B.Sc IV Semester

Unit IV :

Central Orbit

Central force: A force whose line of action always

passes through a fixed point, is called a central force. The

fixed point is known as the centre of force.

Central Orbit: A central orbit is the path described by a

particle moving under the action of a central force.

Example The motion of a planet about the sun.

Differential equation of a central orbit → A particle

moves in a plane with an acceleration which is always

directed to a fixed point O in the plane to obtain differential

equation of the path.

Sol: Let a particle move in a plane with an acceleration P

which is always directed to a fixed point O in the plane.

Take the centre of force O as the pole. Let OX be the initial

line and (r,𝜃) the polar co-ordinates of the position A of the

moving particle at any instant t.

If P be the acceleration of the

A particle is always directed

towards O.

X

We have 𝑑2𝑟

𝑑𝑡2 –r(

𝑑𝜃2

𝑑𝑡= -P [1]

(The –ve sign has been taken because the radial acc, P is in

the direction of r decreasing,)

90

P

L

p

ⱷ

O

⸪ there is no acc. Perpendicular to OA, therefore

1

𝑟{

𝑑

𝑑𝑡 } (r2𝑑𝛉

𝑑𝑡) = 0 → (2)

Integrating, we get r2 ( 𝑑𝜽

𝑑𝑡) = constant= h (say) → (3)

Let r = 1

𝑢 Now from (3)

𝑑𝜃

𝑑𝑡 =

ℎ

𝑟2 = hu2 → (4)

Also, 𝑑𝑟

𝑑𝑡 = -

1

𝑢2(𝑑𝑢

𝑑𝑡) = -

1

𝑢2

𝑑𝑢

𝑑𝜃

𝑑𝜃

𝑑𝑡 = -

1

𝑢2

𝑑𝑢

𝑑𝜃.hu2

= - h𝑑𝑢

𝑑𝜃

And 𝑑2𝑟

𝑑𝑡2 = 𝑑

𝑑𝑡(−ℎ

𝑑𝑢

𝑑𝜃 ) = -h

𝑑

𝑑𝜃(

𝑑𝑢

𝑑𝜃)

𝑑𝜃

𝑑𝑡 = -h

𝑑2𝑢

𝑑𝜃2

𝑑𝜃

𝑑𝑡

= -h𝑑2𝑢

𝑑𝜃2(hu2) = -h2u2𝑑2𝑢

𝑑𝜃2 [from (4) ]

Hence eqn, (1) becomes

-h2u2𝑑2𝑢

𝑑𝜃2 -

1

𝑢(h2u4) = -P

Or h2u2𝑑2𝑢

𝑑𝜃2 + h2u3 = P

Or 𝑑2𝑢

𝑑𝜃2 + u = 𝑃

ℎ2𝑢2 → (5)

Which is the differential eqn. of a central orbit in polar

form.

Pedal form → If p is the length of the perpendicular drawn

from the origin O upon the tangent at the point A.

We have 1

𝑝2 = 1

𝑟2 + 1

𝑟4 𝑑𝑟

𝑑𝜃

2

But u = 1

𝑟 ⸫

𝑑𝑢

𝑑𝜃 = -

1

𝑟2 𝑑𝑟

𝑑𝜃

i.e 𝑑𝑢

𝑑𝜃

2=

1

𝑟4 𝑑𝑟

𝑑𝜃

2

so 1

𝑝2 = u2 + 𝑑𝑢

𝑑𝜃

2 (6)

Differentiating both sides of (6) w.r.t 𝜃 we have

- 2

𝑝3

𝑑𝑝

𝑑𝜃 = 2u

𝑑𝑢

𝑑𝜃 + 2

𝑑𝑢

𝑑𝜃.𝑑2𝑢

𝑑𝜃2 = 2𝑑𝑢

𝑑𝜃 (u +

𝑑2𝑢

𝑑𝜃2 )

Or - 1

𝑝3

𝑑𝑝

𝑑𝜃 =

𝑑𝑢

𝑑𝜃.

𝑃

ℎ2𝑢2 (from 5)

Or - 1

𝑝3

𝑑𝑝

𝑑𝑟 .

𝑑𝑟

𝑑𝜃 = -

1

𝑟2

𝑑𝑟

𝑑𝜃 .

𝑃

ℎ2𝑢2

Or 1

𝑝3

𝑑𝑝

𝑑𝑟 =

1

𝑟2. 𝑃

ℎ2𝑢2

Or 1

𝑝3

𝑑𝑝

𝑑𝑟 = u2.

