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Kepler’s Potential

The Kepler’s potential is given by V (r) = -k/r with k > 0  which is central potential. If Particle of mass m and angular momentum J is interact with potential, then effective potential is given Veffective =Kepler’s System | Mechanics & General Properties of Matter - Physics
If one will plot Veffective against r
Analysis of  Veffective Kepler’s System | Mechanics & General Properties of Matter - Physics
Kepler’s System | Mechanics & General Properties of Matter - Physics the value of Kepler’s System | Mechanics & General Properties of Matter - Physicswhich is minima of the  Veffective.

The Analysis of  Veffective for Kepler’s law
Case: 1 for r = r0 = J2/mk total energy E0 = - mk2/2j2 he motion of path is elliptical
Case: 2  E0 <E< 0  the particle is bounded between turning point a <r< b and the motion is equivalent to small oscillation about stable point r = r0 and shape of orbit is elliptical in nature.
Case: 3  E = 0 the particle is unbounded where  r = c is turning point.
Case: 4  E > 0 the particle is unbounded where  r = d is turning point.

Equation of orbit for Kepler’s

Equation of motion is given by Kepler’s System | Mechanics & General Properties of Matter - Physics
Equation of orbit is given by Kepler’s System | Mechanics & General Properties of Matter - Physics
Kepler’s System | Mechanics & General Properties of Matter - PhysicsKepler’s System | Mechanics & General Properties of Matter - Physics
Kepler’s System | Mechanics & General Properties of Matter - Physics
The equation reduce to d2y/dθ2 + y = 0
The solution of equation reduce to y = A cosθ u - km/J2 = A cos θ ⇒ u = km/J2 + A cos θ
Kepler’s System | Mechanics & General Properties of Matter - Physics
Put l km J2/km = l and e = AJ2/km, then equation reduce to l/r = 1 + e cos θ, which is equation of conics where l is latus rectum and e is eccentricity.

In a central force potential which is interacting with potential V (r) = -k/r can be any conics 

section depending on eccentricity e.

Relationship between Energy and Eccentricity

For central potential V(r) = -k/r, the solution of orbit is l/r = 1 + e cos θ with l = j2/km
The energy is given by E = 1/2Kepler’s System | Mechanics & General Properties of Matter - Physics
The solution of orbit is 1/r = 1 +e cos θ with l = j2/km
Kepler’s System | Mechanics & General Properties of Matter - PhysicsSo Kepler’s System | Mechanics & General Properties of Matter - Physics
After putting the value of l/r = 1 + e cos θ and Kepler’s System | Mechanics & General Properties of Matter - Physicswith l = j2/km in equation of energy, one will get e =Kepler’s System | Mechanics & General Properties of Matter - Physics
the condition on energy for possible nature of orbit for potential
E > 0 e > 1 Hyperbola
E = 0 e = 1 Parabols
E < 0 e< 1 Ellipse
E = -mk2/2J2 e = 0 Circle

Two Body Problem

Reduction of two body central force problem to the equivalent one body problem.
A system of two particles of mass m1 and m2 whose instantaneous position vectors of inertial frame with origin O are r1 and r2 respectively.

Vector m2 relative to m1 is Kepler’s System | Mechanics & General Properties of Matter - Physics
Kepler’s System | Mechanics & General Properties of Matter - Physics

Central Force Motion as a One Body Problem

Consider an isolated system consisting of two particles interacting under a central force f(r). The masses of the particles are m1 and m2 and their position vectors are r1 and r2. We have
r = r1 - r2 ⇒ r = |r| = |r1 - r2|
The equations of motion are
Kepler’s System | Mechanics & General Properties of Matter - Physics
Kepler’s System | Mechanics & General Properties of Matter - PhysicsThe force is attractive for f (r) < 0 and repulsive for f (r) > 0 .Above equation are coupled Equations are coupled together by r ; the behavior of r1 and r2 depends on r = r1 - r2. To find the equation of motion for r we divide equation  by m1 (1) and equation  by m2 (2) and subtract.
This gives
Kepler’s System | Mechanics & General Properties of Matter - Physics
Where Kepler’s System | Mechanics & General Properties of Matter - Physics is identified as reduce mass and equation of motion is written in form of reduced mass asKepler’s System | Mechanics & General Properties of Matter - Physicssystem can be shown as given in figure.

Equation is identical to the equation of the motion for a particle of mass µ acted on by a force Kepler’s System | Mechanics & General Properties of Matter - Physics; no trace of the two particle problem remains.
The two particle problem has been transformed to a one particle problem.

