Chapter 4 Lecture 2 Two body central Problem

Akhlaq Hussain 1 4.3 Equation of for a body under the of and First Integrals

Consider a conservative, where force can be drivable from potential “V(r)”. The problem has spherical symmetry & angular (풍 = 풓 × 푷) conserved. 1 Lagrangian of the system 퐿 = 푇 − 푉 = 휇 푟ሶ 2 + 푟2휃ሶ 2 − 푉 (4.3.1) 2 푟 푑 휕퐿 휕퐿 Using Lagrange’s equation − = 0 푑푡 휕휃ሶ 휕휃 휕퐿 휕퐿 And = 푃 = 휇푟2휃ሶ, and = 0 휕휃ሶ 휃 휕휃

푑 휕퐿 휕퐿 푑 − = 휇푟2휃ሶ = 0 (4.3.2) 푑푡 휕휃ሶ 휕휃 푑푡

2 ⇒ 휇푟 휃ሶ = 푃휃 = 풍 = 푐표푛푠푡푎푛푡 (4.3.3)

Eq. (4.3.3) is first integral of motion 2 4.3 Equation of Motion for a body under the action of central force and First Integrals

푑퐴 푑 1 1 푙 = 푟2휃ሶ = 0⇒퐴 = 푟2휃ሶ = = 푐표푛푠푡푎푛푡 (4.3.4) 푑푡 푑푡 2 2 2휇 Thus, the areal is constant in . (Kepler’s Law)

• Areal velocity conservation is a general property of central force motion • It is not restricted to the inverse-square law force involved in planetary motion. Lagrange’s equation for radial part 푑 휕퐿 휕퐿 푑 휕푉 − = 휇푟ሶ − 휇푟휃ሶ 2 + = 0 푑푡 휕푟ሶ 휕푟 푑푡 휕푟 휕푉 휇푟ሷ − 휇푟휃ሶ 2 + = 0 (4.3.5) 휕푟 휕푉 l Since 푓 = − & 휃ሶ = {from(4.3.3)}, Therefore, Eq. (4.3.5) (푟) 휕푟 휇푟2 퐿2 Since ⇒ 휇푟ሷ − 3 = 푓(푟) (4.3.7) 휇푟 3 4.3 Equation of Motion for a body under the action of central force and First Integrals

l2 휕푉 휕 l2 ⇒ 휇푟ሷ = − = − + 푉 휇푟3 휕푟 휕푟 2휇푟2 휕 l2 Multiplying Both sides with “푟ሶ” ⇒ 휇푟ሶ푟ሷ = −푟ሶ + 푉 휕푟 2휇푟2 푑 1 푑푟 휕 l2 푑 l2 ⇒ 휇푟ሶ 2 = − + 푉 = − + 푉 푑푡 2 푑푡 휕푟 2휇푟2 푑푡 2휇푟2 푑 1 l ⇒ 휇푟ሶ 2 + + 푉 = 0 푑푡 2 2휇푟2 1 l2 ⇒ 휇푟ሶ 2 + + 푉 = 퐶표푛푠푡푎푛푡 (4.3.8) 2 2휇푟2 1 1 l2 퐸 = 휇푟ሶ 2 + + 푉 (4.3.10) 2 2 휇푟2 푟 From eq. (4.3.8) and Eq. (4.3.10), total of a body under the action of central force is constant. 4 4.4 First Integral

The , of the system is l = 휇푟2휃ሶ

l 푑휃 l ⇒ 휃ሶ = ⇒ = 휇푟2 푑푡 휇푟2 l ⇒ 푑휃 = 푑푡 휇푟2 휃 푡 l (Integrating above equation ⇒ ׬ 푑휃 = ׬ 푑푡 (4.4.1 휃표 0 휇푟2 Now the total energy of a body moving under central force is given by

1 1 l2 퐸 = 푇 + 푉 = 휇푟ሶ 2 + + 푉 (4.4.3) 2 2 휇푟2 푟

2 l2 ⇒ 푟ሶ = 퐸 − − 푉 푟 (4.4.4) 휇 2휇푟2 5 4.4 First Integral

푑푟 2 l2 ⇒ = 퐸 − − 푉 푑푡 휇 휇푟2 푟 푟 푑푟 (푡 = ׬ (4.4.5 ⇒ 푟표 2 l2 퐸− −푉 휇 2휇푟2 푟

Eq. (4.4.1) & Eq. (4.4.5) are known as first integral for the motion in central force field. where l , E, 휃표 and 푟표must be known initially.

