Physics+ Newton Laws, Planet Orbits: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Planet Orbits

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 based on many years of observation, that is, experimentally.

On our quest to analytically prove the First Kepler's Law we begin with the proof that the trajectory of a point-mass moving in a central gravitational field produced by another point-mass fixed at some location in space is a flat curve, that is a curve all points of which belong to the same plane.

Consider a gravitational field produced by a point-mass M fixed at some point O in space.
Assume, a point-mass m is moving in this field and at some moment of time t is at point P.

Gravitational field is a central force field, and in the Central Force Field lecture of this chapter of this course we have proven the Angular Momentum Conservation Law.
Here is a short recap.

The point O, the fixed location of the source of gravitational field, is chosen as the origin of coordinates.
Let r(t) be a position vector OP of an object moving in this field at time t.
Let v(t) be the derivative of this position vector by time, that is a vector of velocity of our point-mass m:
v(t) = (dr/dt)(t) = r'(t).
Let a(t) be the derivative of a velocity vector by time, that is a vector of acceleration of our point-mass m, that is the second derivative of a position vector:
a(t) = v'(t) = r"(t).

We assume that our object does not move directly towards or from the center of gravity O, that is vectors r and v are not collinear.
If vectors of position and velocity are collinear, an object would move along a straight line towards or away from the source of gravity, which is a trivial case that we will not consider here.

The Angular Momentum Conservation Law states that vector L(t)=m·r(t)⨯v(t) is a constant of motion, that is L(t) does not depend on time, which is equivalent to its derivative by time is a zero-vector:
dL(t)/dt = L'(t) = 0

The proof is simple and is based on the fact that a vector product of collinear vectors is a zero-vector.
L'(t) = d[m·r(t)⨯v(t)]/dt =
= m·[r(t)⨯v'(t) + r'(t)⨯v(t)] =
= m·[r(t)⨯a(t) + v(t)⨯v(t)]
The first vector product is a zero-vector because, according to the Second Law of Newton acceleration of an object is collinear with the force, which is central, that is collinear with position vector r(t). So, r(t) and a(t) are collinear and, therefore, their vector product is a zero-vector.
The second vector product in the above expression is a zero-vector because the vector product by itself is always a zero-vector, since any vector is collinear to itself.

Since the derivative L'(t) is zero-vector, the Angular Momentum vector L(t) is independent of time and is fixed in space, which allows us to omit (t) from its value.

Recall that the result of a vector product of two vectors is a vector perpendicular to each of them.
Therefore,
L⊥r(t) and L⊥v(t).

Let's construct a plane in space going through vector r(t) (that is, points O and P) and vector v(t) (that is, through endpoint of this vector, thus having three points that define a plane). This plane will be perpendicular to a constant vector L.

Let's define a Cartesian system of coordinates in space with XY-plane to be the plane that we just constructed based on vectors r and v.
The Z-axis in our system of coordinates can be chosen coinciding with vector L that is perpendicular to XY plane.

Both vectors r(t) and v(t) lie in the constructed XY plane, and their Z-coordinate is zero.
Hence, the incremented position of our object at infinitesimally incremented time t+dt, that is a vector
r(t+dt) = r(t) + v(t)·dt
will also lie in the same plane, it Z-coordinate is zero, which means that the movement from point P will continue within the same XY plane.

Repeating this logic to consecutive incremental positions, we see that an entire trajectory will lie in our XY plane.
In other words, Z-coordinate of vector r(t) is zero at any time t.

This concludes the proof that a trajectory of an object in the central gravitational field is a curve lying in one plane.

Physics+ Newton Laws, Gravity Two Objects: UNIZOR.COM - Physics+ - Laws ...

Notes to a video lecture on UNIZOR.COM

Laws of Newton -
Gravitation of Two Objects

We recommend to refresh your knowledge of the gravitational field in three-dimensional space and the concept of field potential (scalar defined at each point of a field) using the material presented in this chapter in lectures
Field & Potential,
Work Lemmas,
Potential Theorem and
Problem 1.

