of its coordinates by that same number. The velocity of a moving point is represented by an arrow emanating from the point itself. The same is true of the acceleration, when you place it at the tail of the velocity vector.

We may simplify the acceleration somewhat by writing it

In this form we see that at each instant is a negative multiple of .

Now, why does an object that moves according to Newton's Law move in a plane? For that we need to understand that each plane passing through the Origin (the Sun) can be described by an equation:

for constants A,B, and C not all 0.

Question 1: Given two points in space,

and

then the points on the line that those (distinct) points generate have the form:

For example,

Show that if P and Q satisfy the equation that is, if,

and

then all the points on this line also satisfy the equation. Explain, in your own words, why this means that the set of points that satisfy the equation (for constant A,B, and C not all 0) must, in fact, be a plane.

End of Question

Now, suppose that our "planet" moves in the curve

under the influence of Newton's Law. This means that

for each instant of time t. Suppose that at some instant of time, , the velocity

and the position

do not lie in the same line through the origin, so that .

So there is a unique plane through the Origin that contains them both. Suppose that plane is defined by the equation

Then it means that

and

Now if we call we are saying that

Therefore is some multiple of say for scalar

We will show then that for all times , we must have

and

And so the planet must remain in this plane forever. To this end, use the cross product to define the vector:

Question 2: Show that

Conclude that since

and of course,

that

and so is a constant vector.

End of Question

But this implies that since

and that means that

and

Thus the planet remains in the plane defined by the equation:

It is also easy to see that if for some time , do not lie in the same line through the origin, so that then for all future (and past) time , it must be true that do not lie in the same line through the origin since is constant. We consider the cases where, for time the velocity is a multiple of position,

or where position is a multiple of velocity

to be singular, or unusual. In such cases, the planet must move in a straight line, and not an ellipse at all.

It is not obvious that if the planet does not satisfy the singular condition for some time then it can never arrive at the center of the Sun. That is, that

for all time t. So far, we have only used the fact that gravitational acceleration is central (towards the Sun) to guarantee that the planet remains in the plane. We actually will need the particular form of Newton's acceleration to show that the planet does not eventually fall into the Sun, even if it does not satisfy the singular condition for some time .

To see this, suppose that at a later time the planet arrived at the center of the Sun, so that

We may assume that the function is continuous, and that restricted to the interval and that it is continuously differentiable. Thus we can argue that is constant

on the interval as we did. Now, if the velocity was also a continuous function of t and if was defined, then since we would have

This means that cannot be defined in such a way that it extends to be a continuous function on all of .

It is not difficult to see in fact that as since remains constant, but .

We will see later however that this strange behavior does not occur for Newton's force law if the planet does not satisfy the singular condition for some time . This is because, as we shall see later, in fact there is a positive number such that for all . If the orbit is not singular at some point, then the planet cannot fall into the Sun (in principle, after all it could still come so close that for all practical purposes it did so).

Under the circumstances that the planet never falls into the Sun, the quantity (essentially the energy, see the discussion of conservation of energy in Satellite Orbits)

is defined for all time in its orbit.

Question 3: Show that if the orbit does not satisfy the singular condition, the function

is constant for all time t. For this, you should differentiate with respect to time, and use the chain rule. This is essentially the "Conservation of Energy" calculation that we have already seen, but done now for planetary (orbital) motion. The earlier calculation we made was for the singular case -- straight up and down motion. Reason carefully how we may conclude that is constant.

End of Question

Plane planetary motion

Now we have accomplished something important. We now know that for a single planet, ignoring all others, if we could assume the Sun did not move, then the planet must move in a plane that contains the Sun. The effect of the planet on the Sun is, however, not entirely negligible. The Sun does wobble just a little because of the planet's pull. Newton therefore imagined a fictitious "center of mass" of the planet-Sun system and showed that his slightly modified equations with that center of mass at the origin (instead of the Sun) gave the Keplerian prediction. Now this wobble is in any case so vanishingly small for Earth that we may safely ignore it in our calculations, and we will. I only mention the correction for completeness.

The task ahead is now the following. We want to show that planets "sweep out equal areas in equal times." Now that we know that the motion is in a plane, we will cast the argument in the polar coordinate plane as we did in Satellite Orbits. We will then reinterpret our conservation of angular momentum result in terms of our new determinant concept and area. Now, we have with the constancy of a powerful analytic tool. So let us put it to use.

We will now restrict attention to the Euclidean plane in which we know the planet moves, and will attach, as needs be, an appropriate metric structure inherited from . In this calculation, we assume that the motion is not singular at some time, and that the planet does not "fall into the Sun" so that it is not singular for all time. We will justify the latter assumption in the next Chapter. Recall the polar coordinate description of a plane curve in Satellite Orbits: Polar Curves and their Moving Frames, and Motion in a Gravitational Field.

We defined a polar curve, in the following way. For each angle , there is a point along the curve at that angle. Its polar coordinates are: , and its Cartesian coordinates are:

where

If we connect this point to the origin (the Sun) by an imaginary line, that line intersects the unit circle, the set of points at distance 1 from the origin, in a unique point (since, as we observed, and will show later, is never 0). This point, for each , is denoted . So the curve is the projection of the curve onto the unit circle.

