Chapter 4

Lecture 3

Two body central

Force Problem
Akhlaq Hussain 1
4.8 Prove that for central force field the equation of motion can be written as;
𝒅𝟐 𝒖 𝝁𝒇 𝒖 𝒅𝟐 𝒖 𝒇𝒖
+𝒖 = − 𝟐 𝟐 And +𝒖 = − 𝟐 𝟐
𝒅𝜽𝟐 𝒍 𝒖 𝒅𝜽𝟐 𝝁𝒉 𝒖

where 𝒉 = 𝒓𝟐 𝜽ሶ
Solution: Consider a particle of mass “𝜇” is at a distance “𝒓” from the origin. The
acceleration of the particle can have two components in the polar coordinates
𝑎𝑟 = 𝑟ሷ − 𝑟𝜃ሶ 2 (4.8.1)
𝑎𝜃 = 𝑟𝜃ሷ + 2𝑟ሶ 𝜃ሶ (4.8.2)
Since the central force is always directed along the radial vector “𝒓”. The radial force
is responsible for the motion. Therefore;
𝑓 𝑟 = 𝜇 𝑟ሷ − 𝑟𝜃ሶ 2 (4.8.3)
𝑓𝜃 = 0 (4.8.4)
1 1
Let us consider a function “𝑢” such that 𝑢 = ⇒ 𝑟 =
𝑟 𝑢
2
 𝑑𝑟 1 𝑑𝑢 1 𝑑𝑢 𝑑𝜃
=− =−
𝑑𝑡 𝑢2 𝑑𝑡 𝑢2 𝑑𝜃 𝑑𝑡
𝑑𝑢
⇒ 𝑟ሶ = −𝑟 2 𝜃ሶ
𝑑𝜃
𝑑𝑢
⇒ 𝑟ሶ = −ℎ (4.8.5)
𝑑𝜃
Differentiating above equation with respect to t
𝑑 𝑟ሶ 𝑑 𝑑𝑢 𝑑 𝑑𝑢
= −ℎ = −ℎ
𝑑𝑡 𝑑𝑡 𝑑𝜃 𝑑𝜃 𝑑𝑡
𝑑 𝑑𝑢 𝑑𝜃
⇒ 𝑟ሷ = −ℎ
𝑑𝜃 𝑑𝜃 𝑑𝑡
𝑑 𝑑𝑢 𝑑2 𝑢
⇒ 𝑟ሷ = −ℎ𝜃ሶ = −ℎ𝜃ሶ 2 (4.8.6)
𝑑𝜃 𝑑𝜃 𝑑𝜃
Since ℎ2 = 𝑟 2 𝜃ሶ ⇒ 𝜃ሶ = ℎൗ𝑟 2 𝑜𝑟 𝜃ሶ = ℎ𝑢2 , Putting in Eq. (4.8.6)
𝑑 𝑢 2
𝑟ሷ = −ℎ2 𝑢2 2 (4.8.7)
𝑑𝜃
3
 1
Putting 𝑟 = , 𝜃ሶ = ℎ𝑢2 and Eq. (4.8.7) in Eq. (4.8.3)
𝑢
2
𝑑 𝑢 1
𝑓 𝑟 = 𝜇 𝑟ሷ − 𝑟𝜃ሶ 2 ⇒ 𝑓𝑢 = 𝜇 2
−ℎ 𝑢 2 −𝜇 ℎ𝑢2 2
𝑑𝜃2 𝑢
𝑑 𝑢 2
⇒ 𝑓𝑢 = −𝜇ℎ2 𝑢2 2 − 𝜇ℎ2 𝑢3
𝑑𝜃