The Problem 1 in this chapter of this course presented a proof that gravitation field of a point-mass M located at some point Q in three-dimensional space is conservative.

The work performed by this field when it moves some test object of mass m from point A to point B depends only on the location of these endpoints of its trajectory.

The potential of the field at point P is an amount of work needed by a field force to move a test object of a unit mass from point P to infinity.

Also, the gradient of a potential equals to the field intensity - the force (vector) acting on a test object of unit mass at any point.

Earlier in this course the gravitational force F(P) at some point P of the field was presented as
F(P) = F(r) = −G·M·m·r/r³
where r=QP is the relative position vector from the source of gravity Q to point P,
r is the magnitude (scalar) of a relative position vector r,
M is a mass of an object that is the source of gravitation,
m is a mass of a test object and
G is the Universal Gravitational Constant.

Of course, r²=x²+y²+z²
where (x,y,z) are Cartesian coordinates of vector QP (point Q is the source of gravity).

In coordinate form, components of the gravitational force vector F(P)=F(r) are:
F_x(P)=−G·M·m·x/(x²+y²+z²)^3/2
F_y(P)=−G·M·m·y/(x²+y²+z²)^3/2
F_z(P)=−G·M·m·z/(x²+y²+z²)^3/2

The potential of a gravitational field was determined as a scalar function defined at any point P of the field as
U(P) = U(r) = G·M/r =
= G·M/(x²+y²+z²)^1/2 =
= G·M·(x²+y²+z²)^−1/2

Its gradient, a vector of partial derivatives by coordinates, is
∇U(r)={∂U/∂x, ∂U/∂y, ∂U/∂z}
which should be equal to a vector of gravitational field intensity (the force per unit of test object's mass).

Indeed,
∂U(P)/∂x =
= −½G·M·(x²+y²+z²)^−3/2·2x =
= −G·M·x/(x²+y²+z²)^3/2 =
= (1/m)·F_x(P)

Similarly, partial derivatives by y and z produce the corresponding components of the gravitational field intensity - the force acting on a unit of mass of a test object.
∂U(P)/∂y =
= −G·M·y/(x²+y²+z²)^3/2 =
= (1/m)·F_y(P)
∂U(P)/∂z =
= −G·M·z/(x²+y²+z²)^3/2 =
= (1/m)·F_z(P)

We came to conclusion that
∇U(P)={∂U/∂x,∂U/∂y,∂U/∂z}=
= (1/m)·(F_x(P),F_y(P),F_z(P)) =
= E(P)
where E(P) is a vector of field intensity (the force per unit of test object's mass) at point P.

Let's determine a potential of the gravitational field produced by two point-masses M_a and M_b fixed at positions A and B correspondingly within some inertial frame of reference.

At any point P in space we can consider two positional vectors:
r_a=AP originated at A (first source of gravitation) and ending at point P and
r_b=BP originated at B (second source of gravitation) and ending at the same point P.

Assume, coordinates of all points involved are
A(x_a,y_a,z_a)
B(x_b,y_b,z_b)
P(x_p,y_p,z_p)
Then the coordinates of vectors r_a and r_b are
r_a = (x_p−x_a, y_p−y_a, z_p−z_a)
r_b = (x_p−x_b, y_p−y_b, z_p−z_b)

According to a principle of superposition, the combined gravitational force of our two sources acting on a test object of mass m positioned at point P should be a vector sum of two separate forces, that is
F(P) = −(G·M_a·m)·(r_a/r_a³) −
− (G·M_b·m)·(r_b/r_b³)

The field intensity (the force per unit of mass of a test object) at point P, therefore, is
E(P) = −(G·M_a)·(r_a/r_a³) −
− (G·M_b)·(r_b/r_b³)
Notice, the above formula involves the addition of vectors.

