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On this page, we will simulate satellite orbits. You will be able to examine a variety of orbits in the simulation, and will discover a number of surprising facts. These take the form of "conservation laws" and we will use our moving frame representation with polar coordinates 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 satellite to the center of the Earth, that line sweeps out equal areas in equal times. In the Galilean language of the Introduction, "The area swept out increases uniformly with time."

Now there is an essential complication in this analysis. The orbit is a curve in a plane, and so it has two components of motion. Our polar coordinate representation will resolve that complication momentarily. But before we take up that problem, let us digress to examine a simpler, 1-dimensional problem: the 1-dimensional form of a different conservation law called the "conservation of energy." We will return to the full 2-dimensional form in the next microworld: Conservation of Energy.

Conservation of Energy (Dimension 1) and Escape Velocity

A rocket that places a satellite into orbit is usually a multistage rocket with "boosters" that provide the initial propulsion, and are discarded after their burn. They take the satellite to some altitude at some velocity, and, once they are discarded, the satellite is essentially a "projectile". We will always assume that the satellite is a projectile, that it has no rocket power, but only coasts with the energy and momentum with which it was supplied when the rockets were discarded.

In the 1-dimensional problem that we want to consider, the satellite moves in a straight line, radially away from the center of the Earth. Since the motion is in a line, we may use that line as a "ruler" to coordinatize its motion. We mentioned earlier that if we assume the satellite is not moving on a straight line away from or towards the Earth, at a single point in time, then it will never move that way at any instant of the past or future, and in fact, r(q) will never be 0. But it will also follow that if it is moving on a straight line away from or towards the Earth, at a single point in time, then it will always move on that line at any instant of the past or future. Whether it turns back and falls into the Earth will be entirely determined by its "energy." If it is not destined to return to the Earth, we will say that it has "escaped."

Suppose then that the satellite projectile, launched vertically upward with some velocity , moves in such a way that its altitude at time t is given by the function . Then we know that the velocity . We follow the convention in physics and represent time derivatives with "dots" above the variable. So we shall often write .

Consider the function of r and v (essentially the energy):

We see that for a constant the equation generally defines a curve in the plane. Not always, but generally. We will show that the vertically projected satellite will move in such a way that there is a constant C such that

One of our future tasks in Conservation of Energy will be to determine the velocity required for the rocket to "escape" the Earth's own gravitational pull. We take the mass of the Earth to be

And we assume that the Earth has radius

Then, according to Newton's Universal Law, there is a universal constant G (actually only known to about 3 decimal places!)

  
And the gravitational acceleration that pulls the rocket back to Earth is, at height h km above the surface of the Earth,

Here, we measure distance to surface of the Earth in km=kilometers. And in the appropriate units.

Question 1: Assuming that there is a constant C such that for our vertically projected satellite,

use the numbers above to calculate the gravitational acceleration in at the surface of the Earth (h = 0). You know the answer already, so this will verify what you know. Also calculate the constant in units

End of Question

Solution: We took the mass of the Earth to be

And we assumed that the Earth has radius

Then, according to Newton's Universal Law, there is a universal constant G

  
And the gravitational acceleration that pulls the rocket back to Earth is, at height h km above the surface of the Earth,

Here, we measure distance to surface of the Earth in km=kilometers. And in the appropriate units.

Thus, this acceleration is

This is roughly

The constant

End of Solution

Now we will prove that our projectile must move in such a way that for all times  , the points lie in one such curve, for some constant . This is the 1-dimensional version of conservation of energy.

To see that it is true, consider the function of time:

Then according to the chain rule:

Now (by definition), and, according to Newton's law,

. So

for all times . This means that is constant. And that means that there is a constant C such that for all times . Therefore the "energy" is conserved

We will come back to this point in a moment.

Now let us calculate escape velocity at 200 km up using the conservation of energy. We know that for our projectile rocket, that the quantity

is constant. Suppose that at the moment the boosters are discarded, at altitude 200 Km, the velocity is . Then this constant will be

Suppose that . Then it is not difficult to see that at all future times

So

From this it is clear that (as long as the projectile travels on a straight line from the center of the Earth, which is the

case we are considering here) v(t) must always be positive, in fact, it must always be larger than

Thus the rocket "escapes" the Earth's gravitational pull, because it never "stops" climbing.

What is the smallest value for for which (at altitude 200 Km)? We want 

This means that

So calculating, using the value of K from the previous problem:

This is "escape velocity" at height 200 km.

But perhaps we spoke too quickly when we concluded that is constant above. How do we know that if for all , then is constant?

The Mean Value Theorem

We have certainly assumed that is differentiable for all . And the derivative of a constant function is 0, but it is not immediately obvious that if a differentiable function has 0 derivative, then the function is constant. The proof depends on a very important and useful theorem in Calculus called the Mean Value Theorem.

Let us start by seeing what needs to be proved. Suppose that and are numbers and that .

And suppose that . Then the average rate of change from a to b is

And this is, by our assumption, not equal to 0.

Suppose we knew that this nonzero average rate of change was equal to the instantaneous rate of change at some number c between a and b. We shall prove that this is true momentarily. It is essentially the Mean Value Theorem, but for now, we can see that it is at least plausible. Then

since for that c,

Thus if the derivative of was 0 at every point, then it could not happen that for some , . So must in that case be equal to some constant for all t.  The reasoning is intricate, but it is worth your time to think it through.

So let us see what the Mean Value Theorem says, exactly.

Mean Value Theorem: If is a continuous function on a closed interval and is differentiable on the open interval then there is a number c such that  

and that

There may be more than one such number, but there is at least one.