2 2 𝑑2 𝑢
⇒ 𝑓𝑢 = −𝜇ℎ 𝑢 +𝑢
𝑑𝜃 2
𝑓𝑢 𝑑2 𝑢
⇒ − 2 2 = +𝑢
𝜇ℎ 𝑢 𝑑𝜃 2
𝑑2 𝑢 𝑓𝑢
⇒ +𝑢 = − 2 2 (4.8.8)
𝑑𝜃 2 𝜇ℎ 𝑢
As required.
Since 𝑙 = 𝜇𝑟 2 𝜃ሶ = 𝜇ℎ putting in Eq. (4.8.8)
𝑑2 𝑢 𝜇𝑓 𝑢
+𝑢 = − 2 2 (4.8.9)
𝑑𝜃2 𝑙 𝑢
4
As desired.
 𝒅𝒖 𝟐
4.9 Show That: a) 𝒗𝟐 = 𝒓ሶ 𝟐 + 𝒓𝟐 𝜽ሶ 𝟐 = 𝒉𝟐 + 𝒖𝟐
𝒅𝜽

b) Using results from part “a” also prove that the conservation of energy equation will be
𝒅𝒖 𝟐 𝟐 𝑬−𝑽 𝟏
+ 𝒖𝟐 = if 𝒖 =
𝒅𝜽 𝝁𝒉𝟐 𝒓

Solution: Let us consider a particle of mass “𝜇” and position vector “𝒓”.
1 1
Since 𝑢 = ⇒ 𝑟 =
𝑟 𝑢
𝑑𝑟 1 𝑑𝑢 1 𝑑𝑢 𝑑𝜃
= − 2 − 2
𝑑𝑡 𝑢 𝑑𝑡 𝑢 𝑑𝜃 𝑑𝑡
𝑑𝑢 𝑑𝑢
⇒ 𝑟ሶ = −𝑟 2 𝜃ሶ ⇒ 𝑟ሶ = −ℎ
𝑑𝜃 𝑑𝜃
2 2 2 ሶ2
Therefore, 𝑣 = 𝑟ሶ + 𝑟 𝜃
𝑑𝑢 2 1 𝑑𝑢 2
⇒ 𝑣2 = −ℎ + 2 ℎ𝑢2 2 = ℎ2 + ℎ2 𝑢 2
𝑑𝜃 𝑢 𝑑𝜃

𝑑𝑢 2
⇒ 𝑣2 = ℎ2 + 𝑢2 (4.9.1)
𝑑𝜃 5
Since 𝐸 = 𝑇 + 𝑉 ⇒ 𝑇 = 𝐸 − 𝑉
1
⇒ 𝜇𝑣 2 = 𝐸 − 𝑉
2

1 𝑑𝑢 2
⇒ 𝜇ℎ2 + 𝑢2 = 𝐸 − 𝑉
2 𝑑𝜃

𝑑𝑢 2 2 𝐸−𝑉
⇒ + 𝑢2 = (4.9.2)
𝑑𝜃 𝜇ℎ2
Eq. (4.9.1) and Eq. (4.9.2) are as desired.

6
Problem (Page 293, Classical Mechanics by Marion)
Find the force law for a central force field that allows a particle to move in a
logarithmic spiral orbit given by 𝒓 = 𝒌𝒆𝜶𝜽 , where “k” and “𝜶” are constants.
Also find value of 𝜽 𝒕 and 𝒓 𝒕 . Also find Energy of the orbit.
Solution. Since we have verified that
𝑑2 𝑢 𝜇𝑓 𝑢 𝜇𝑟 2 𝑓 𝑟
+𝑢 = − 2 2 = − 2 (1)
𝑑𝜃2 𝑙 𝑢 𝑙
Now using
1 1 −𝛼𝜃
𝑟= 𝑘𝑒 𝛼𝜃 ⇒u = = 𝑒
𝑟 𝑘
Differentiating Twice with respect to θ
𝑑2 𝑢 𝛼2 −𝛼𝜃
= 𝑒 = 𝛼2𝑢 (2)
𝑑𝜃 2 𝑘