From the fact that field potential is an amount of work needed to move a test object of a unit mass from its initial location P to infinity and that the work is, generally speaking, an additive quantity intuitively follows that potential of two fields should be equal to a sum of potentials.

More rigorously, work is an integral of a scalar product of force by differential of distance
W = ∫[P,∞](F·dl)

If F=F_a+F_b, the work of a combined force is the sum of work of its components
W = ∫[P,∞]((F_a+F_b)·dl) =
= ∫_[P,∞](F_a·dl) + ∫_[P,∞](F_b·dl) =
= W_a + W_b
which proves additive characters of work and, consequently, of field potential in case a field is generated by more than one source.

The sum of potentials involves the addition of scalars, which is easier than addition of vectors of forces, so dealing with potentials is a preferable way.

Now consider a different scenario.

Assume, our two objects of mass M_a and M_b are not fixed at particular points A and B in space, but can move freely, and their corresponding position vectors are r_a and r_b.
Assume further that there are no external forces that can influence the movement of our system of two objects, and the only force that somehow affects their movement is the force of gravitation of one to another.
We will prove that in this case a center of mass of these two objects moves uniformly in space along a straight line with constant velocity.

The only force acting on the a-object with mass M_a is the gravitation force sourced at b-object with mass M_b.
This force is directed along a straight line connecting these two objects and its magnitude is
F_ab(t) = G·M_a·M_b/d²(t).
where d=|r_a − r_b| is a time-dependent distance between these objects.

In a more appropriate vector form it looks like
F_ab(t) =
= G·M_a·M_b·[r_a(t)−r_b(t)]/d³(t).

Similarly, the only force acting on the b-object with mass M_b is the gravitation force sourced at a-object with mass M_a.
This force is directed along a straight line connecting these two objects and its magnitude is
F_ba(t) = G·M_b·M_a/d²(t).

In a more appropriate vector form it looks like
F_ba(t) =
= G·M_a·M_b·[r_b(t)−r_a(t)]/d³(t).

So, the magnitudes of these forces, F_ab(t) and F_ba(t), are the same, but directions are opposite to each other, F_ab(t) is directed from a-object to b-object, while F_ba(t) is directed from b-object to a-object:
F_ab(t) = −F_ba(t)

Using the Newton's Second Law, knowing the forces and masses, we obtain equations for accelerations of these two objects, second derivatives of position vectors r_a" and r_b".
r_a" · M_a = F_ab
r_b" · M_b = F_ba
In the equations above and following it is assumed that all variables except masses are time-dependent functions. We just skip (t) for brevity.

If masses M_a and M_b are positioned at r_a and r_b, the center of mass is at position
r = (M_a·r_a + M_b·r_b)/(M_a+M_b)
The acceleration of this point is it's second derivative of position
r" = (M_a·r_a" + M_b·r_b)"/(M_a+M_b) =
= (F_ab + F_ba)/(M_a+M_b) = 0
because, as we stated above, F_ab(t)=−F_ba(t)

Since acceleration of the center of mass is zero, it velocity is a constant vector, which means that the center of mass in the absence of external forces moves along straight line with the same velocity.

Physics+ Newton Laws, Problem 5: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 5

Problem A

A point mass M is uniformly rotating within a plane with XY Cartesian coordinates on radius R and angular speed of rotation ω on a weightless unstretchable thread around a point that is the center of coordinates on this plane.
Determine the force F with which a thread holds this point mass on its fixed orbit (centripetal force) and express this force in Cartesian coordinates [F_x,F_y].
What is the direction and magnitude of this force?