Transformation of two body on center of mass reference frame and calculation of energy

We shall show that the problem is easier to handle if we replace rand r2 by r = r1 - rand the center of mass vectorKepler’s System | Mechanics & General Properties of Matter - PhysicsThe equation of motion for R is trivial since there are no external forces. The equation for r turns out to be like the equation of motion of a single particle and has a straight forward solution.
The equation of motion for R is Kepler’s System | Mechanics & General Properties of Matter - Physics, which has the simple solution R = R+ vt
The constant vectors R0 and v depend on the choice of coordinate system and the initial conditions. If we are clever enough to take the origin at the center of mass, R0 = 0 and v = 0.
Kepler’s System | Mechanics & General Properties of Matter - PhysicsKepler’s System | Mechanics & General Properties of Matter - Physics
Solving for r1 and r2 gives
Kepler’s System | Mechanics & General Properties of Matter - Physics
For a simple trick lets fixed center of mass at origin ie for put R = 0
Kepler’s System | Mechanics & General Properties of Matter - Physics
Total energy of system is given by
Kepler’s System | Mechanics & General Properties of Matter - Physics
Put value of Kepler’s System | Mechanics & General Properties of Matter - Physics
The energy of system is
Kepler’s System | Mechanics & General Properties of Matter - Physics 
where μ = Kepler’s System | Mechanics & General Properties of Matter - Physicsis identified as reduced mass.
Hence V (r) is central potential so motion is then motion is confided in the plane    
Kepler’s System | Mechanics & General Properties of Matter - Physics   
Kepler’s System | Mechanics & General Properties of Matter - Physics

Kepler’s Law

Kepler discuss the orbital motion of the sun and Earth system under the potential, V(r) = -k/r where k = Gmsme, it is given ms and me is mass of Sun and Earth. Although Kepler discuss sun and earth system but method can be used for any system which is interacting with potential V(r) = - k/r
The reduce mass for sun and earth system is µ = Kepler’s System | Mechanics & General Properties of Matter - Physics
Let us assume mass of earth me = m

Kepler’s First Law

Every planet (earth) moves in an elliptical orbit around the sun, the sun is being at one of the foci. Where sun and earth interact each other with potential V(r) = k/r We solve equation of motion in center of mass reference frame with reduce mass μ = me = m

Equation of Motion

Now, we discuss the case specially of elliptical orbit as Kepler discuss for planetary motion.
Total energy, E = 1/2 mr2  + J2/2m2 - k/r where Veffective = J2/2mr2 - k/r with constant angular momentum J.
If one will plot Veffective vs r , it is clear that for negative energy the orbit is elliptical which is shown in figure.

Earth is orbiting in elliptical path with sun as focus as shown in figure.
Kepler’s System | Mechanics & General Properties of Matter - PhysicsLet equation of this ellipse is Kepler’s System | Mechanics & General Properties of Matter - Physics
where b = Kepler’s System | Mechanics & General Properties of Matter - Physics
Minimum value of r is (a - ae) and maximum value of r is (a + ae), rmax + rmin = 2a.
From plot of effective potential, it is identified rmax and rmin is the turning point, so at these points radial velocity is zero.
Kepler’s System | Mechanics & General Properties of Matter - Physics given equation is quadratic in term of r,  for their root at rmax and rmin. Using theory of quadratic equation sum of root, rmax + rmin = -2mk/2mE
⇒ E = -k/2α, which is negative.

Kepler’s Second Law

Equal Area will swept in equal time or Areal velocity is constant.
dA/dt = J/2m = (which is derived earlier)

Kepler’s Third Law

The square of time period (T) of revolution in elliptical orbit is proportional to cube of semi major axis a i.e., T2 ∝ a3
Kepler’s System | Mechanics & General Properties of Matter - Physics(πab is the area of ellipse)
Kepler’s System | Mechanics & General Properties of Matter - Physics
Kepler’s System | Mechanics & General Properties of Matter - Physics
If k = Gmsm then ⇒ T= Kepler’s System | Mechanics & General Properties of Matter - Physicswhere ms is mass of the Sun.

Example 4: Given a classical model of tritium atom with nucleus of charge +1 and a single electron in a circular orbit of radius r0, suddenly the nucleus emits a negatron and changes to charge +2 (the emitted negatron escapes rapidly and we can forget about it) the orbit suddenly has a new situation.
(a) Find the ratio of the electron’s energy after to before the emission of the negatron
(b) Describe qualitatively the new orbit
(c) Find the distance of closest and the farthest approach for the new orbits in units of r0

(a) As the negatron leaves the system rapidly, we can assume that its leaving has no effect on the position and kinetic energy of the orbiting electron.
From the force relation for the electron, Kepler’s System | Mechanics & General Properties of Matter - Physics
And we find its kinetic energy Kepler’s System | Mechanics & General Properties of Matter - Physics
And its total mechanical energy,
Kepler’s System | Mechanics & General Properties of Matter - Physics
Before the emission of the negatron. After the emission the kinetic energy of the electron is still Kepler’s System | Mechanics & General Properties of Matter - Physics while its potential energy suddenly changes toKepler’s System | Mechanics & General Properties of Matter - Physics
Thus after the emission the total mechanical energy of the orbiting electron is
Kepler’s System | Mechanics & General Properties of Matter - Physics
In other words, the total energy of the orbiting electron after the emission is three times as large as that before the emission.
(b) As E2 = Kepler’s System | Mechanics & General Properties of Matter - Physicsthe condition for circular motion is no longer satisfied and the new orbit is an ellipse.
(c) Conservation of energy gives Kepler’s System | Mechanics & General Properties of Matter - Physics
At positions where the orbiting electron is at the distance of closest or farthest approach to the atom, we have Kepler’s System | Mechanics & General Properties of Matter - Physics 
for which Kepler’s System | Mechanics & General Properties of Matter - Physics
Then with, j2 = Kepler’s System | Mechanics & General Properties of Matter - Physics
The above becomes 3r2 - 4r0r+ r02 = 0 with solutions r = r0/3 = r = r0
Hence the distances of closest and farthest approach in the new orbit are respectively
rmin = 1/3, rmax = 1