7
 𝑑2 𝑢
Putting value of u and in equation 1
𝑑𝜃 2
𝑑2 𝑢 𝜇𝑟 2 𝑓 𝑟
+𝑢 = − 2
𝑑𝜃2 𝑙

2 𝜇𝑟 2 𝑓 𝑟
⇒𝛼 𝑢+𝑢 = − 2
𝑙
𝑙2
⇒ 𝑓𝑟 = − 3 𝛼2 + 1 (3)
𝜇𝑟
Eq. 3 represents the force responsible for motion.
Now the central potential responsible for the motion of the particle will be
𝑙2
𝑉 = − ‫= 𝑟𝑑 𝑟 𝑓 ׬‬ − 2 𝛼2 + 1 (4)
2𝜇𝑟
1 𝑙2
Total energy of the system is 𝐸 =𝑇+𝑉 = 𝜇𝑟ሶ 2 + + 𝑉 (5)
2 2𝜇𝑟 2
𝑑𝑟 𝑑𝜃
Now 𝑟ሶ =
𝑑𝜃 𝑑𝑡 8
 𝑑𝑟 𝑑𝑟 𝑙
𝑟ሶ = 𝜃ሶ =
𝑑𝜃 𝑑𝜃 𝜇𝑟 2
𝑙 𝑙
𝑟ሶ = 𝑘𝛼𝑒 𝛼𝜃 = 𝑟𝛼
𝜇𝑟 2 𝜇𝑟 2
𝑙
𝑟ሶ = 𝛼 (6)
𝜇𝑟
1 𝑙2
Now 𝐸 =𝑇+𝑉 = 𝜇𝑟ሶ 2 + +𝑉
2 2𝜇𝑟 2

1 𝑙𝛼 2 𝑙2 𝑙2
⇒𝐸= 𝜇 + − 𝛼2 + 1
2 𝜇𝑟 2𝜇𝑟 2 2𝜇𝑟 2
𝑙2 𝑙2
⇒𝐸= 𝛼2 +1 − 𝛼2 + 1 = 0 (7)
2𝜇𝑟 2 2𝜇𝑟 2
Eq. 7 gives the total energy of the system. Zero value represent a bound system. Now
we will determine of 𝜃 𝑡 and 𝑟 𝑡
9
 𝑙
Since 𝜃ሶ =
𝜇𝑟 2
𝑑𝜃 𝑙 2𝛼𝜃 𝑑𝜃 𝑙
= ⇒𝑒 = 𝑑𝑡
𝑑𝑡 𝜇𝑘 2 𝑒 2𝛼𝜃 𝜇𝑘 2

𝑒 2𝛼𝜃 𝑙𝑡
Integrating both sides we get = +𝐶
2𝛼 𝜇𝑘 2

𝑙𝑡
𝑒 2𝛼𝜃 = 2𝛼 2
+𝐶
𝜇𝑘
1 𝑙𝑡
⇒ 𝜃 𝑡 = ln 2𝛼 +𝐶 (9)
2𝛼 𝜇𝑘 2

𝛼𝜃 𝑟2
Now 𝑟 = 𝑘𝑒 ⇒ = 𝑒 2𝛼𝜃
𝑘2

𝑟2 𝑙𝑡
⇒ 2 = 2𝛼 2
+𝐶
𝑘 𝜇𝑘
𝑙𝑡
⇒𝑟 𝑡 = 2𝛼𝑘 2 +𝐶 (10)
𝜇𝑘 2 10
 4.10 Equation of motion for a body under central force
(inverse square law force)
𝒍
𝒅𝟐 𝒖 𝝁𝒇 ൗ𝒓𝟐
Solve + 𝒖 = − 𝟐 𝒖𝟐 and 𝜽 = 𝜽𝒐 + ‫׬‬ 𝒅𝒓 and prove that the solution is the
𝒅𝜽𝟐 𝑳 𝒖
𝒍𝟐
𝟐𝝁 𝑬−𝑽 𝒓 −
𝟐𝝁𝒓𝟐

