Let's derive the satellites' orbit by Newtonian physics.
2023/04/13 Kenlo Nasahara
You played with an animation page of Satellite's orbit simulation (Kepler's laws). Now let's learn how these phenomena take place under the laws of physics.
Satellite's orbit is controlled by a gravity force. We treat the satellite as a particle without size and shape. We are showing an important fact (theorem):
If the gravity force comes from a mass with a perfect spherical shape, the orbit is mathematically solved and expressed by a quadratic curve (二次曲線) on a plane.
In fact, all the orbit lines appearing on the animation was quadratic curves on a plane. It is the simplest model of a satellite's orbit. The goal of this page is to derive this from the law of physics.
Note that this theory is about a model (approximation) which is not the real orbit. The real orbit is not a true quadratic curve or on a true plane, because the satellite gets more complicated force than the gravity force coming from just a single source with a perfect sphere.
We start with Newton's equation:
First, we rewrite in to a form using a plane polar coordinate . Namely, any vectors are represented by a linear combination of two unit vectors:
which directs from the origin to the point we are considering, and
which directs perpendicular to and anti-clockwise direction in a rotational movement whose center is at the origin. Remember that and change in response to the change of the position we consider, namely, the position of the satellite.
By definition, it is obvious that the position vector of the satellite is described by
Then the velocity of the satellite is:
Note that the dot at the head of a letter means derivative with respect to time, nemaly, .
By using and , we get more specific form of :
Because we have decided to describe all vectors with and , we must describe by them , too. Let's try using :
By substituting it to , we obtain:
Then we proceed to satellite's acceleration :
Here we can describe by:
By substituting it to , we obtain:
Meanwhile, by gravity's law,
By using , , and , we obtain:
Because and are linearly independent,
By multiplying to ,
The left side of is equal to . Therefore,
It means is a constant (independent of time ). We represent it as a constant . Namely,
or
By substituting in with ,
Now we are going to rewrite this equation by changing the variable from to . First, we rewrite and as follows:
(We used at the final step of both reforms.)
Now we define
Then
Therefore, becomes
By replacing in with , we get:
This is an easy ordinary differential equation, whose general solution is:
where, is an arbitrary constant. By introducing the following two parameters,
and we take the origin of angle so as to make (for example, by assuming becomes minimum when )), becomes to
This is in coincident with a general form of a quadratic curve with as the "eccentricity". Now we have arrived at our goal!
Let's appreciate some more about our theory.
Suppose if . Then becomes . It means the orbit is a circle with a radius equal to the constant .
Suppose if . Then the denominator of is always a positive value. It is in between (when ) and (when ). Therefore, the maximum and the minimum of are:
The satellite moves always in a distance between these two limits. Such curve is (intuitively) an allipse. These two limits are on a common straight line (because is 0 and ). It means +
is the length of the long axis (major axis). The half of it is called "semi major axis" and described by a constant conventionally. Namely,
This is an important formula.
Now let's derive the length of the "minor axis", which is perpendicular axis to the major axis. It is the maximum of the coordinate of the satellite. If you remember the polar coordinate, you can easily derive
All we need is to find the maximum of this function in terms of . Note that depends on . Using ,
By differentiating it with , we get:
It becomes zero when . In such a circumstance (especially the max of ),
Then takes a valu of
This is the half length of the minor axis, or "semi-minor axis" represented by conventionally. Namely,
Some peoplemay wonder if \eref{eq:orbit_quadratic04} truely draw an ellipse, even if we have known the length of two main axis. Then I suggest them to try deriving \eref{eq:orbit_quadratic04} by starting from an assumption that the orbit is an ellipse with given values of and \(b). Here is a summary of how to do it:
Because the focal point F is at the origin of the coordinate, we need to obtain the position of the center of the ellipse. The distance between the focus and the center is (please consider why by yourself) . The ellipse's center is at left with this distance from the origin. Then the equation of the ellipse should be
You reform this equation so as to make it only with by using
all of which are already given formula. Then a little complicated calculation (but not very difficult) will derive .
So far, we have discussed the shape of the orbit. But not mentioned how quick (or slow) the satellite moves along it. Now let's discuss it.
According to the Kepler's second law (or conservation of angular momentum in Newtonian dynamics), the area swept by the line between the focus (the gravity source) and the satellite per unit time (i.e. "aerial speed") is a constant. It is . Using , it is . (Truely a constant!)
Then the time the satellite takes to evolve one time around is, the area of the ellipse divided by this constant aerial speed . Namely,
This is a very important formula, which shows the Kepler's third law: The period of the satellite is proportional to (semi major axis)^(3/2). It does not explicitly depend on the minor axis.
But this formula is still not enough. We further want to know at any time where is the satellite on the orbit. We use the following figure. We assume that the satellite is at the point A in the figure at . We represent the area of a shape X as A(X). The time when the satellite arrive at P is determined by A(the fan-like shape FPA), divided by the areal speed.
A(the fan-like shape FPA) is A(the fan-like shape FP'A) .
A(the fan-like shape FP'A) is A(fan OP'A) - A(OP'F).
A(fan OP'A) is .
A(OP'F) is .
Therefore,
A(the fan-like shape FPA) is .
By combining all these together, we obtain:
Here we define a variable called "mean anomaly" as:
Then becomes:
This is another important formula, called Kepler's equation.
If you want to know the position of the satellite at time , the following procedures will give you the answer:
- Convert to by .
- Solve to obtain .
- Obtain P' from and .
- Obtain P from P' by making of the coordinate of P'.
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By FranFranz - Own work CC BY-SA 4.0, Link