Physics+ Functional, Variation: UNIZOR.COM - Physics+ 4 All - Lagrangian

Notes to a video lecture on UNIZOR.COM

Lagrangian -
Functional and Variation

To introduce concepts of Functional (a noun, not an adjective) and Variation which happen to be very important mathematical tools of Physics, let's consider the following problem.

Imagine yourself on a river bank at point A.
River banks are two parallel straight lines with distance d between them.
You have a motor boat that can go with some constant speed V relative to water.
The river has a uniform current with known speed v which we assume to be less than the speed of a boat V.
You want to cross a river to get to point B exactly opposite to point A, so segment AB is perpendicular to the river's current.

Problem:

How should you navigate your boat from point A to point B to reduce the time to cross the river to a minimum?

It sounds like a typical problem to find a minimum of a function (to minimize time). But this resemblance is only on a surface.

In Calculus we used to find minimum or maximum of a real function of real argument by differentiating it and checking when its first derivative equals to zero.

In our case the problem is much more complex, because we are not dealing with a function (time to cross the river) whose argument is a real number. The argument to our function (time to cross the river) is a trajectory of a boat from point A to point B.
And what is a trajectory of a boat?

Trajectory is a set of positions of a boat, which is, in its own rights, can be a function of some argument (trajectory can be a function of time, of an angle with segment AB or a distance from line AB in a direction of a river's current).
Trajectory is definitely not a single real number.

In our case the trajectory is determined by two velocity vectors:
velocity vector of a boat V and
velocity vector of a river's current v.

The boat's velocity vector, while having a constant magnitude V can have variable direction depending on navigation scenario.
The current's velocity vector has constant direction along a river bank and constant magnitude v.

So, the time to cross the river is not a function in our traditional meaning as a function of real argument, it's "a function of a function", which is called Functional (a noun, not an adjective).
Examples of Functionals as "functions of functions" are
- definite integral of a real function on some interval,
- maximum or minimum of a real function on some interval, - average value of a real function on some interval,
- length of a curve that represents a graph of a real function on some interval,
etc.

It is impossible to determine minimum or maximum of a Functional by differentiating it by its argument using traditional Calculus, because its argument is not a real number, it's a function (in our case, it's a trajectory as a function of time or some other parameter).
We need new techniques, more advanced Calculus - the Calculus of Variations to accomplish this goal.

We have just introduced two new concepts - a Functional (a noun, not an adjective) as a "a function of a function" and Calculus of Variations as a new technique (similar but more advanced than Calculus) that allows to find minimum or maximum of a Functional.
These concepts are very important and we will devote a few lectures to address these concepts from purely mathematical point before starting using them for problems of Physics.

Before diving into a completely new math techniques, let's mention that in some cases, when solving a problem of finding minimum or maximum of a functional, we can still use classic approach of Calculus.
This can be done if an argument to a functional (a function in its own rights) can be defined by a single parameter. In this case a functional can be viewed as a regular function of that parameter and, as such, can be analyzed by classic Calculus techniques.

Here is an example that is based on a problem above, but with an additional condition about trajectories.
Instead of minimizing the time to cross the river among all possible trajectories, we will consider only a special class of trajectories achieved by a specific scenario of navigation that allows one single real number to define the whole trajectory.

Assume, your navigation strategy is to maintain a constant angle φ between your course and segment AB with positive φ going counterclockwise from segment AB.

Obviously, angle φ should be in the range (−π/2,π/2).

With an angle φ chosen, a boat will reach the opposite side of a river, but not necessarily at point B, in which case the second segment of a boat's trajectory is to go along the opposite bank of a river up or down a stream to get to point B.

The problem now can be stated as follows.
Find the angle φ to minimize traveling time from A to B.
For this problem a functional (time to travel from A to B), which depends on trajectory from A to B, can be considered as a regular function (time to travel from A to B) with real argument (an angle φ).