equation of conic. i.e. the motion under the inverse square law force represent motion on conic
path. Also discuss the possibilities of bound and unbound system.
Let us consider a particle of mass “𝜇” in under inverse square law force. The equation of motion can be
written as
𝑑2 𝒖 𝜇𝑓 𝑢
+𝒖 = − 2 2 (4.10.1)
𝑑𝜃2 𝑙 𝑢
Since the inverse square attractive force
𝑘
𝑓𝑟 = − 2 = −𝑘𝑢2
𝑟
𝑑2 𝒖 𝜇𝑘𝑢2
+𝒖=
𝑑𝜃2 𝑙 2 𝑢2
𝒅𝟐 𝒖 𝝁𝒌
+𝒖 = (4.10.2) 11
𝒅𝜽𝟐 𝑙𝟐
4.10 Equation of motion for a body under central force
(inverse square law force)
Starting with equation Eq. (4.10.2)
𝑑2 𝒖 𝜇𝑘 𝑑2 𝒖 𝜇𝑘
+𝒖= ⇒ + 𝒖− =0 (4.10.2)
𝑑𝜃2 𝐿2 𝑑𝜃 2 𝑙2
Consider a function
𝜇𝑘
𝑦=𝒖 − 2 (4.10.3)
𝑙
Differentiating above equation Twice
𝑑2 𝑦 𝑑2 𝒖
= (4.10.4)
𝑑𝜃 2 𝑑𝜃 2
Now
𝑑2 𝑦 𝑑2 𝑢 𝜇𝑘
+𝑦 = +𝒖− =0
𝑑𝜃 2 𝑑𝜃 2 𝑙2
𝑑2 𝑦
+𝑦 =0 (4.10.5) 12
𝑑𝜃 2
 4.10 Equation of motion for a body under central force
(inverse square law force)
It is a second order differential equation where “𝑦” is a function of “𝜃”
And 𝑦 = 𝐴𝑐𝑜𝑠 𝜃 − 𝜃𝑜 (4.10.6)
𝜇𝑘
𝑦=𝒖− = 𝐴𝑐𝑜𝑠 𝜃 − 𝜃𝑜
𝑙2
1 𝜇𝑘
= + 𝐴𝑐𝑜𝑠 𝜃 − 𝜃𝑜
𝑟 𝑙2
𝑙2
𝜇𝑘 𝐴𝑙 2
⇒ =1 + 𝑐𝑜𝑠 𝜃 − 𝜃𝑜 (4.10.7)
𝑟 𝜇𝑘
𝛼
Equation of conic. = 1 + 𝑒𝑐𝑜𝑠 𝜃 − 𝜃𝑜 (4.10.7)a
𝑟
𝑙2
where 𝛼 = Semi latus rectum.
𝜇𝑘
𝐴𝑙 2
and 𝑒 = is eccentricity which is defined as the measure
𝜇𝑘
of deviation from circular shape. 13
 4.10 Equation of motion for a body under central force
(inverse square law force)
Now consider the first integral for the motion under central force
𝑙ൗ
𝑟2
𝜃 = 𝜃𝑜 + ‫׬‬ 𝑑𝑟 (4.10.8)
𝑙2
2𝜇 𝐸− 𝑉 𝑟 −
2𝜇𝑟2

1 𝑘
Since 𝑑𝑢 = − 2 𝑑𝑟 & 𝑉= − = −𝑘𝑢 (4.10.9) & (4.10.10)
𝑟 𝑟
Putting Eq. (4.10.4) and Eq. (4.10.5) in Eq. (4.10.3)
𝑑𝑢
𝜃 = 𝜃𝑜 − ‫׬‬ (4.10.11)
2𝜇𝐸 2𝜇𝑘
2 + 2 𝑢 − 𝑢2
𝑙 𝑙