Solution A

Position of a rotating point mass, as a function of time t, is a vector r(t) with the following coordinates
X(t) = R·cos(ω·t)
Y(t) = R·sin(ω·t)

Coordinates of the velocity vector V=[V_x,V_y] are, by definition, derivatives of coordinates of position
V_x(t) = −R·ω·sin(ω·t)
V_y(t) = R·ω·cos(ω·t)

Vector of velocity is perpendicular to a position vector. To confirm it, let's form a scalar product of these two vectors and check if it's equal to zero.
X(t)·V_x(t) + Y(t)·V_y(t) =
= −R²·ω·cos(ω·t)·sin(ω·t) +
+ R²·ω·sin(ω·)·cos(ω·t) = 0

Coordinates of the acceleration vector a=[a_x,a_y] are, by definition, derivatives of coordinates of the velocity vector
a_x(t) = −R·ω²·cos(ω·t)
a_y(t) = −R·ω²·sin(ω·t)

Since the centripetal force and acceleration are related, according to the Newton's Second Law, as F=M·a,
F_x(t) = M·a_x = −M·R·ω²·cos(ω·t)
F_y(t) = M·a_x = −M·R·ω²·sin(ω·t)

As we see, the vector of force is collinear and opposite in direction to the vector of position, that is, it's directed towards the center of rotation.

At the same time, the vector of force, being collinear to the vector of position, is perpendicular to the vector of velocity.

The magnitude of the centripetal force is
|F(t)| = M·R·ω²


Problem B

In the context of the Problem A above, consider that the plane of rotation is made of some frictionless material and positioned perpendicularly to the force of gravity. There is a hole in the center of rotation and a thread that holds the rotating point mass M goes down through that hole and is held by hand in a fixed position.
The radius of rotation is the same as above R and the angular speed is ω.
What mass m₀ should be attached to the bottom of a thread to replace the holding hand to maintain original radius and angular speed of rotation?
How the rotation will change (in terms of radius and angular speed) if we add another mass m₁ to the mass m₀ needed to maintain the original rotation?

Solution B

The results of the Problem A above regarding the direction and magnitude of the centripetal force F are:
1. The centripetal force is directed towards the center of the rotation, opposite to a position vector of the rotating object;
2. The magnitude of this force is |F(t)| = M·R·ω²

The condition of the Problem B is that the source of the centripetal force holding an object on a fixed orbit of rotation is the holding hand. If we replace it with the gravitation force of an object of mass m₀, the magnitude of the gravitational force must be equal to the one calculated above:
m₀·g = M·R·ω²
Therefore,
m₀ = M·R·ω²/g

Adding another mass m₁ to the above will change the picture. Both radius and angular frequency of rotation will change to correspond to a changed magnitude in the centripetal force.
At the same time, all forces acting on a rotating object are central, and that is a sufficient condition for the Law of Conservation of Angular momentum.
The above is sufficient to calculate a new radius and angular speed of a rotating object.

New magnitude of a centripetal force is (m₀+m₁)·g.
Therefore,
(m₀+m₁)·g = M·R_new·ω²_new
This is one equation with two variables to find:
R_new and ω_new.

Taking into account the Law of Conservation of the Angular Momentum will produce another equation, and that should be sufficient to find both unknown variables.

The magnitude of the original angular momentum of an object rotating on radius R with the angular speed ω is
L = |L| = |r⨯ p|
where
r is a position vector of a rotating object and
p is a linear momentum of a rotating object.

Linear speed v of an object rotating on a radius R with an angular speed ω is R·ω.
Since r is a radial vector and p=M·v is tangential to an orbit, these two vectors are perpendicular to each other, and the magnitude of their vector product is a product of their magnitudes.
Therefore,
L = M·R·v = M·R²·ω

After we changed the centripetal force the angular momentum must be the same. Therefore,
L = M·R²·ω =
= L_new = M·R²_new·ω_new
or
R²·ω = R²_new·ω_new

This produces the second equation we need to determine new radius and angular velocity. Hence, the system of two equations with two unknows is
(m₀+m₁)·g = M·R_new·ω²_new
R²·ω = R²_new·ω_new

Resolving the second equation for ω_new and substituting it into the first produces
ω_new = R²·ω/R²_new
(m₀+m₁)·g =
= M·R_new·(R²·ω/R²_new)² =
= M·R⁴·ω²/R³_new