Solution for Constant Angle φ:

If you maintain this constant angle φ, you can represent the velocity vector of a boat going across a river before it reaches the opposite bank as a sum of two constant vectors
V = V_⊥ + V_||
where
V_⊥ is a component of the velocity vector directed across the river (perpendicularly to its current) and
V_|| is a component of the velocity vector directed along the river (parallel to its current).
The magnitudes of these vectors are
|V_⊥| = V·cos(φ)
|V_||| = V·sin(φ)

The time for a boat to reach the opposite bank across a river is
T_⊥(φ)=d / |V_⊥|=d/ [V·cos(φ)]

In addition to moving in a direction perpendicular to a river's current across a river with always positive speed V_⊥=V·cos(φ), a boat will move along a river because of two factors: a river's current v and because of its own component V_|| of velocity.
The resulting speed of a boat in a direction parallel to a river's current is V·sin(φ)−v, which can be positive, zero or negative.

Therefore, when a boat reaches the opposite side of a river, depending on angle φ, it might deviate from point B up or down the current by the distance
h(φ) = [V·sin(φ)−v]·T_⊥(φ)
This expression equals to zero if the point of reaching the other bank coincides with point B, our target.
The condition for this is
V·sin(φ)−v=0 or
sin(φ)=v/V or
φ=arcsin(v/V)=φ₀.
So if we choose a course with angle φ₀=arcsin(v/V), we will hit point B, and no additional movement will be needed.

Positive h(φ) is related to crossing a river upstream from point B when the angle of navigation φ is greater than φ₀ and negative h(φ) signifies that we crossed the river downstream of B when the angle of navigation φ is less than φ₀.

In both cases, after crossing a river we will have to travel along a river's bank up or down the current to cover this distance h(φ) to get to point B.

Our intuition might tell that an angle φ₀ of direct hit of point B at the moment we reach the opposite side of a river, when h(φ)=0, should give the best time because we do not have to cover additional distance from a point we reached the opposite bank to point B.

It's also important that the actual trajectory of a boat in this case will be a single straight segment - segment AB - the shortest distance between the river banks.

In general, the time to reach the other side of a river depends only on component V_⊥ of the boat's velocity and it equals to
T_⊥(φ) = d / [V·cos(φ)]

If we choose a course with angle φ=φ₀ to reach the opposite side exactly at point B, the following equations take place
sin(φ₀)=v/V
V²·sin²(φ₀) = v²
V²·cos²(φ₀) = V²−v²
V·cos(φ₀) = √V²−v²
T_⊥(φ₀) = d / √V²−v²
This is the total time to get to point B.
If we choose some other angle φ≠φ₀, we have to add to the time of crossing a river T_⊥(φ) the time to reach point B going up or down a stream along the opposite river bank.

Let's prove now that the course with angle φ=φ₀ results in the best travel time from A to B.

The distance from a point where we reach the opposite bank to point B is
h(φ) = [V·sin(φ)−v]·T_⊥(φ) =
= [V·sin(φ)−v]·d / [V·cos(φ)]

This distance must be covered by a boat by going down (if h(φ) is positive) or up (if h(φ) is negative) the river's current.
Let's consider these cases separately.

Case h(φ) is positive

This is the case of angle φ is greater than φ₀.
Obviously, the timing to reach point B in this case will be worse than if φ=φ₀ with h(φ₀)=0.
First of all, with a greater than φ₀ angle φ the river crossing with speed V_⊥(φ)=V·cos(φ) will take longer than with speed V_⊥(φ₀)=V·cos(φ₀) because cos() monotonically decreases for angles from 0 to π/2.
Secondly, in addition to this time, we have to go downstream to reach point B.
So, we should not increase the course angle above φ₀.

Case h(φ) ≤ 0 because φ ≤ φ₀

This scenario is not so obvious because crossing the river with angle φ smaller than φ₀ but greater than 0 takes less time than with angle φ₀.
But it adds an extra segment to go upstream after a river is crossed.

The extra distance h(φ) is negative because V·sin(φ is less than v, which allows a current to carry a boat below point B.
Since the point of crossing the river is below point B, the distance |h(φ)| should be covered by going upstream with speed V−v, which will take time
T_h(φ) = |h(φ)| / (V−v)

Using the same expression for h(φ) but reversing its sign to deal with its absolute value, we get the additional time to reach point B after crossing a river

Th(φ) =

d·[v−V·sin(φ)] V·cos(φ)·(V−v)

The total travel time from point A to point B in this case is T(φ)=T_⊥(φ)+T_h(φ), which after trivial simplification looks like

T(φ)=

d[1−sin(φ)] (V−v)cos(φ)

This function is monotonically decreasing by φ because its derivative

T'(φ) =

d[sin(φ)−1] (V−v)cos²(φ)

is negative.
Therefore, its minimum is when its argument is the largest, that is if φ=φ₀, h=0 and the time to get to point B is
T_AB = T_⊥ = d / [V·cos(φ)]

So, the answer to our simplified problem, when we managed to solve it using the classic methodology, is to choose the course of navigation from A to B at angle φ₀=arcsin(v/V).
The minimum time of traveling is T_AB=d / [V·cos(φ)]

As you see, in some cases, when a set of functions that are arguments to a functional can be parameterized by a single real value (like with an angle φ in the above problem), optimization problems can be solved using classic Calculus.