Eq. (4.4.1) & Eq. (4.4.5) gives “푟” and “휃” in terms of t. We are often interested to find “휃” in terms of “푟” which will determine the shape of the of the body.

6 4.4 First Integral

푑휃 l 푑휃 푑푟 l Since = ⇒ = 푑푡 휇푟2 푑푡 푑푟 휇푟2 l ⇒ 푑휃 = 푑푟 (4.4.6) 휇푟2푟ሶ From Eq. (4.4.4) we know that

2 l2 푟ሶ = 퐸 − − 푉 (4.4.7) 휇 휇푟2 푟 l ⇒ 푑휃 = 푑푟 2 l2 휇푟2 퐸− −푉 휇 휇푟2 푟

lൗ 푟 푟2 (휃 = 휃표 + ׬푟 푑푟 (4.4.8 ⇒ 표 l2 2휇 퐸− −푉 2휇푟2 푟

Eq. 47 gives “휃” in terms of “푟” which determine the shape of the orbit of the body 7 under the action of central force field. 4.5 General Features of Motion Under Central Force

휇 푟ሷ − 푟휃ሶ 2 = 퐹 ቐ 푟 (4.5.1) 휇 푟휃ሷ + 2푟ሶ휃ሶ = 퐹휃

The tangential component “퐹휃” is zero because the force is radial 2 휇 푟ሷ − 푟휃ሶ = 퐹푟 2 ⇒ 휇푟ሷ = 퐹푟 + 휇푟휃ሶ l2 ⇒ 휇푟ሷ = 퐹 + (4.5.2) 푟 휇푟3

퐿2 is known as . It is a pseudo or false force since it does not arise from the 휇푟3 interaction between the particles in the orbit. It appears due to accelerated motion of the body. Since l2 = 휇2푟4휃ሶ 2 l2 휇 푟2휃ሶ 2 휇푣2 푚푣2 ⇒ = 휇푟휃ሶ 2 = = or 휇푟3 푟 푟 푟 8 4.5 General Features of Motion Under Central Force

“휇푟ሷ” is the effective force responsible for the motion and can be derived from potential “Veff” 푑푉 휇푟ሷ = − 푒푓푓 (4.5.3) 푑푟 Therefore Eq. (4.5.2) can be written as

2 2 푑푉푒푓푓 l 푑푉 l 퐹 + ⇒ 푉 = − ׬ − + 푑푟 = − 푑푟 푟 휇푟3 푒푓푓 푑푟 휇푟3 l2 ⇒ 푉 = 푉 + (4.5.4) 푒푓푓 2휇푟2

For an inverse square law (gravitational or electrostatic force) 푘 푘 퐹 = − ⇒ 푉 = − 푟 푟2 푟 푘 l2 Therefore, 푉 = − + (4.5.5) 푒푓푓 푟 2휇푟2 Note that the centrifugal potential reduces the effect of the inverse square law 9 4.5 General Features of Motion Under Central Force

Not the total energy of the system is 1 퐸 = 휇푟ሶ 2 + 푉 2 푒푓푓

2 ⇒ 푟ሶ = 퐸 − 푉 (4.5.6) 휇 푒푓푓

The centrifugal part gives a repulsive potential while the inverse square law part gives an attractive potential.

Centrifugal part decreases much faster with “r” as compared to the inverse square attractive part. The combine potential is given as the Veff which decrease sharply from positive value to negative and then increase with r. The Veff approaches to zero value at infinite value of r.

10 4.6 Motion in arbitrary potential Field

Let an arbitrary potential Veff which may or may not be same as the real problem and it might appear in different problems. The Energy and potential curves intersect at “푟1”, “푟2” and “푟3”.

퐸 = 푉푒푓푓 (4.6.1) 1 And 휇풓ሶ 2 = 0 & 푟ሶ = 0 2 The curve can be divided into three regions.

Region for 풓 < 풓ퟏ

퐸 < 푉푒푓푓 (4.6.2) 1 & 푇 = 휇풓ሶ 2 < 0 2 & velocity has imaginary value. Hence motion in this region is not possible. 11 4.6 Motion in arbitrary potential Field

Region for 풓ퟏ < 풓 < 풓ퟐ

In this region 퐸 > 푉푒푓푓

for 푟 < 푟1and 푟2 < 푟, 1 The kinetic energy T = 휇풓ሶ 2< 0 2 Which is not possible therefore the body will back on 푟1and 푟2.