From the above follows the value for a new radius of rotation
R³_new = M·R⁴·ω²/[(m₀+m₁)·g]
R_new =
= {M·R⁴·ω²/[(m₀+m₁)·g]}^⅓
The new angular speed can be derived from the above and the condition ω_new=R²·ω/R²_new
ω_new = R²·ω·
·{(m₀+m₁)·g/[M·R⁴·ω²]}^⅓

This rather complex formula needs to be verified somehow.
Let's consider a simple case of a radius shortened by half and check what additional mass m₁ is needed.
In a formula for R³_new above we will set R_new=½R.
(½R)³ = M·R⁴·ω²/[(m₀+m₁)·g]
from which follows
1 = 8M·R·ω²/[(m₀+m₁)·g]
(m₀+m₁)·g = 8M·R·ω² =
= 8m₀·g
Therefore,
m₁ = 7m₀.
In other words, to decrease the radius by half we have to add 7 times the mass that used to hold a rotating object on its original distance from a center of rotation, that is we have to increase the mass by factor 8.

As we know (see Problem A above), the magnitude of the centripetal force is
|F(t)| = M·R·ω²
Also known is that, when the radius of rotation shortens by half, the angular velocity increases by factor 4.
From these two statements follows that a centripetal force in this case increases by a factor of 8, which corresponds to our calculations above.

Physics+ Newton Laws, Problem 2: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 2

Problem

A chain of infinitesimally small links with overall length L and mass M is stretched on a table with its one end hanging off the edge of a table.
Half of a chain is lying on the table, while another half is hanging down.
The chain is stretched on a table along a straight line perpendicular to the edge of a table.

Initially, the chain is held at rest. Then at time t=0 it is let go, so the hanging part of a chain pulls the rest of it off the table.

The experiment ends at time t=T when the chain's endpoint B will be at the table's edge, where endpoint A was in the beginning of movement.

Assume that table legs are long enough, so chain, when it slides all the way down, does not reach the floor until it is fully off the edge of a table.
There is no friction between a chain and a table.
The acceleration of free falling is g.

What will be the speed of a chain at time t=T?
In other words, what will be the speed of the endpoint B, when this endpoint will be at the table's edge?

Solution

Let's try to solve this problem using only the Newton's Laws.

Assume that at a moment in time t the chain slides off the table by the length s(t).
Then the mass of the part of a chain that is not supported by a table will be
(½L+s(t))·M/L
It will pull down the part of a chain still on the table with the force
F = (½L+s(t))·(M/L)·g
This force pulls the chain of mass M with acceleration
a(t) = d²s(t)/dt² = s"(t).

Knowing the force, the mass and the acceleration, we can use Newton's Second Law getting a differential equation for s(t):
(½L+s(t))·(M/L)·g = M·s"(t)
or
s"(t) − s(t)·g/L − ½·g = 0

Initial conditions are
s(0)=0; s'(0)=0

To solve the above differential equation is not easy, which makes the whole approach less practical.

Let's try to solve it differently.
Assume, the length s(t) is an independent variable.
We know the force of gravity F(s(t)) in terms of s(t):
F(s(t)) = (½L+s(t))·(M/L)·g

Multiplied by infinitesimal increment of the length of the chain ds(t) that slides from the table during the infinitesimal increment of time dt, it will give an infinitesimal increment of work performed by this force
dW = F·ds

Integrating this from s(0)=0 to s(T)=½L, we will get the total work performed by the force of gravity
W_[0,T] = ∫_[0,½L]F(s)·ds =
= ∫_[0,½L](½L+s)·(M/L)·g·ds =
= ∫_[0,½L]½L·(M/L)·g·ds +
+ ∫_[0,½L]s·(M/L)·g·ds =
= ½M·g·s|_[0,½L] +
+ (M/L)·g·(s²/2)|_[0,½L] =
= ¼M·L·g + ⅛M·L·g =
= ⅜M·L·g

All this work was done by a gravitational field. The potential energy of a chain in a gravitational field has decreased by this amount, and this work was converted into kinetic energy of the chain.