The subject of a few future lectures is Calculus of Variations that allows to solve problems of optimization in more complicated cases, when parameterization of arguments to a functional is not possible.

Physics+ Kepler Third Law: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Kepler's Third Law

Kepler's Third Law states that for all objects moving around a fixed source of gravitational field along elliptical orbits the ratio of a square of their period of rotation to a cube of a semi-major axis is the same.

As in the case of the Kepler's First Law, this Third Law has been based on numerous experiments and years of observation.

Based on all the knowledge conveyed in previous lectures on Kepler's Laws, we will derive this Third Law theoretically.

Let's make a simple derivation of Kepler's Third Law in case of a circular orbit.

In this case the velocity vector of an object circulating around a central point is always perpendicular to a position vector from a center to an object.
Since the gravitational force is collinear with a position vector, it is also perpendicular to velocity, which is tangential to a circular orbit. Therefore, gravitational force makes no action along a velocity vector which makes the magnitude of the velocity vector constant.

Let's introduce the following characteristics of motion:
t - absolute time,
r - radius of a circular orbit of a moving object,
F - vector of gravity,
M - mass of the source of gravitational field,
m - mass of object moving in the gravitational field,
r - position vector from the source of gravitational field to a moving object,
r'=v - velocity vector of a moving object,
r"=v'=a - acceleration vector of a moving object,
T - period of circulation,
ω=2π/T - scalar value of angular velocity,
Here bold letters signify vectors, regular letters signify scalars and magnitudes of corresponding vectors, single and double apostrophes signify first and second derivative by time.

Constant magnitude v of velocity vector means constant angular velocity ω and obvious equality v=r·ω.

Magnitude a of an acceleration vector can be simply found by representing a position vector as a pair of Cartesian coordinates (x,y):
x = r·cos(ωt)
y = r·sin(ωt)
x' = −r·ω·sin(ωt)
y' = r·ω·cos(ωt)
x" = −r·ω²·cos(ωt)=−ω²·x
y" = −r·ω²·sin(ωt)=−ω²·y
and, therefore,
a = r" = −ω²·r
(collinear with r and F)
from which follows
a = |a| = |−ω²·r| = ω²·r

According to the Newton's Second Law,
F = m·a
According to the Universal Law of Gravitation,
F = G·M·m/r²
Therefore,
a = ω²·r = G·M/r²
from which follows
ω² = G·M/r³

Since ω=2π/T,
4π²/T² = G·M/r³
T²/r³ = 4π²/(G·M) - constant
End of proof for circular orbit.

Let's prove it in a more complicated general case of any elliptical orbit.
We will use the First and the Second Kepler's Laws as well as the results presented in the previous lecture Planet Orbit Geometry to derive this Third Law.

Recall the Kepler's Second Law (see the lecture Kepler's Second Law in this course).
We have introduced a function A(t) that represents an area of a sector bounded by r(0), r(t) and a trajectory from a planet's position P(0) at time t=0 to its position at any moment of time P(t).
Then the area swept by position vector r(t) during the object's motion from time t₁ to t₂ equals
ΔA_[t1,t2] = A(t₂) − A(t₁)

Using the above symbols, the Kepler's Second Law can be formulated as
If t₂−t₁ = t₄−t₃ then
A(t₂)−A(t₁) = A(t₄)−A(t₃)
The above condition is equivalent to a statement that
dA(t)/dt is constant or, equivalently, that A(t) is a linear function of time t with A(0)=0.

In the same lecture we have proven that
dA(t)/dt = ½|L|/m
where L is an angular momentum of a moving object (constant in a central force field) and m is object's mass.

