Physics+ Kepler's 1st Law: UNIZOR.COM - Physics+ 4 All - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Kepler's First Law

Kepler's First Law states that all planets move around the Sun on elliptical orbits with the Sun in one of the two focus points of their orbits.
Kepler had come up with this law experimentally based on many years of observations.

Based on our knowledge of Physics, we will prove analytically a more general theorem about objects moving in the central gravitational field:
The trajectory of an object moving in the central gravitational field is a flat curve of the second order (ellipse, parabola or hyperbola) with the source of the gravitational field in the focal point of this curve.

The previous lecture Planet Orbits of this section of this course was about why the trajectory of an object in a central gravitational field is a flat curve with the center of a gravitational field lying within this plane.
So, we assume this topic is covered.

Also covered were the properties of ellipse, hyperbola and parabola, including their canonical equations in Cartesian and polar coordinates (see UNIZOR.COM > Math+ 4 All > Geometry).

Assume, a point-mass M is the source of a gravitational field and is located at point O in space.

Assume further that a point-mass m is moving in the gravitational field of point-mass M with no other forces involved, and at time t it's located at point P(t). Then its position relative to the source of gravity is vector OP(t)=r(t).

The velocity vector of an object is v(t)=r'(t) - a derivative of the position vector by time (here we will use a single apostrophe to signify the first derivative of a function by time and a double apostrophe - to signify its second derivative by time).

Let's choose some moment of time as the beginning of motion and consider it as an initial time t=0.
The initial position of our object moving in a central gravitational field will be r(0) and initial velocity will be v(0).

As we have proven in the previous lecture, an entire trajectory of the motion of point-mass m will be within the plane of motion going through the center of gravity O and vectors r(0) and v(0).
Consequently, at any moment of time t vectors of position r(t) and velocity v(t) will lie in the same plane.

Let's introduce a system of coordinates OXY on this plane with an origin O at the center of gravity and axis OX directed to a position of our object r(0) at time t=0.
Let's fix the directions of axes OX and OY within the plane of motion for now, but keep in mind that we might rotate axes in the future to simplify the final equations of a trajectory.

In such a system of XY-coordinates with point O as an origin the vector OP(t)=r(t) can be represented as a pair
r(t) = {x(t),y(t)}.

In these XY-coordinates the velocity vector v(t) can be represented as
v(t) = r'(t) = {x'(t),y'(t)}.

Our task is to analytically describe the trajectory of an object in a gravitation field as a curve inside a plane OXY.

The first foundation for theoretical derivation of the Kepler's First Law is the Second Law of Newton.

Provided the force F (a vector) acts on an object of mass m, this Law states that F = m·a
where a is a vector of object's acceleration, that is the first derivative from the object's velocity vector v or, equivalently, the second derivative of object's position vector r
a(t) = v'(t) = r"(t)

The second component is the Newton's Universal Law of Gravitation.

It assumes that an object of mass M is fixed in space at the origin of coordinates and is the only source of gravitation.

It further assumes that an object of mass m is moving in the gravitational field produced by an object of mass M and at time t is at position defined by a position vector r(t) that stretches along the line between the source of gravitation (the origin of coordinates) and the object's location.

Then the Law of Gravitation states that
(a) the vector of gravitational force F(t) produced by object of mass M acting on an object of mass m at any time t is collinear with the position vector r(t) and directed towards the object of mass M and
(b) the magnitude F(t) of the gravitational force F(t) is
F(t) = G·M·m/r²(t)
where
G is the universal gravitational constant whose value is 6.67430·10⁻¹¹ N·m²/kg² and
r(t) is the magnitude of the position vector r(t) - the distance of an object from the origin of coordinates.