Since the chain, when it completely slides from the table at time T has kinetic energy E=½M·v², where v its speed, ⅜M·L·g = ½M·v²
Therefore,
v² = ¾L·g
from which we derive the value of the chain's speed at the moment it completely slides off the table
v = √¾L·g
Notice that the speed of a chain at time t=T does not depend on the mass of a chain.

Yet another, even simpler approach to this problem is to use only a loss of potential energy as a reason for an increase in kinetic energy.

Let's take the surface of a table as the level zero of potential energy of a chain.

In the beginning, at time t=0 half a chain is on this surface, the potential energy P₁ of this half is zero:
P₁ = 0

Another half of a chain is below the level zero and, therefore, its potential energy is some negative value P₂.

The chain is at rest and, therefore, its total energy is
E = P₁ + P₂ = P₂

At the end of motion at time t=T the first half of a chain that had zero potential energy P₁ takes the place of the second one in the beginning of motion and its potential energy will be
P'₁ = P₂
The second half of a chain moves down by half the length of chain ½L, decreasing its potential energy to level P'₂.

Now the chain is moving and has some kinetic energy K.
Therefore, the total energy of a chain is
E = P'₁ + P'₂ + T =
= P₂ + P'₂ + T

From the Law of Energy Conservation
E = P₁ + P₂ = P'₁ + P'₂ + T

Knowing that
P₁ = 0 and P'₁ = P₂
we conclude
E = P₂ = P₂ + P'₂ + T

We see now that
P₂ = P₂ + P'₂ + T
Hence, canceling P₂,
0 = P'₂ + T
which allows to find kinetic energy T
T = −P'₂

Therefore, the loss of potential energy is the negative potential energy P'₂ of the second half of a chain at the end of motion.

It can be calculated by intuitively obvious method that considers all the mass concentrated in the middle of this half of a chain hanging below the surface of a table by ¾L, which leads to its potential energy
P = −½M·g·¾L = −⅜M·g·L.

Alternatively, we can calculate the potential energy of this half of a chain directly integrating potential energy of each infinitesimal part of it.
Let x be a distance of the point of the chain's second half from the table (level zero).
Then
P'₂ = −∫_[½L,L](M/L)·g·x·dx =
= −(M/L)·g·(x²/2)|_[½L,L] =
= −(½−⅛)M·g·L =
= −⅜M·g·L

In all cases the loss of potential energy is ⅜M·g·L.
From the Law of Conservation of Energy follows that kinetic energy of a chain must increase by the same amount, which leads to same value of a speed of a chain at time t=T
v = √¾L·g

Physics+ Newton Laws, Problem 3: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 3

Problem

A chain of infinitesimally small links with overall length L and mass M is hung vertically by its top end A.
Its bottom end B touches a flat platform with a spring under it.
Assume that a platform and a spring are weightless, so the spring is in neutral position.
The elasticity coefficient of a spring is k.
Gravity acceleration is g.


If the chain is let go from point A, it will fall down on a platform, which will squeeze the spring.
What is the maximum distance the platform will go down, shortening the spring?

Solution

We will compare the potential energy of a chain in its initial position to potential energy of a squeezed spring.
As the chain falls down to a platform, its potential energy is gradually transformed into kinetic energy, which, in turn, is transformed into an energy of a spring.

When chain is fully on a platform, the energy of an oscillating spring (kinetic plus potential) should be equal to an initial potential energy of a chain.
At that time a spring will oscillate between maximum squeeze and maximum elongation with its total energy at the end points be only potential and equal to initial potential energy of a chain before it started to fall down.