2𝜇𝐸 2𝜇𝑘
Let = 𝑎, = 𝑏 and −1 = 𝑐
𝑙2 𝑙2
𝑑𝑢 𝑑𝑢
Then 𝜃 − 𝜃𝑜 = − ‫׬‬ = −‫׬‬
2𝜇𝐸 2𝜇𝑘
+ 2 𝑢 − 𝑢2 𝑎+𝑏𝑢+𝑐𝑢2
𝑙 2 𝑙

14
 4.10 Equation of motion for a body under central force
(inverse square law force)
1 𝑏+2𝑐𝑢
𝜃 − 𝜃𝑜 = − 𝑐𝑜𝑠 −1 −
−𝑐 𝑏2 −4𝑎𝑐

𝜇𝑘
− 2 +𝑢
𝜃𝑜 − 𝜃 = 𝑐𝑜𝑠 −1 𝑙
𝜇𝑘 2 2𝜇𝐸
2 + 2
𝑙 𝑙

𝜇𝑘 𝜇𝑘 2𝑙 2 𝐸
𝑢= + 1+ 𝑐𝑜𝑠 𝜃𝑜 − 𝜃
𝑙2 𝑙2 𝜇𝑘 2

𝑙2
𝜇𝑘 2𝑙 2 𝐸
⇒ = 1+ 1+ 𝑐𝑜𝑠 𝜃𝑜 − 𝜃
𝑟 𝜇𝑘 2

𝛼
⇒ = 1 + 𝑒𝑐𝑜𝑠 𝜃𝑜 − 𝜃 = 1 + 𝑒𝑐𝑜𝑠 𝜃 − 𝜃𝑜 (4.10.12)
𝑟

We have shown that the solution of the first integral is an equation of conic
𝑙2 2𝑙 2 𝐸
𝛼= semi latus rectum and 𝑒 = 1+ is the eccentricity 15
𝜇𝑘 𝜇𝑘 2
 4.10 Equation of motion for a body under central force
(inverse square law force)
For Eq.(4.10.11) & Eq.(4.10.12) if we assume 𝜃𝑜 = 0 &𝜃 = 0𝑜 & 180𝑜
𝛼 𝛼 𝛼 𝛼
𝑟1 = = & 𝑟2 = = (4.10.13) & (4.10.14)
1+𝑒 2𝑙2 𝐸 1−𝑒 2𝑙2 𝐸
1+ 1+ 1− 1+
𝜇𝑘2 𝜇𝑘2

For 𝑒 > 1 of 𝐸 > 0, 𝑟2 is negative

And 𝑒 = 1, 𝐸 = 0, 𝑟2 is infinity
Both cases ⇨ motion is unbound
Therefore 𝑒 < 1 and 𝐸 < 0 is necessary to keep a bounded motion.
The finite and positive values of 𝑟1 and 𝑟2 represents the turning points.
𝝁𝒌 𝟐𝒍𝟐 𝑬
Comparing the equation of eccentricity 𝑨= 𝟏+ (4.10.15)
𝒍𝟐 𝝁𝒌𝟐

16
 4.10 Equation of motion for a body under central force
(inverse square law force)
Nature of the Orbit
The nature of orbit is determined by eccentricity 𝑒 which depend on energy
Value of E Value of eccentricity Nature of orbit
𝐄>𝟏 𝐞>𝟏 Hyperbola
𝐄=𝟎 𝐞=𝟏 Parabola
𝐕𝐞𝐟𝐟 𝐦𝐢𝐧 < 𝐄 < 𝟎 𝟎<𝐞<𝟏 Ellipse
𝐄 = 𝐕𝐞𝐟𝐟 𝐦𝐢𝐧 𝐞=𝟎 Circle
1 𝐿2 1
we can always set 𝜃𝑜 = 0 And = 𝛼 = ⇒ = 𝐶 1 + 𝑒𝑐𝑜𝑠 𝜃 − 𝜃𝑜
𝑐 𝜇𝑘 𝑟

• Bound motion is possible only for Ellipse or circle.