Therefore,

A(t) = t·½|L|/m

Assume, we want to know how much area is swept by position vector r(t) during a period T of a complete round movement of a planet around the Sun.
Obviously, it's
A(T) = T·½|L|/m
At the same time, A(T) is an area of an elliptical orbit of a planet, and we know that the area of an ellipse along which a planet moves equals to
A(T) = π·a·b
where a is semi-major and b is semi-minor axes (see the lecture More on Ellipse in this course).

Therefore,
π·a·b = T·½|L|/m

Let's assume that at time t=0 a planet is at the furthest from the Sun point (aphelion), it's initial position vector is r₀ and its velocity vector is v₀.
At this initial point on an orbit the position vector r₀, lying along the major axis of an ellipse, and a tangential to an ellipse vector of velocity v₀ are perpendicular to each other.

Therefore, the magnitude of the angular momentum |L| equals to a product of planet's mass, a magnitude of its position vector (that is, a distance of the Sun) and a magnitude of its velocity vector:
|L| = m·|r₀|·|v₀| or L/m = r₀·v₀

Now the above formula for a period T of a planet's rotation around the Sun is
π·a·b = ½T·r₀·v₀

In the previous lecture Planet Orbit Geometry we have derived the expressions for major and minor axes of an elliptical orbit of a planet in terms of its initial position and velocity at aphelion:

a =

r0 2−β

b =	r0√β·(2−β) 2−β

where β=r₀·v₀²/(G·M)

Let's substitute these expression into a formula connecting a period T with an area of an ellipse:
πr₀²√β·(2−β)/(2−β)²=½T·r₀·v₀

Let's square both sides to get rid of a radical:
π²r₀⁴β/(2−β)³=¼T²·r₀²·v₀²

Next is just technicality.
Cancel one r₀ from both sides
π²r₀³β/(2−β)³=¼T²·r₀·v₀²

Replace r₀³/(2−β)³ with a³ (see formula above)
π²a³β=¼T²·r₀·v₀²

Replace β with r₀·v₀²/(G·M) (see formula above)
π²a³r₀·v₀²/(G·M)=¼T²·r₀·v₀²

Cancel r₀·v₀² on both sides
π²a³/(G·M)=¼T²

Final result:

T² a3

=

4π² G·M

The right side is a constant that contains no planet-specific parameters (like initial position and velocity), which means that any planet has the following property (Kepler's Thirt Law).
The ratio of a square of the period of a planet's rotation around the Sun to a cube of a semi-major axis of its elliptical orbit is constant that depends only on a mass of the Sun.

Physics+ Orbit Geometry: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Planet Orbit Geometry

Before studying this material we strongly recommend to study geometrical aspects of ellipse (for example, in Math+ 4 All => Geometry => Ellipse) and physical aspects of movement in the gravitational field in lectures of this course Central Force Field, Planet Orbits, More on Ellipse and Kepler's First Law.

The purpose of this lecture is to determine the geometric characteristics of a planet's orbit based on its initial position and velocity relative to the Sun.

Before addressing these issues let's recall the characteristics of an object of mass m circulating on a circular orbit around a fixed in space central object of mass M.
In particular, we are interested in some relationship between the moving object's linear speed v and a radius r of rotation around a central source of gravitation of mass M in order to stay on a circular orbit.

According to the Newton's Universal Law of Gravitation, the force of gravity F is
F = G·M·m/r²
where G is the gravitational constant.
The Newton's Second Law connects this force to a mass m and centripetal acceleration a of a moving object
F = m·a
Therefore,
G·M·m/r² = m·a
from which follows
a = G·M/r²

We can express the linear speed along an orbit v and centripetal acceleration a in terms of constant radius of uniform rotation r on a circular orbit with constant angular speed of rotation ω using time-dependent Cartesian coordinates {x(t),y(t)} of a moving object as follows.

Position vector:
x(t) = r·cos(ω·t)
y(t) = r·sin(ω·t)

Velocity vector:
x'(t) = −r·ω·sin(ω·t)
y'(t) = r·ω·cos(ω·t)
v = √x'(t)²+y'(t)² = r·ω

Acceleration vector:
x"(t) = −r·ω²·cos(ω·t)
y"(t) = −r·ω²·sin(ω·t)
a = √x"(t)²+y"(t)² = r·ω² = v²/r

Therefore, returning to the Universal Law of Gravitation,
v²/r = G·M/r²
and we conclude that for a circular rotation of an object in a gravitational field

β =

r·v² G·M

= 1

In the above formula we have introduced a symbol β with which we expressed the main condition of object to stay on a circular orbit.
This symbol will be used in our analysis of an elliptical orbit of an object in a central gravitational field.