Let's take the level of a platform in its neutral position as level zero of potential energy.
The mass density of a chain per unit of length is
μ = M/L
An infinitesimal piece of a chain of length dy has mass μ·dy.
Its potential energy on a height y above the surface of a platform in its neutral position, which is level zero, is
dP = μ·g·dy·y = (M/L)·g·y·dy
Potential energy of the whole chain is
P = ∫_[0,L](M/L)·g·y·dy =
= (M/L)·g·y²/2|_[0,L] = M·g·L/2
Check units:
(kg/m)·(m/sec²)·m² =
= (kg·m/sec²)·m = watt

According to the Hook's Law, the force of resistance of a spring F is proportional to the length of its shortening x.
F = k·x
When a spring is squeezed from the length x to x−dx, the work performed by an external force that caused it is
dW = F·dx = k·x·dx
To shorten a spring by the length S we have to do the work
W = ∫_[0,S]k·x·dx = k· S²/2
Check units:
[(kg·m/sec²)/m]·m² =
= (kg·m/sec²)·m = watt

From the Law of Energy Conservation follows that the potential energy of a chain in its initial position should be equal to the work performed by the force of a falling chain to squeeze a spring under a platform, that is
P = W
from which we derive
M·g·L/2 = k·S²/2
and
S = √M·g·L/k

Answer:
A spring will shorten its length by
S = √M·g·L/k
Check units:
√kg·(m/sec²)·m/(kg/sec²) = m

Physics+ Newton Laws, Central Field: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Central Force Field

The subject of this lecture is a theory behind practical problems of space travel around some gravitating mass or a movement of an electron around an electrically charged nucleus, or other types of movement in a central force field, where the force at any point A is always directed towards or away from some central point O and its magnitude is the same at all points that lie on the same distance from the center O.

Consider an inertial reference frame in our three-dimensional space with origin at point O and a force field F defined in all points with the following conditions satisfied:
(a) for any point A in our space the vector of force F(A) at this point is always colinear with position vector r=OA;
(b) for any two points A and B lying on the same distance from the center O vectors F(A) and F(B) have the same magnitude and both directed either towards or away from the center O.

Obviously, from condition (b) follows that all force vectors at all points lying on the same distance around center O have the same magnitude and all of them are directed the same way relatively to center O - either towards it or away from it.

Defined in such way, the force field is called central force field.

The vector of central force at any point A is defined by position vector r=OA and can be expressed as a scalar f(r) that depends only on the distance r=|r| of a point A from the origin of coordinates (positive or negative to differentiate between the directions towards or away from center O) multiplied by the vector of position r.
F(A) = F(r) = f(r)·r.

Imagine that at time t=0 some object of point mass m is at position defined by a vector r₀ and its initial velocity is v₀.

Velocity vector, by definition, is the first derivative of position
v(t) = r'(t)

Vector of acceleration is the first derivative of velocity by time or the second derivative of position
a(t) = v'(t) = r"(t).

Now we can express the Newton's Second Law as
F = f(r)·r = m·a = m·r"
and express an acceleration in terms of a central force and mass
a = f(r)·r/m

Consider a vector of angular (rotational) momentum of motion of our object L defined as the vector (cross) product of the position vector r(t) by the momentum of motion p=m·v(t):
L = r⨯(m·v) = m·r⨯v
and analyze how this angular momentum changes with the time.
More precisely, let's prove that the angular momentum of an object in a central field is a constant (the Law of Conservation of Angular Momentum in Central Field).

The first derivative of an angular momentum L is
L'(t) =
= m·[r(t)⨯v'(t) + r'(t)⨯v(t)]
Take into account that the derivative of a position is a velocity and the derivative of a velocity is an acceleration.
Then L'(t) =
= m·[r(t)⨯a(t) + v(t)⨯v(t)]

The second component in the above expression is a zero-vector because the vector product of collinear vectors is always a zero-vector.
Substituting acceleration for its expression in terms of force, position and mass a=f(r)·r/m, we obtain
L'(t) = f(r)·r(t)⨯r(t) = 0
because, again, a vector (cross) product of two collinear vectors is a zero-vector.