𝑃

ℎ2𝑢2

⸫ P = ℎ2

𝑝3

𝑑𝑝

𝑑𝑟

Which is the differential eqn. of a central orbit in pedal

form.

Que1: A particle describe the curve an = rn𝐜𝐨𝐬 𝒏𝜽

under a force to the pole, Find the law of force.

Sol: → The curve is rncos 𝑛𝜃 = an

Or an un= cos 𝑛𝜃 (1) Where r = 1

𝑢

Taking log from both sides, we get

n log a+ n log u = log cos 𝑛𝜃

Now differentiating w.r.t 𝜃 we, get

𝑛

𝑢

𝑑𝑢

𝑑𝜃 = -

𝑛𝑠𝑖𝑛𝑛𝜃

𝑐𝑜𝑠𝑛𝜃

Or 𝑑𝑢

𝑑𝜃 = - u tan𝑛𝜃 (2)

Or 𝑑2𝑢

𝑑𝜃2 = - 𝑑𝑢

𝑑𝜃 tan𝑛𝜃 – unsec2n𝜃

Or 𝑑2𝑢

𝑑𝜃2 = utan2n𝜃 – unsec2n𝜃 from (2)

The differential eqn. of the path is

P = h2u2 [ 𝑑2𝑢

𝑑𝜃2 + u]

⸫ P = h2u2 [ u + utan2n𝜃 – unsec2n𝜃 ]

= h2u3 [1 + tan2n𝜃 – nsec2n𝜃 ]

= h2 u3 [sec2n𝜃 - nsec2n𝜃 ]

= h2u3 (1- n) sec2n𝜃 [from (1) ⸪ an un = cos 𝑛𝜃 ]

= h2u3 (1- n) 𝑟𝑛

𝑎𝑛

2

= ℎ 2

𝑟 3(1- n)

𝑟 2𝑛

𝑎 2𝑛

= ℎ 2

𝑎 2𝑛 (1 – n) r 2n – 3

Thus P 𝛼 r 2n – 3 i.e The force is directly proportional to

the (2n -3)th power distance from the pole. ans.

(2): rn = A cosn𝜽 +B sinn𝜽

Soln: rn = A cosn𝜃 + B sinn𝜃

Let A= k cos𝛼 and B = k sin𝛼

Where k and 𝛼 are constants

Then rn = k cos𝛼 cosn𝜃 + k sin𝛼 sinn𝜃

Replacing r by 1

𝑢 we have

rn = 1

𝑢

𝑛= k cos (n𝜃 − 𝛼) (1)

Taking log from both sides

-n log u = log k + log cos (n𝜃 − 𝛼)

Differentiating w.r.t 𝜃 we have

- 𝑛

𝑢

𝑑𝑢

𝑑𝜃 = -

sin (𝑛𝜃−𝛼)

cos (𝑛𝜃−𝛼) .n

Or 𝑑𝑢

𝑑𝜃 = u tan (n𝜃 − 𝛼)

⸫ 𝑑2𝑢

𝑑𝜃2 = 𝑑𝑢

𝑑𝜃 tan (n𝜃 − 𝛼) + u n sec2 (n𝜃 − 𝛼)

The differential eqn. of the path is

P = h2u2 [ 𝑑2𝑢

𝑑𝜃2 + u]

⸫ P = h2u2 [ u + u tan2(n𝜃 − 𝛼) + u n sec2(n𝜃 − 𝛼) ]

= h2u3 [1 + tan2 (n𝜃 − 𝛼) + n sec2 (n𝜃 − 𝛼) ]

= h2 u3 [sec2 (n𝜃 − 𝛼) + n sec2 (n𝜃 − 𝛼) ]

= h2u3 (1+ n) sec2 (n𝜃 − 𝛼)

[from (1) ⸪k un = sec (n𝜃 − 𝛼) ]

= h2u3 (1+n) (k u n) 2 = ℎ 2

𝑟 3(1+n) (k u n) 2

= ℎ 2

𝑟 2𝑛+3 (1 + n) k 2 = (1 + n) ℎ 2

𝑟 2𝑛+3 (A2 + B2)

Thus P 𝛼 1

𝑟 2𝑛+3 ans.

