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 = |
|
= 1 |
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)= |
|
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) = |
|
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) = |
|
| r(π) = |
|
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 = |
|
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 = |
|
b² = a² − c² =
= [r0² − r0²·(1−β)²] /(2−β)² =
= r0²·β·(2−β) /(2−β)²
Therefore,
| b = |
|
| A= |
|
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