Combining conditions (a) and (b) together, we can express the force of gravity in vector form as
F(t) = −G·M·m·r(t)/r³(t)

The Newton's Second Law, which connects a force and an acceleration, and the Universal Law of Gravitation, which defines the value of a gravitational force, allow to establish the differential equation that defines the motion of our object in a gravitational field
F(t) = m·a(t) = m·r"(t) =
= −G·M·m·r(t)/r³(t)
or, canceling mass m,
r"(t) = −G·M·r(t)/r³(t)

The fact that mass m can be canceled is quite remarkable. It says that different objects would follow the same trajectory in the gravitational field, if at some moment their positions and velocities are the same.
This is not supposed to be a surprise. Recall Galileo's experiments with different objects dropped from the Tower of Pisa. They would spend the same time to hit the ground regardless of their mass.
If, in addition to dropping them down, you give them the same horizontal speed, they would spend the same time going horizontally and fall on the same distance from a vertical, thus going along the same trajectory, regardless of mass.

The above equation is a vector differential equation that can be transformed into a system of two differential equations for Cartesian coordinates {x(t),y(t)} of vector r(t) with
r"(t) = {x"(t),y"(t)} and
r(t) = √x²(t)+y²(t)

In coordinate form this system of two differential equations with two unknown functions x(t) and y(t) looks like
x"(t)=−G·M·x(t)/r³
y"(t)=−G·M·y(t)/r³

Directly solving this system of differential equations is not an easy task to do.
But let's keep in mind that its solution will give us more than we asked for. We wanted to prove that objects in a gravitational field move along ellipses, hyperbolas or parabolas. We just needed a shape of trajectories, not a time-dependent position at every moment in time.

Having this smaller task in mind, we do not have to derive explicit functions x(t) and y(t) that are solutions to the above system of equations, we just have to concentrate on geometrical characteristics of trajectories.

Consider a position vector r from a fixed center of gravity (point O in our system of coordinates) to a position of an object at any moment of time.
Apparently, having a dependency of its length r=|r| on its angle θ from OX-axis is sufficient to determine geometric characteristics of its trajectory.
Basically, we can derive the geometric characteristics of a trajectory by analyzing its equation in polar coordinates {r,θ}. Deriving a function r=r(θ) is a much easier task than solving the above system of differential equations.
Let's concentrate on this smaller task.

Consider an angle ∠θ(t) from the positive direction of X-axis to position vector r(t).
Using this, coordinates of position are
x(t) = r(t)·cos(θ(t))
y(t) = r(t)·cos(θ(t))

At this point let's concentrate not on time-dependency of coordinates, but on their dependency on the angle θ. That will allow to derive the shape of a trajectory without knowing exactly the coordinates of an object moving along this trajectory at any moment of time.
Now the coordinates of an object's position, as functions of θ, are
x(θ(t)) = r(θ(t))·cos(θ(t))
y(θ(t)) = r(θ(t))·cos(θ(t))
Now we are not interested in function θ(t) and the above equations can be viewed as x(θ) = r(θ)·cos(θ)
y(θ) = r(θ)·cos(θ)

Using this representation, the original system of differential equations is
x"(t) = −G·M·cos(θ)/r²(θ)
y"(t) = −G·M·sin(θ)/r²(θ)

Recall from the lecture Planet Orbits of this part of the course that vector of Angular Momentum L=m·r·v of an object moving in a central gravitational field is a constant.
This vector L is perpendicular to both r and v and, therefore, is directed along the third dimension OZ-axis that is perpendicular to a plane of object motion.

The length L=|L| (a constant) of an Angular Momentum vector equals to L=m·r²·θ'.

Proof

In three dimensional XYZ space the position vector r and velocity vector v=r' have Z-coordinate equal to zero:
r = {r·cos(θ), r·sin(θ), 0}.
v = {r'·cos(θ)−r·sin(θ)·θ', r'·sin(θ)+r·cos(θ)·θ', 0}.

According to the rules of vector product in three-dimensional Cartesian coordinates, their vector product is
L/m = r⨯v = {0, 0, r·cos(θ)·[r'·sin(θ)+r·cos(θ)·θ'] −
− r·sin(θ)·[r'·cos(θ)−r·sin(θ)·θ']} =
= {0, 0, r²·[cos²(θ)+sin²(θ)]·θ'} =
= {0, 0, r²·θ'}
Therefore,
L = m·r²·θ'

Now we can express r² in terms of θ' and constants
r² = L/(m·θ')
Therefore, the system of our differential equations can be expressed only in terms of θ(t) as follows
x"(t) = −θ'·cos(θ)·(G·M·m/L)
y"(t) = −θ'·sin(θ)·(G·M·m/L)
In these equations θ(t) is an unknown function of time, while x"(t) and y"(t) are components of an acceleration vector and are the first derivatives of the components of a velocity vector {x'(t),y'(t)}.