Let's switch now to a more complicated case of elliptical orbit.

Kepler's First Law states that all planets move around the Sun along elliptical orbits with the Sun in one of the two focal points of their orbits.
Kepler had come up with this law experimentally based on many years of observations.
In the lecture Kepler's First Law we have proven it based on the Newton's Second Law and the Universal Law of Gravitation.

The geometric properties of a planet's elliptical orbit are completely defined by two parameters: its major axis of the length 2a and eccentricity e, from which we can derive a minor axis of the length 2b and focal distance 2c using equations
c = a·e
b²+c²=a² => b=a·√1−e²

Assume, a point-mass M (the Sun) is the source of a gravitational field and is fixed in our space.
Assume further that a point-mass m (a planet) is moving relatively to this point-mass M in its gravitational field with no other forces involved.
Two major characteristics of this motion are the planet's position and velocity.

In the lecture Planet Orbits we have proven that the trajectory of a planet's movement around the Sun is a plane that we called the plane of motion.



This plane goes through the source of gravity and contains two vectors - the vector of initial position r₀=r(0) from the source of gravity to a moving object at some chosen moment in time t=0 (from the Sun to a planet) and the vector of object's initial velocity v₀=v(0) (tangential to a trajectory) at the same time.

In the lecture Central Force Field we have proven that Angular Momentum vector L=m·r(t)⨯v(t) of a planet moving around the Sun is independent of time, is a constant vector perpendicular to a plane of motion.

In this lecture we will examine the geometric properties of an elliptical trajectory of a planet moving around the Sun and see how these geometric properties relate to a planet's initial position and velocity relative to the Sun at some chosen moment in time t=0.

We know that the trajectory of a planet is an ellipse lying within a plane of motion with the Sun at one of this ellipse focal points.
Using this, let's choose the coordinate system and the initial moment in time t=0 to simplify the equation of an orbit.

We will use the polar coordinate systems lying within a plane of motion.
The pole of the polar system will be at the Sun.
The polar axis will coincide with the major axis of an elliptical orbit and will be directed from the Sun towards the furthest from it point on a major axis (aphelion).

As the initial time t=0 we will choose a moment in time when a planet is at this furthest point from the Sun.
At this point of intersection of an ellipse and its major axis the initial velocity, being tangential to an ellipse, is perpendicular to a major axis and, therefore, perpendicular to a position vector.

Therefore, vectors r(0) and v(0) are perpendicular to each other with their magnitudes, correspondingly, r₀ and v₀.

So, at time t=0 a planet is at the polar angle θ=0 and distance from a pole (the Sun) r(0)=r₀.
The velocity vector is perpendicular to a major axis and its magnitude is v(0)=v₀.
Parameters r₀ and v₀ are given. Based on them, we have to determine the characteristics of an elliptical orbit - its semi-major axis a and the eccentricity e.

As described in the Math+ 4 All - Geometry - Ellipse lecture, the canonical equation of our elliptical orbit in these polar coordinates is

r(θ)=

a(1−e²) 1−e·cos(θ)

Setting θ=0 gives the an equation
r(0) = r₀ = a·(1+e)
It's one equation with two variables a and e. We need more equations to find these characteristics of an elliptical orbit.

Setting θ=π in the above equation of an ellipse (that is, considering the opposite point on an ellipse along the major axis called perihelion) gives
r(π) = a·(1−e)
This adds the second equation for a and e but adds another unknown r(π).

While r(π) and v(π) (the magnitudes of the position and the velocity vectors at the perihelion) are unknown, we can use two Laws of Conservation between points θ=0 and θ=π to determine them - the Law of Conservation of Angular Momentum and the Law of Conservation of Energy.

Recall the Conservation of Angular Momentum Law for an object in a central force field
L = m·r⨯v

The velocity and position vectors are perpendicular to each other at point θ=0 as well as at point θ=π.
Therefore, the magnitude of the vector product of these two vectors at both points equals to the product of their magnitudes and is the same because of conservation of angular momentum.
L(0) = m·r₀·v₀ =
= m·r(π)·v(π) = L(π)
From this follows
v(π) = r₀·v₀/r(π)

Next we will use the conservation of energy - the sum of potential energy of an object in the gravitational field and its kinetic energy is constant.