Region for 풓ퟐ < 풓 < 풓ퟑ

In this region 퐸 < 푉푒푓푓 1 & 푇 = 휇풓ሶ 2 < 0 Therefore, the motion 2 in this region is not possible. 12 4.6 Motion in arbitrary potential Field

Region for 풓 > 풓ퟑ

Turning point is 푟 = 푟3.

The particle approaches to 푟3 and rebounded.

퐸 = 푇 + 푉푒푓푓 = 0 ⇒ 푇 = −푉푒푓푓

2 푟ሶ = −푉 (4.6.3) 휇 푒푓푓

푟ሶ = escape velocity; the initial velocity required to escape from the potential field 푉푒푓푓. The nature of motion of the particle discussed earlier with help of arbitrary potential will help to understand the nature of orbit. 13 4.7 Motion in Inverse Square Law Force Field

푘 퐹 = (4.7.1) 푟 푟2 푘 ⇒ 푉 = (4.7.2) 푟 푟

Therefore, the effective potential 푉푒푓푓 is given by 푘 l2 푉 = + (4.7.3) 푒푓푓 푟 2휇푟2 The value of “k” depends on the nature of physical problem. For example, i) gravitational force between two spherical bodies of m1 and m2 푘 = −퐺푚1푚2 (4.7.4) ii) Electrostatic force 푞 푞 푘 = 1 2 (4.7.5) 4휋휖표 The nature of the orbit depends on sign of “k”. If k > 0 ⇒repulsive & for k < 0 ⇒ attractive. 14 4.7 Motion in Inverse Square Law Force Field

If effective potential Veff is plotted against “r” for different values of “k” and “L” following curves are obtained. Case I 풌 > ퟎ, l > ퟎ Case II 풌 > ퟎ, l = ퟎ Case III 풌 = ퟎ, l > ퟎ Case IV 풌 < ퟎ, l > ퟎ Case V 풌 < ퟎ, l = ퟎ

These curves can be very helpful in understanding the nature of the orbit.

A body with total energy E > Veff approaching to the centre of force from infinite distance. The particle will be 15 deflected as given in figure. 4.7 Motion in Inverse Square Law Force Field

(1)For E1 at r = r1 푘 l2 퐸 = 푉 = − + 1 푒푓푓 푟 2휇푟2

Turning point at 푟 = 푟1. Motion represents scattering, where body is not bound to the centre and deflected away.

(ii) For E2 = 0 ′ Possible roots are 푟 = 푟1 and 푟 = ∞. The particle moves away & fall continuously.

(iii) For E3 < 0

Two roots 푟 = 푟2 and 푟 = 푟3 of equation are real and distinct. 16 4.7 Motion in Inverse Square Law Force Field

(iv) For E4 = 푽풆풇풇, which is tangent of the curve. 푑푉 Therefore 푒푓푓 = 0 푑푟 푑푉 푙2 ⇒ − = 0 푑푟 휇푟3 푑푉 푙2 ⇒ 퐹 = − = − = −휇푟휃ሶ 2 푟 푑푟 휇푟3 휇푟2휃ሶ 2 휇푣2 ⇒ 퐹 = − = − 푟 푟 푟

Thus 퐹푟 is equal to the centrifugal force required to maintain of the body around the centre of the force. Thus 퐹푟 is that maintain the orbit. 17 풅풖 ퟐ 4.9 Show That: a) 풗ퟐ = 풓ሶ ퟐ + 풓ퟐ휽ሶ ퟐ = 풉ퟐ + 풖ퟐ 풅휽 b) Using results from part “a” also prove that the conservation of energy equation will be

풅풖 ퟐ ퟐ 푬−푽 ퟏ + 풖ퟐ = if 풖 = 풅휽 흁풉ퟐ 풓

Solution: Let us consider a particle of mass “휇” and vector “풓”. 1 1 Since 푢 = ⇒ 푟 = 푟 푢 푑푟 1 푑푢 1 푑푢 푑휃 = − − 푑푡 푢2 푑푡 푢2 푑휃 푑푡 푑푢 푑푢 ⇒ 푟ሶ = −푟2휃ሶ ⇒ 푟ሶ = −ℎ 푑휃 푑휃 Therefore, 푣2 = 푟ሶ 2 + 푟2휃ሶ 2