Example 5: A satellite of mass m = 2000 kg is in elliptical orbit about earth. At perigee it has an altitude of 1,100 km and at apogee it has altitude 4,100 km. assume radius of the earth is Re = 6, 400 km. it is given GmMe = 8 × 1017 j m
Kepler’s System | Mechanics & General Properties of Matter - Physics(a) What is major axis of the orbit?
(b) What is eccentricity of the orbit?
(c) What is angular momentum of the satellite?
(d) How much energy is needed to fix satellite in to orbit from surface of earth?

rmax = 4100 + 6400 = 10500 km
rmin= 1100 + 6400 = 7500km
(a) rmax + rmin = 2α ⇒ 18000 = 2a ⇒ a = 9000km
(b) Kepler’s System | Mechanics & General Properties of Matter - Physics
It is given k = 8 × 1017 J.m
(c) Kepler’s System | Mechanics & General Properties of Matter - Physics
Kepler’s System | Mechanics & General Properties of Matter - Physics
(d) When satellite is at surface of earth,
R = 6400 Km
Kepler’s System | Mechanics & General Properties of Matter - Physics


Example 6: For circular and parabolic orbits in an attractive 1/r potential having the same angular momentum, show that perihelion distance of the parabola is one-half the radius of the circle.

For Kepler’s problem, l/r = 1 + e cos θ, for circular orbit e = 0 ⇒ l/rc = 1
And for parabola e = 1, l/rp = 1 + cos θ, ris minimum when cosθ is maximum
Kepler’s System | Mechanics & General Properties of Matter - Physics


Example 7: A planet of mass m moves in the inverse square central force field of the Sun of mass M . If the semi-major and semi-minor axes of the orbit are a and b , respectively, the find total energy of the planet by assuming sun is at center of ellipse.

Assume Sun is at the centre of elliptical orbit.
Kepler’s System | Mechanics & General Properties of Matter - PhysicsConservation of energy Kepler’s System | Mechanics & General Properties of Matter - Physics
Conservation of momentum L =mv1a= mv2b
Kepler’s System | Mechanics & General Properties of Matter - Physics
Kepler’s System | Mechanics & General Properties of Matter - Physics

The document Kepler’s System | Mechanics & General Properties of Matter - Physics is a part of the Physics Course Mechanics & General Properties of Matter.
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FAQs on Kepler’s System - Mechanics & General Properties of Matter - Physics

1. What is Kepler's potential?
Ans. Kepler's potential refers to the gravitational potential energy associated with a two-body system, such as a planet orbiting a star. It is named after Johannes Kepler, a German astronomer who formulated the laws of planetary motion. The potential describes the attractive force between two objects and helps to understand the dynamics of their motion.
2. What is the significance of Kepler's potential in the Two Body Problem?
Ans. Kepler's potential plays a crucial role in solving the Two Body Problem, which involves determining the motion of two objects under their mutual gravitational attraction. It provides a mathematical framework to calculate the potential energy of the system, which is related to the positions and masses of the two bodies. By analyzing the potential, scientists can derive the equations of motion and accurately predict the orbits and trajectories of celestial bodies.
3. How does Kepler's System relate to Kepler's potential?
Ans. Kepler's System refers to the set of laws formulated by Johannes Kepler that describe the motion of planets around the Sun. These laws are derived from the observations of planetary motion and are based on the concept of Kepler's potential. The laws state that planets move in elliptical orbits with the Sun at one of the foci and that the area swept by the planet's radius vector is proportional to the time taken. Thus, Kepler's System is intimately connected to the understanding of Kepler's potential.
4. Can Kepler's potential be applied to other celestial objects apart from planets?
Ans. Yes, Kepler's potential can be applied to other celestial objects apart from planets. It is a general concept that applies to any two-body system where gravity is the dominant force. For example, it can be used to study the motion of moons around a planet, satellites around a star, or even binary star systems. By analyzing the potential energy, scientists can determine the gravitational interactions and predict the behavior of these objects in space.
5. How does Kepler's potential contribute to understanding the stability of planetary systems?
Ans. Kepler's potential plays a crucial role in understanding the stability of planetary systems. By analyzing the potential energy of a two-body system, scientists can determine the equilibrium points, where the gravitational forces between the objects are balanced. These equilibrium points, such as Lagrange points, provide insights into the stability of planetary orbits and the presence of stable regions in the system. Understanding the stability of planetary systems is essential for studying the long-term behavior and evolution of celestial bodies.
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