Central Force - Mechanics & General Properties of Matter - Physics

Central Force

In classical mechanics, the central-force problem is to determine the motion of a particle under the influence of a single central force. A central force is a force that points from the particle directly towards (or directly away from) a fixed point in space, the center, and whose magnitude only depends on the distance of the object to the center.

In central force potential V is only function of r, a central force is always a conservative force; the magnitude F of a central force can always be expressed as the derivative of a time independent potential energy

And the force F is defined as F = (force is only in radial direction)

External torque and angular momentum of system
but for central force,
External torque τ = 0 , so angular momentum is conserved.
Now, if we calculate hence position vectoris perpendicular to angular momentum vector, henceis conserved. Its magnitude and direction both are fixed, The central forceis along to r and can exert no torque on the mass m. Hence the angular momentum J of  m is constant. However, J is fixed in space, and it follows that r can only move in the plane perpendicular to J through the origin.

Equation of Motion for central force

The equation of motion in polar coordinate is given by

For central force problem F(r) = - ∂V/∂r and F_θ = 0

Equation of Motion and Condition for Circular Orbit Condition of circular orbit From equation of motion in radial part m
For circular orbit of radius r₀, r = r₀ and
And  is identified as angular frequency in circular orbit.

For circular orbit angular frequency ω₀ is given by ω₀ =

Conservation of Angular Momentum and Areal Velocity

= F₀, but for central force, F_θ = 0 ⇒
Angular momentum =  it is also seen
So motion due to central force is confine into a plane and angular momentum is perpendicular to that plane.
For the central force problem, now A = 1/2 r.rdθ
Areal velocity =
. It is given that
Which means equal area will swept in equal time

Total Energy of the System

Hence total energy is not explicitly function of time t so ∂E/∂r = 0. One can conclude that total energy in central potential is constant.
E = 1/2 mv² + V(r) and Velocity
So total energy, E = 1/2

The sum rotational kinetic energy as a function of only r and potential energy as a function of only r i.e. V(r) are identified as effective potential
The concept of effective potential allow to two dimensional system in one system as V_effective is only function  of r .

Analysis of effective potential
The effective potential is define as The nature of orbit will dependent on nature of effective potential and total energy of a system which is shown in figure.
And lets discuss how variation of energy will leads to shape of orbit
Case 1: if energy E < V₀ there is turning point at r = a the classical region as r > a , the particle is unbounded so nature of orbit may be either parabolic or hyperbolic

Case 2: If energy E = V₀ there r = r₀  is identified as stable equilibrium point because > 0 so possible orbit is V_effective circular in nature with radius r₀ for region, the  angular frequency in circular orbit is ω₀ =
There is turning point at r = a. r > a is also identified as classical region. The particle is unbounded so nature of orbit may be either parabolic or hyperbolic

Case 3: If energy V₀ < E < V_max the particle is bounded between turning point a and b and shape of orbit is elliptical in region.
Radius r = r₀ of circular orbit is also identified as stable equilibrium point, so . Then new orbit is identified as elliptical orbit. The angular frequency in new elliptical orbit is identified as oscillatory motion so for small oscillation the angular frequency is
Another turning point is r = c , region r > c is identified as classical region the particle is unbounded so nature of orbit may be either parabolic or hyperbolic.

Case 4: E = V_max  hence r = b is unstable equilibrium point. , the orbit can be unstable circular of radius b .there is another possible shape as it is elliptical between turning point a and  unstable equilibrium point b .unbounded  orbit is also possible for r > b
Case 5: E > V_max  there is only one turning point r = a and region r > a is identified as classical region the particle is unbounded so nature of orbit may be either parabolic or hyperbolic

Differential Equation of Orbit

From equation of motion in radial part,
where J = mr



Example 1: Consider that the motion of a particle of mass m in the potential field V(r) = kr²/2 If l is angular momentum,
(a) What is effective potential (V_eff) of the system. plot V_eff vs r
(b) Find value of energy such that motion is circular in nature.
(c) If particle is slightly disturbed from circular orbit such that its angular remain constant. What will nature of new orbit? Find the angular frequency of new orbit in term of m, l, k.

(a)  
 
(b) + kr = 0 at r = r₀ so r₀ =
For circular motion, mω₀²r₀ = k_r0, where r₀ is radius of circle ω₀ =
Total energy, E =
(c) orbit is elliptical in nature
 

Example 2: A particle of mass m moves under the influence of an attractive central force f (r).
(a) What is condition that orbit is circular in nature if J is the angular momentum of particle
(b) If force is in form of f(r) = -k/rⁿ determine the maximum value of n for which the circular  orbit can be stable.

(a) if v_eff, for circular stable orbit
(b) f(r) = -k/rⁿ, for circular motion
It is given ∂V/∂r = - f(r) if f(r) = -k/rⁿ ⇒ ∂V/∂r = k/rⁿ

Example 3: A particle of mass m and angular momentum l is moving under the action of a central force f(r) along a circular path of radius a as shown in the figure. The force centre O lies on the orbit.
(a) Given the orbit equation in a central field motion.

Determine the form of force in terms of l ,m, a and r.
(b) Calculate the total energy of the particle assuming that the potential energy V(r) ⇢ 0 as r⇢ ∞.

(a) from the figure, 2cosθ,



(b)

Hence,