푑푢 2 1 푑푢 2 ⇒ 푣2 = −ℎ + ℎ푢2 2= ℎ + ℎ2푢2 푑휃 푢2 푑휃

푑푢 2 ⇒ 푣2 = ℎ2 + 푢2 (4.9.1) 푑휃

18 Since 퐸 = 푇 + 푉 ⇒ 푇 = 퐸 − 푉 1 ⇒ 휇푣2 = 퐸 − 푉 2

1 푑푢 2 ⇒ 휇ℎ2 + 푢2 = 퐸 − 푉 2 푑휃

푑푢 2 2 퐸−푉 ⇒ + 푢2 = (4.9.2) 푑휃 휇ℎ2 Eq. (4.9.1) and Eq. (4.9.2) are as desired.

19 Problem (Page 293, Classical by Marion) Find the force law for a central force field that allows a particle to move in a logarithmic spiral orbit given by 휶휽 풓 = 풌풆 , where “k” and “휶” are constants. Also find value of 휽 풕 and 풓 풕 . Also find Energy of the orbit. Solution. Since we have verified that

휇푓 1 푑2푢 + 푢 = − 푢 푑휃2 푙2푢2

푑2푢 휇푟2푓 + 푢 = − 푟 (1) 푑휃2 푙2 Now using 1 1 푟 = 푘푒훼휃 ⇒ = 푒−훼휃 푟 푘 Differentiating Twice with respect to θ

푑2 1 훼2 = 푒−훼휃 푑휃2 푟 푘 푑2푢 훼2 ⇒ = 푒−훼휃 = 훼2푢 (2) 푑휃2 푘 푑2푢 Putting value of u and in equation 1 푑휃2 푑2푢 휇푟2푓 + 푢 = − 푟 푑휃2 푙2 20 휇푟2푓 ⇒ 훼2푢 + 푢 = − 푟 푙2 푙2 ⇒ 푓 = − 훼2 + 1 (3) 푟 휇푟3 Eq. 3 represents the force responsible for motion. Now the central potential responsible for the motion of the particle will be

푙2 (푉 = − ׬ 푓 푑푟 = − 훼2 + 1 (4 푟 2휇푟2 Total energy of the system is

1 푙2 퐸 = 푇 + 푉 = 휇푟ሶ 2 + + 푉 (5) 2 2휇푟2 Now

푑푟 푑휃 푟ሶ = 푑휃 푑푡 푑푟 푑푟 푙 푟ሶ = 휃ሶ = 푑휃 푑휃 휇푟2 푙 푙 푟ሶ = 푘훼푒훼휃 = 푟훼 휇푟2 휇푟2 푙 푟ሶ = 훼 (6) 휇푟 21 1 푙2 Now 퐸 = 푇 + 푉 = 휇푟ሶ 2 + + 푉 2 2휇푟2

1 푙훼 2 푙2 푙2 ⇒ 퐸 = 휇 + − 훼2 + 1 2 휇푟 2휇푟2 2휇푟2

푙2 푙2 ⇒ 퐸 = 훼2 + 1 − 훼2 + 1 = 0 (7) 2휇푟2 2휇푟2 Eq. 7 gives the total energy of the system. Zero value of the system represent a bound system.

Now we will determine of 휃 푡 and 푟 푡 푙 푑휃 푙 Since 휃ሶ = ⇒ = 휇푟 푑푡 휇푟 푑휃 푙 푙 = ⇒푒2훼휃푑휃 = 푑푡 푑푡 휇푘2푒2훼휃 휇푘2

푒2훼휃 푙푡 Integrating both sides we get = + 퐶 2훼 휇푘2 푙푡 1 푙푡 푒2훼휃 = 2훼 + 퐶 ⇒ 휃 = ln 2훼 + 퐶 (9) 휇푘2 푡 2훼 휇푘2 Now 푟 = 푘푒훼휃

푟 푟2 ⇒ = 푒훼휃 ⇒ = 푒2훼휃 푘 푘2

2 푟 푙푡 2 푙푡 ⇒ 2 = 2훼 2 + 퐶 ⇒ 푟 푡 = 2훼푘 2 + 퐶 (10) 푘 휇푘 휇푘 22