Proof:

The proof depends on an observation called Rolle's Theorem. Let us state and prove that first.

Rolle's Theorem

(Lemma) Rolle's Theorem:

If g is a continuous function on a closed interval and is differentiable on the open interval , and if then there is a number c between a and b such that .

Proof of Lemma: This depends on the crucial fact that a function such as g is either constant, in which case
for all c between a and b, or it takes on either a maximum or a minimum value between a and b. This in turn follows from the fact that g is continuous at each point between a and b.

There is a deep theorem about continuous functions defined on closed intervals that guarantees that they attain maximum and minimum values on . We cannot prove this last fact here, and so we will have to accept it as true. Given that g attains either a maximum or a minimum at some point c such that , then .

Indeed, from what we said about differential approximation in Art of Approximation, we see that

Then if we could find small such that and . The argument is similar if . This would contradict the assertion that g attained either a maximum or a minimum at c.

End of Lemma

Question 3: Returning to the Mean Value Theorem, consider again the function  . Construct a new function:

Observe that

Conclude from Rolle's Theorem that for some c such that , that

Now show that if c is such a number, then for that c,

Explain how you may conclude this.

End of Question

End of Theorem

Now, let us return from our digression to the 2-dimensional satellite problem. The way it works is this. Typically, the rockets are discarded, and the satellite begins to "coast" at an altitude, say h km above the surface of the Earth. At that height, once the boosters have been discarded, the whole future of the satellite depends on its velocity. We will formulate an appropriate notion of energy in the next microworld, and establish that it also is conserved.

Conservation of Angular Momentum

Here, we want to use our moving frames to show that a different quantity (called angular momentum) is conserved. Recall that on the previous page: Polar Curves and their Moving Frames, we asked you to prove that when the acceleration of the satellite is represented in the coordinates of the frame then

or simply

We call the first coordinate

and the second coordinate

Newton's Gravitational Law

Let us represent Newton's law in quantitative form. If represents the distance of a satellite from the center of the Earth, then the downward acceleration due to gravity is:

where G is a universal constant and M is the mass of the Earth. We discussed the units above.

The negative sign means that the "upward velocity decreases," which means that the satellite is attracted toward the Earth in the sense that the instantaneous rate of change of its velocity is a vector that "points" from its position

towards the center of the Earth at each instant of time t.

We saw that in the coordinates of our moving frame:

Now if we describe the acceleration also in terms of our moving frame, we see that we must have

That is, the angular acceleration must be 0! This is the basic reason for introducing these moving frames.

So it follows that

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

Clearly,

We may thus apply the Mean Value Theorem as we did above to conclude 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 satellite, we call the constant quantity

the angular momentum of the satellite. And from this, we see that the angular momentum is always conserved.

Later, we will interpret as the rate at which an imaginary line connecting the satellite to the Earth sweeps out area. This also is constant. So area is swept out uniformly with time.

Notice that the only property of gravity that we needed to prove that angular momentum is conserved is the fact that the acceleration is radial and depends only on position (not velocity). The precise magnitude of gravity was not needed. We say that gravity is a central force (always attracting to the center), and we conclude that for any central force, the angular momentum must be conserved.

In the next microworld, Conservation of Energy, we will show that we can define another conserved quantity for satellite orbits: the energy, just as we did in the 1-dimensional case. Together these two conserved quantities will be enough to allow us to deduce Kepler's laws from Newton's single hypothesis of universal gravitation.

Exploration: Satellite orbits following Newton

Now in this exploration, you may "launch" satellites around the Earth and view their orbits. They will appear to be ellipses, and you may check whether they are. To launch your satellite, press Begin Orbit on the control panel below:

Pause or stop the simulation by pressing the "p" key (or as a last resort, pressing Esc). You will see something like:

As the system draws the orbit on the right, it represents the position with the light blue arrow, the velocity with the green arrow, and the acceleration with the red arrow. The red arrow (acceleration) is always parallel to the light blue arrow (position) but points in the opposite direction. Newtonian gravitational acceleration is central. These arrows are scaled for visibility here.

You may continue after you press p to pause by pressing the Continue button. If you press Reset, the system clears everything so that you may draw a new orbit. For that, change the value in the Direction of Launch slider and press Begin Orbit again.

Now the window on the left traces the velocity vectors as they are seen with reference to our moving frame. Notice that for simulated satellite orbits the track of the velocity vectors seems to be a circle in this view.

The system also draws a blue line in the left hand window. The points on that line are the points

as they are represented in our moving frame.

The fact that the second coordinate of these points have constant value: simply reflects the fact that is conserved. The fact that the red curve is a circle depends on the particular form of Newtonian gravitation and is related to the important principle of conservation of energy that we will study in the next microworld.

Now you may check that the orbit drawn using Newton's equations in the right window is indeed an ellipse. For that, once you draw an orbit, use the control panel

Every conic section has the general analytic form in Cartesian coordinates

Since the section is unchanged by multiplication of the equation by a non-zero constant, the section corresponds to a line in the 6-dimensional space of coefficients, or a point in 5-dimension projective space. Thus there are 5 degrees of freedom in the choice of coefficients. The choice of 5 points in the conic section entirely determines it, since substitution gives 5 equations in 6 unknowns, determining the line associated with the conic.

Thus, press Select 5 points and click carefully any 5 points along the orbit. The system will then draw a yellow curve that is the conic section determined by those 5 points. For the orbit

you will see (if you click very close to points on the orbit)

The yellow curve was determined by only 5 of its points! On the other hand, if you "missed" one of the orbit points when you clicked, you can press Clear last curve, and the curve (but not the orbit) will be erased, so that you can try again.

Experiment!