Notice that expressions on the right side of both differential equations are full derivative by time of simple functions of θ, which is easy to integrate.
Therefore, integrating by time left and right sides of equations, we obtain
x'(t) = −sin(θ)·(G·M·m/L) + c₁
y'(t) = cos(θ)·(G·M·m/L) + c₂
or, substituting for simplicity a constant K for G·M·m/L,
x'(t) = −sin(θ)·K + c₁
y'(t) = cos(θ)·K + c₂
where constants c₁ and c₂ are defined by initial conditions at time t=0, when θ(0)=0:
x'(0) = −sin(0)·K + c₁
y'(0) = cos(0)·K + c₂
c₁ = x'(0)
c₂ = y'(0)−K
The initial components of velocity at x'(0) and y'(0) are given and K=G·M·m/L is a known constant.

The remaining task is to find the length of a position vector r(θ), as a function of angle θ from the above expressions for components of the velocity.

The components of a position vector, as functions of θ, are
x = r·cos(θ)
y = r·sin(θ)
differentiating these by time, we obtain
x'(t) = r'·cos(θ) − r·sin(θ)·θ'
y'(t) = r'·sin(θ) + r·cos(θ)·θ'

Comparing these with the expressions for components of a velocity vector above, give
r'·cos(θ) − r·sin(θ)·θ' =
= −sin(θ)·K + c₁
r'·sin(θ) + r·cos(θ)·θ' =
= cos(θ)·K + c₂

The next step is to get rid of r' and θ'.
The way to do it is to multiply the first equation by sin(θ), the second - by cos(θ) and subtract the first from the second.
r·cos²(θ)·θ' + r·sin²(θ)·θ' =
= cos²(θ)·K + c₂·cos(θ) +
+ sin²(θ)·K − c₁·sin(θ)
Now we can use trigonometric identity sin²(θ)+cos²(θ)=1 getting
r·θ' = K + c₂·cos(θ) − c₁·sin(θ)

Also, recall that L=m·r²·θ' and, therefore, θ'=L/(m·r²).
Using this result in the following
L/(m·r)=K+c₂·cos(θ)−c₁·sin(θ)

Resolving for r results in the expression of r as a function of θ - exactly what we are looking for
r = 1/[A+B·cos(θ)−C·sin(θ)]
where all coefficients are known constants:
A = K·m/L
B = c₂·m/L = [y'(0)−K]·m/L
C = c₁·m/L = x'(0)·m/L
K = G·M·m/L

The above looks like a complete solution r(θ), but we can improve it a little.
The expression B·cos(θ)−C·sin(θ) can be transformed into D·cos(θ+φ) as follows
B·cos(θ)−C·sin(θ) =
= √B²+C²·[cos(θ)·B/√B²+C² −
− sin(θ)·C/√B²+C²]
We can always find an angle φ such that
B/√B²+C² = cos(φ)
C/√B²+C² = sin(φ)
which results in
B·cos(θ)−C·sin(θ) =
= √B²+C²·[cos(θ)·cos(φ)−
+sin(θ)·sin(φ)] =
= √B²+C²·cos(θ+φ)

Therefore,
r = 1/[A+√B²+C²·cos(θ+φ)]
Let's do a small correction in the direction of the X-axis.
Originally, we placed the beginning of coordinates point O at the point where the source of gravitation is located. Then we chose OX axis directed to the initial position of our object at time t=0.
Now, knowing angle φ we can change the direction of the X-axis by turning it by an angle φ.
As a result, in this new coordinate system the equation of the trajectory looks even simplier:
r = 1/[A+√B²+C²·cos(θ)]
which, depending on the values of all constants involved, is an equation in polar coordinates of either an ellipse or a hyperbola, or a parabola, as explained in the previous lecture dedicated to 2-nd order curves.

End of proof.