We represented the point on the unit circle with the symbol: , which is, apparently, nothing but a "wrapping

function" discussed in most trigonometry texts,

so finally:

The important point here is that has two Cartesian components for each angle . It is a point on the unit circle.

We also assumed that we can actually use to parameterize the curve . It is important to know that as time increases, increases. Since our quantity is constant, it follows that as long as the planet does not fall into the Sun, we may assume that this is so. The planet does not "back up" as Ptolemy thought. So we observed that differentiation was with respect to the variable , rather than with respect to time. In particular, we calculated the derivative of .

by differentiating each coordinate separately as we do for all curves. This gave a new curve that we called

We know that

and also that for each ,  is perpendicular to .

Summing up Newton's observations so far, we have plane curves, parameterized now by the angle ,

all associated with our planet.

The pair of vectors are perpendicular to each other, and each has length 1. They defined a "moving frame" along the curve . Imagine that you ride along the orbit, and at each point , you construct the perpendicular pair: (blue) and (dark red) at that point, as pictured below:

The bright red vector is the "velocity" as we traverse the curve by covering equal angles in equal times in the counterclockwise sense.

We let itself be an increasing function of time t. That is, for the planet, the angle increases with t. So we may write the curve that describes the planet's position also as

Differentiation with respect to time, t, is denoted by a "dot" rather than a "prime", but remember that

First, we calculated the velocity,

using the chain rule and the observation that

Thus, in the coordinates of our moving frame:

Now we differentiate again to see that the acceleration is given by

or, in the coordinates of our moving frame:

We called the first coordinate

and the second coordinate

We saw that in the coordinates of our moving frame:

Now when we described the acceleration also in terms of our moving frame, we saw using Newton's Law that we must have

That is, the angular acceleration must be 0! This was the basic reason for introducing these moving frames. So it follows that

Now consider the quantity associated with the planet at each point in time:

Clearly,

We concluded that the quantity is constant throughout the motion. It is a conserved quantity for the orbit in the plane. If m is the mass of the planet, we called the constant quantity

the angular momentum of the planet. And from this, we saw that the angular momentum was always conserved.

Interpretation of Angular momentum in terms of cross product

We are now in a position to interpret in terms of the cross product: .

Recall that

So

Thus from properties of the cross product,

So it follows from the fact that are orthogonal unit vectors,

In this way, we rediscover the conserved quantity . The quantity

is the angular momentum of the planet, which does not change with time.

We now wish to interpret the constancy of as the statement of Kepler's Second Law:

"If an imaginary line is drawn from the sun to the planet, the line will sweep out equal areas in space in equal periods of time for all points in the orbit."

We do not know yet that the orbit is an ellipse. That is Kepler's First Law, but we will know that soon. So the following picture is suggestive of the situation.

For that, we shall approximate the area of the ellipse using the time, t, as the parameter. And the parameter should vary from 0 to T where T is the amount of time required for the planet to make one complete revolution.

Now if this seems strange to you, it should. What do ellipses have to do with planets? Well, Kepler and Newton showed us. And we can (and often do) take such serendipitous connections with the physical world as clues about the deeper structure (and significance) of purely mathematical objects.

Having chosen the parameter interval, we must next find the appropriate "approximating function." For that, recall our diagram

and recall that the quantity Q that we associate with this interval is the "area swept out" by as t varies from 0 to T. This is the area of a growing wedge. We want to measure the value of _{using integration.}

We argued that in fact the area of this sector is approximately equal to the area of the approximating triangle

We called the area of the triangle DA. You showed that the area of this triangle, DA, is exactly equal to

for small positive angles Dq . Thus, calculating

we saw that this was

since

Now the quantity

is the constant for the planet.

Now I claim that if we choose two angles sufficiently close, say , the area swept out between those angles between is exactly equal to

for some angle between . To see this, consider the new picture, where we choose an interval for on which is increasing or decreasing:

where

and

then the area of the sector of circle between is

and the area of the sector of circle between is

and so there is an angle between such that

by the Intermediate Value Theorem.

This means that if we use the angle as the parameter for the quantity Q, we have a simple approximating function for Q.

So we know that

It would be difficult, however, to find an anti-derivative for .

Instead, we observe that is a strictly increasing function of the time t if we traverse the ellipse as a planet would using Newton's gravitational law. So we write and now use time t as the parameter.

We see that for each partition,

of the parameter interval, with all small enough, there is a corresponding partition

and we may choose a sequence of parameter values:

such that

For each i, in the partition, there is another parameter point
such that

so that finally,

The problem is that will not equal . The best we can do with this construction is to say that there are two times

for each i such that

Thus, we do not have a simple approximating function for Q for the time parameterization, but we do have a general approximating function f, where

and so

Recall that the function of t

is constant for the planet. This fact was proven independently of the present considerations. It only requires the we assume that the acceleration is central, or towards the Sun.

Deduction of Kepler's Second Law

According to what we have said, is an anti-derivative for the constant function . Put another way, we have finally justified the assertion that . The time rate of change of area is the constant .