As we see, the derivative by time of the vector of angular momentum of an object in a central force field equals to zero-vector, which means that the vector of angular momentum in a central force field is a constant.

Notice that in our proof of the Angular Momentum Conservation Law we relied only on the central character of the force field, not its specific form for gravitational or electrostatic fields.

Physics+ Newton Laws, Problem 4: UNIZOR.COM - Physics+ - Laws of Newton

Notes to a video lecture on UNIZOR.COM

Laws of Newton - Problem 4

Problem

This problem is related to an effect on rotation of a figure skater when she pulls the hands closer to her body, thereby increasing the angular speed of rotation.

The physical problem that models this situation is as follows.
A point mass m initially rotates on a thread of length r₀ with initial angular speed ω₀.
We would like to shorten the radius of rotation and see how this is related to a speed of rotation.

To accomplish this, let's put the thread that holds a rotating mass through a thin tube positioned perpendicularly to a plane of rotation with its one end used as a center of rotation, so we can shorten the radius of rotation by pulling a thread from the other end of a tube.


Assuming the radius of rotation is changing with time and, finally, equals to r, what then will be the angular speed of rotation ω?

Solution

The solution is based on the Rotational (Angular) Momentum Conservation Law.
This law states that vector
L = r⨯ p
where r is a position vector relative to a center of rotation
and p is a momentum of motion of an object,
is preserved and does not change in direction or magnitude.

Consider the beginning and the end of the process of reducing the radius of rotation. In both these cases the motion of our object is rotational, and vectors of position r and velocity v are perpendicular to each other.
Therefore, their vector product has a magnitude
L = m·r·v
Since v=r·ω, L = m·r²·ω
This value is the same at the beginning and at the end of the process.
Based on the Law of Conservation of Angular Momentum, dependency of angular speed ω from radius r can be easily derived as
L = m·r²·ω = L₀ = m·r₀²·ω₀
r²·ω = r₀²·ω₀
ω = ω₀·r₀²/r²

Incidentally, the expression I=m·r² is called moment of inertia, so the Law of Conservation of Angular Momentum can be expressed as
L = I·ω = L₀ = I₀·ω₀

For example, if the radius is reduced by half, the angular speed will increase by a factor of four.

Let's consider a more practical case of a rotating figure skater, when she, to increase the speed of rotation, brings her arms close to a body.
This is a rotation of a solid object of variable geometry, and we will simplify our job by considering the following model.

Assume, there is a rotating object that has two parts.
One part, modeling arms initially stretched, but later brought tightly to the body, has mass m and rotates on initial radius r₀, which will be changed to a smaller radius r.
Another part, modeling the rest of a figure skater's body, has mass M and a constant radius of rotation R.

Assume, initial angular speed of rotation of a figure skater with arms stretched is ω₀.

The total angular momentum with arms stretched is
L₀ = m·r²₀·ω₀ + M·R²·ω₀.

Bringing the arms close to the body of a figure skater in our model will be reducing the radius of rotation of mass m from initial r₀ to final R which is smaller.
This will change the angular speed of rotation to ω.

The final angular momentum with arms close to the body and a new angular speed of rotation ω is.
L = (M+m)·R²·ω.

According to the Law of Conservation of Angular Momentum, the final angular momentum L is equal to the initial one L₀
L = L₀
Therefore,
m·r²₀·ω₀ + M·R²·ω₀ =
= (M+m)·R²·ω.
from which follows that the final angular momentum is
ω =
= ω₀·(M·R²+m·r²₀)/[(M+m)·R²]

Since r₀ (radius of rotation of stretched arms) is greater than R (radius of rotation of the rest of the body), the numerator of the above fraction is greater than denominator, which makes final angular momentum ω greater than the initial one ω₀.