Kepler’s System | Mechanics & General Properties of Matter - Physics PDF Download

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 V_effective =
If one will plot V_effective against r
Analysis of  V_effective 
 the value of which is minima of the  V_effective.

The Analysis of  V_effective 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
Equation of orbit is given by


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

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 = 1/2
The solution of orbit is 1/r = 1 +e cos θ with l = j²/km
So
After putting the value of l/r = 1 + e cos θ and with l = j²/km in equation of energy, one will get e =
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

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

The 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

Where  is identified as reduce mass and equation of motion is written in form of reduced mass assystem 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 ; 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 r₁ and r₂ by r = r_{1 -} r₂ and the center of mass vectorThe 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 , which has the simple solution R = R₀ + 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.

Solving for r₁ and r₂ gives

For a simple trick lets fixed center of mass at origin ie for put R = 0

Total energy of system is given by

Put value of
The energy of system is
 
where μ = is identified as reduced mass.
Hence V (r) is central potential so motion is then motion is confided in the plane    
   

Kepler’s Law

Kepler discuss the orbital motion of the sun and Earth system under the potential, V(r) = -k/r where k = Gm_sm_e, 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 =
Let us assume mass of earth m_e = 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 V_effective = J2/2mr2 - k/r with constant angular momentum J.
If one will plot V_effective 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.
Let equation of this ellipse is
where b =
Minimum value of r is (a - ae) and maximum value of r is (a + ae), r_max + r_min = 2a.
From plot of effective potential, it is identified r_max and r_min is the turning point, so at these points radial velocity is zero.
 given equation is quadratic in term of r,  for their root at r_max and r_min. Using theory of quadratic equation sum of root, rmax + r_min = _{-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
(πab is the area of ellipse)


If k = Gm_sm then ⇒ T² = where m_s 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,
And we find its kinetic energy
And its total mechanical energy,

Before the emission of the negatron. After the emission the kinetic energy of the electron is still  while its potential energy suddenly changes to
Thus after the emission the total mechanical energy of the orbiting electron is

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 = the condition for circular motion is no longer satisfied and the new orbit is an ellipse.
(c) Conservation of energy gives
At positions where the orbiting electron is at the distance of closest or farthest approach to the atom, we have  
for which
Then with, j² =
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
r_min = 1/3, r_max = 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 R_e = 6, 400 km. it is given GmMe = 8 × 10¹⁷ j m
(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?

r_max = 4100 + 6400 = 10500 km
r_min= 1100 + 6400 = 7500km
(a) r_max + r_min = 2α ⇒ 18000 = 2a ⇒ a = 9000km
(b)
It is given k = 8 × 10¹⁷ J.m
(c)

(d) When satellite is at surface of earth,
R = 6400 Km

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, r_p is minimum when cosθ is maximum

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.
Conservation of energy
Conservation of momentum L =mv₁a= mv₂b