This is Kepler's Second Law.

Later, when we come to Kepler's Third Law, we will be interested in calculating the actual area of the elliptical orbit. Thus, since the function is another obvious anti-derivative of , we may conclude that

or

Therefore, we have measured using integration. We see that

where T is the time required for one revolution, the planetary year.

We also now know why the (constant) function , which is not a simple approximating function for Q, for the time parameterization, but is a general approximating function, may still be integrated to yield Q.

Exploration: Testing Kepler's Second Law

How does the moon "fall around the Earth?" Why do astronauts float in the Space Shuttle even when they are only a few hundred Km above the Earth? The answer to these questions is in Newton's Law of Universal Gravitation. Here, we apply it to deduce Kepler's second law. It is often cloaked in an obscure language, so that Newton's clear geometric intuition, when he explained this is often missed. We will attempt to recover that picture in the simulation on this page.

We will also "test" Kepler's Second Law. The second law states that if an imaginary line is drawn from the sun to the planet, the line will sweep out equal areas in space in equal periods of time for all points in the orbit.

You may launch a "solar probe" into orbit from the Earth when it is at the position (1,0) on the screen. That is where we assume the Earth is at launch. In terms of absolute coordinates this is a point 150 million km from the Sun. For simplicity, we use a grid for which one unit on the screen is one Astronomical Unit: that is, the mean distance from the Earth to the Sun.

When you launch the probe, you may choose its absolute direction. For that, use the "Direction" scroller.

Choose also the duration of a time interval during which you will track the probe each time you press "Continue." Type the number (between 0 and 1) in the field:

The smaller the number, the more accurate the tracking will be. Then press the button to start the probe.

The probe leaves the Earth radially at "escape" velocity, and so the absolute velocity will be obtained by combining the velocity it has because it moves with the Earth with the velocity it has with respect to the Earth. This combination is what is depicted with the green vector. This indicates the absolute direction and magnitude of motion. The position at any time of the probe is represented by the blue vector. The length of the green vector is magnified by a factor of 20 so that you may see it on the screen.

Now the probe will travel for the duration of time you set, then it will stop. Each time you want to track it for the next leg of its journey, press the Continue button:

The first few steps may look like:

As the probe traverses its orbit, a dark red ray is drawn from the Sun after each leg of the journey (not to be confused with "Step Size" which you should probably leave alone). These rays define sectors of the orbit whose areas ought to be equal, since they are computed for simulation time equal to the "Duration" you chose. That duration has nothing to do with real time, by the way. It is simply a convenient measure of time. Above, you see 5 such sectors.

At the end of each leg, the system prints an approximate area for the sector just drawn. It does the approximation using

and so is the area of a triangle, not a sector. These numbers should be approximately equal for each leg. They are the more closely, the smaller the duration, as you will see.

You will notice that the lengths and directions of the green and blue vectors change with time. This is to be expected. As the probe nears the Sun (as in the illustration), the green vector, the velocity, becomes quite large. When it is far from the Sun, its length is quite small. The blue vector behaves in the opposite way.

Now, the green vector is drawn from the probe rather than from the Sun. This is for picturesque reasons. It highlights the fact that this is the vector of motion. The vector of motion of the probe is "tangent" to the orbit it traces at each point of the motion. But notice that at each point, the two vectors: the blue vector and the green vector together determine a parallelogram. This parallogram has the following simple interpretation. If units for distance are taken as Astronomical Unit (AU), and if units for velocity are taken as: then the area of the parallelogram is twice the rate of change of area of the ellipse being swept out at any point in time, where the rate of change of area is measured in units of .

Notice that we have "magnified" for the screen the appearance of the "green" vector. So that these units do not exactly coincide with what appears on the screen.

Now this brings us to the red vector that shows up in our pictures. It is drawn, like the green one, from the probe itself. And, like the green one, it is magnified to enable us to view it. The factor of magnification that we use for it here is 100. It is the vector of acceleration, that is, of time rate of change of velocity, that the Sun exerts on the probe. Notice that it is always pointing towards the Sun.

As the probe nears the Sun, the acceleration vector becomes larger, and as it recedes, that vector becomes smaller. Notice, however, that it always points towards the Sun! Now the interpretation of that vector is a little more complicated than the interpretation of the others. It acts continually to change the velocity vector, to "pull" it towards the Sun. Since it changes the velocity, and, (recalling Galileo's observation) because it changes the velocity vector, it causes a change in position, also.

This explains how an object can "fall" around the Sun. The probe "wants" to move in the direction of the velocity vector. As it does, that vector changes direction towards the Sun in accordance with its acceleration towards the Sun. Thus, it is always "falling", but it is also "moving sideways". Now any object (like an astronaut) that is moving with the probe is subject to exactly the same accelerations. The mass does not matter. Thus, the astronaut "falls" in exactly the same way as the probe, and therefore does not accelerate towards any of its walls. (To be honest, there are "tidal accelerations" that would eventually cause some small drift, since the center of mass of the astronaut is not the same as that of the probe, but we can ignore these.)