• The motion of planets is either circular of elliptical.
• The variation of length of the day and seasonal
changes suggest that the path of the planet is elliptical.
17
 4.10 Equation of motion for a body under central force
(inverse square law force)
Elliptic Orbit
The ellipse is a curve traced out by a particle moving in such a way that the sum of
its distance from two fixed foci 𝑂 and 𝑂′ is always constant.
𝑂𝑃 + 𝑂′ 𝑃 = 𝑟 + 𝑟 ′ = 2𝑎
a is the semi-major axis
𝑟1 + 𝑟2 = 2𝑎 (1)
𝛼 𝛼
𝑟1 = & 𝑟2 = (2 & 3)
1+𝑒 1−𝑒
2𝛼
From (1), (2) and (3) ⇒ 𝑟1 + 𝑟2 = = 2𝑎
1−𝑒 2
𝛼
𝑎= ⇒ 𝛼 = 𝑎 1 − 𝑒2 (4.10.15)
1−𝑒 2
𝑎 1−𝑒 2 𝑎 1−𝑒 2
𝑟= or 𝑟 = This is the polar equation of the orbit.
1+𝑒𝑐𝑜𝑠𝜃 1+𝑒𝑐𝑜𝑠 𝜃−𝜃𝑜 18
 4.10 Equation of motion for a body under central force
(inverse square law force)
Elliptic Orbit
Not the distance between two foci 𝑂𝑂′ = 𝑟2 − 𝑟1
2𝑒𝛼
⇒ 𝑂𝑂′ = = 2𝑎𝑒
1−𝑒 2
𝑂𝑂′
⇒ = 𝑎𝑒 (4.10.16)
2
From the figure it is clear that 𝑂𝑃′ = 𝑂′ 𝑃′ and
𝑂𝑃′ + 𝑂′ 𝑃′ = 2𝑎 & 𝑂𝑃′ = 𝑎
2
𝑂𝑂′
Now from figure 𝑏 2 = 𝑂𝑃′ 2 −
2

⇒ 𝑏 2 = 𝑎2 − 𝑎2 𝑒 2 = 𝑎2 1 − 𝑒 2
⇒𝑏=𝑎 1 − 𝑒2
19
4.10 Equation of motion for a body under central force
(inverse square law force)

If 𝒆 ≠ 𝟎 then,
2𝑙 2 𝐸
Since 𝑒= 1+
𝜇𝑘 2

2𝑙 2 𝐸
Therefore, 𝑏=𝑎 1−1−
𝜇𝑘 2

2𝑙 2 𝐸
𝑏=𝑎 −
𝜇𝑘 2

The energy of the bounded system is less than zero therefore

it will give a real value solution.
20
 4.10 Equation of motion for a body under central force
(inverse square law force)
If 𝒆 = 𝟎 ellipse will become circle 𝒃 = 𝒂
(Note in the region when body passes through closest distance the curve is arc of
a circle)
2𝑙 2 𝐸
And 1+ =0
𝜇𝑘 2
𝜇𝑘 2
𝐸=−
2𝑙 2
Eq. (4.10.15) will be reduced to 𝑎 = 𝛼, therefore the radius of the circle is;
𝜇𝑘2
𝑙2 − 2𝐸
𝑟𝑜 = 𝑎 = =
𝜇𝑘 𝜇𝑘
𝑘
And 𝑟𝑜 = −
2𝐸
𝑘
And 𝐸=−
2𝑎 21
4.10 Equation of motion for a body under central force
(inverse square law force)
Putting this value in equation of eccentricity we get
𝑙2
𝑒= 1−
𝜇𝑘𝑎

Using this value the semi-minor axis b can be written as.

𝑙2
𝑏=𝑎 1−1+
𝜇𝑘 2 𝑎

𝑙2
𝑏=𝑎
𝜇𝑘 2 𝑎
1Τ 𝑙
𝑏=𝑎 2
𝑘 𝜇

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