Potential energy of an object of mass m (a planet) in the gravitational field of an object of mass M (the Sun) located at a distance r from it is
U = −G·M·m/r
Kinetic energy of this object, when its speed is v, is
K = ½m·v²
Full energy is E = K + U

Therefore,
E(0) = ½m·v₀² − G·M·m/r₀ =
= ½m·v²(π) − G·M·m/r(π) =
= E(π)

From the equation for conservation of angular momentum we get the value of v²(π) in terms of r(π) and initial parameters r₀ and v₀ as follows
v²(π) = r₀²·v₀²/r²(π)
and put it in the equation for energy conservation getting an equation to determine r(π)
½m·v₀²−G·M·m/r₀ =
= ½m·r₀²·v₀²/r²(π)−G·M·m/r(π)

For brevity, let's temporarily use variable x instead of r(π).
To simplify this equation, let's reduce all members by mass m: ½v₀²−G·M/r₀ =
= ½r₀²·v₀²/x²−G·M/x
and multiply by 2x² gathering all members to the left side of an equation getting
x²·(2G·M/r₀−v₀²) −
−x·(2G·M) + r₀²·v₀² = 0

Recall that earlier in this lecture we analyzed the circular rotation in a central gravitational field and introduced a symbol β=r·v²/(G·M), which was supposed to be equal to 1 in order for an object to stay on a circular orbit.
Multiplying our equation for x by r₀/(G·M) and using this symbol β=r₀·v₀²/(G·M), the equation looks much simpler:
x²·(2−β) −x·(2r₀) + β·r₀² = 0

Canonical quadratic equation
P·x² + Q·x + R = 0
where
P = 2−β
Q = −2r₀
R = β·r₀²
has solutions

x = r(π) =

−Q±√Q²−4P·R 2P

with all components described above known.

The expression under a square root can be simplified.
Q²−4P·R = 4r₀²−4(2−β)βr₀² =
= 4r₀² − 8βr₀² +4β²r₀² =
= 4r₀²·(1−2β+β²) =
= 4r₀²·(1−β)²

Therefore,

x = r(π) =

2r0 ± 2r0·(1−β) 2(2−β)

or

r(π) =

r0·(1±(1−β)) 2−β

If we choose a plus sign in the numerator, the value for r(π) will be equal to r₀, which only possible if our ellipse is a circle with coinciding focal points.
This is a trivial case and we will not consider it here.

In all other cases the solution is
r(π) = r₀·β/(2−β) where β=r₀·v₀²/(G·M), which contains only known values - initial distance from the Sun and speed of a planet at aphelion and other known variables.

Therefore, we have a simple system of two equations with two unknowns a and e:
r(0) = r₀ = a·(1+e)
r(π) = a·(1−e)

This system has solutions
a = ½[r₀+r(π)]
e = 2r₀ / [r₀+r(π)] − 1 =
= [r₀−r(π)] / [r₀+r(π)]
From these we can determine the focal distance
c = a·e = ½[r₀−r(π)]
and semi-minor axis
b = √a²−c² = [r₀·r(π)]^½

Putting the obtained expression for r(π) into above formulas we have the geometric properties in terms of initial distance of a planet from the Sun and its linear speed at aphelion:
a = ½[r₀+r(π)] =
= ½[r₀+r₀·β/(2−β)]
which can be transformed into

a =

r0 2−β

Let's do the same with eccentricity e:
e = [r₀−r(π)] / [r₀+r(π)] =
= [r₀−r₀·β/(2−β)] /
/ [r₀+r₀·β/(2−β)]
which can be transformed into

e = 1 − β

The focal distance:
c = a·e
which can be expressed as

c =	r0(1−β) 2−β

The semi-minor axis:
b² = a² − c² =
= [r₀² − r₀²·(1−β)²] /(2−β)² =
= r₀²·β·(2−β) /(2−β)²
Therefore,

b =	r0√β·(2−β) 2−β

To conclude geometric properties of an elliptical orbit, since we know both semi-axes, we can determine an area of an ellipse A=π·a·b

A=	πr0²√β·(2−β) (2−β)²