(3). r = a sin n𝜃

Soln. r = a sin n𝜃

r = a sin n𝜃

or 1

𝑢 = a sin n𝜃

Or u= 1

𝑎𝑐𝑜𝑠𝑒𝑐𝑛𝜃 (1)

Differentiating w.r.t 𝜃

𝑑𝑢

𝑑𝜃 = -

𝑛

𝑎 𝑐𝑜𝑠𝑒𝑐𝑛𝜃𝑐𝑜𝑡𝑛𝜃 = -nu cot𝑛𝜃

And 𝑑2𝑢

𝑑𝜃2 = n2 u cosec2𝑛𝜃 - n 𝑑𝑢

𝑑𝜃 cot𝑛𝜃

= n2 u cosec2𝑛𝜃 – n (-n u cot𝑛𝜃) cot𝑛𝜃

= n2 u cosec2𝑛𝜃 + n2ucot2𝑛𝜃

The differential eqn. of the central orbit is

P = h2u2 [ 𝑑2𝑢

𝑑𝜃2 + u]

P = h2u2 [n2 u cosec2𝑛𝜃 + n2ucot2𝑛𝜃 + u]

= h2u3 [n2 cosec2𝑛𝜃 + n2 (cosec2𝑛𝜃 -1) + 1]

= h2u3 [2n2 cosec2𝑛𝜃 – (n2 – 1)]

⸪ from (1) 𝑐𝑜𝑠𝑒𝑐𝑛𝜃= ua

⸫ P = h2[ 2n2u3.u2. a2 – u3( n2 – 1)]

= h2 [2𝑛 2𝑎 2

𝑟 5 – (

𝑛 2−1

𝑟 3)] ⸪ r =

1

𝑢

⸫ P 𝛼 h2 [2𝑛 2𝑎 2

𝑟 5 – (

𝑛 2−1

𝑟 3)] ans.

Q 2: Find the law of force towards the pole under which

the following curves are described.

(1) r2 =2ap (2) p2 =ar

Sol 1: The eqn. of the curve is r2 =2ap

⸫ 1

𝑝 =

2 𝑎

𝑟 2

Or 1

𝑝 2 =

4 𝑎

𝑟 4

2

Differentiating w.r.t ‘r’ we have

- 2

𝑝 3𝑑𝑝

𝑑𝑟 = -

16𝑎

𝑟 5

2

⸫ ℎ2

𝑝3 𝑑𝑝

𝑑𝑟 =

8𝑎2

𝑟 5ℎ2 → (1)

𝑛𝑜𝑤 𝑓𝑟𝑜𝑚 𝑝𝑒𝑑𝑎𝑙 𝑒𝑞𝑛, 𝑤𝑒 ℎ𝑎𝑣𝑒

P = ℎ2

𝑝3 𝑑𝑝

𝑑𝑟 =

8𝑎2

𝑟 5ℎ2

⸫ P 𝛼 1

𝑟 5 i.e the force varies inversely as the fifth

power of the distance from the pole.

Sol 2: The equation of the curve is p2=ar, which is the

pedal equation of a parabola referred to the focus as the

pole

⸫ 1

𝑝2 = 1

𝑎 1

𝑟

Differentiating w.r.t. ‘r', we get

- 2

𝑝 3𝑑𝑝

𝑑𝑟 = -

1

𝑎

1

𝑟2 ...(1)

⸫ ℎ2

𝑝3 𝑑𝑝

𝑑𝑟 =

ℎ2

2𝑎

1

𝑟2 [from (1)]

From the pedal equation of a central orbit, we have

P = ℎ2

𝑝3 𝑑𝑝

𝑑𝑟 =

ℎ2

2𝑎

1

𝑟2 [from (1)]

⸫ P 𝛼 1/r2 i.e., the force varies inversely as the square of

the distance from the pole.

Theorem:

Rate of description of the sectorial area: → In every

central orbit the sectorial area traced out by the radius

vector to the centre of force increases uniformly per unit

time and the linear velocity varies inversely as the

perpendicular from the centre upon the tangent to the

path.

Q(r + 𝜹𝒓, 𝜽 + 𝜹𝜽)

P(r,𝜽)

∅

𝛿𝜃

𝜽

X

.

M

90

O

p

Let O be the origin and ox as the initial line. Let P(r,𝜽)

and Q (r + 𝛿𝑟, 𝜃 + 𝛿𝜃) be the positions of a particle

moving in a central orbit at times t and t + 𝛿𝑡 respectively

Sectorial area OPQ described by the particle in time 𝛿𝑡 =

area of the ∆𝑂𝑃𝑄

[⸪ the point Q is very close to P and ultimately we have

to take limit as Q→P]

= 1

2.OP.OQ sin< 𝑂𝑃𝑄 =

1

2𝑟(r + 𝛿𝑟)sin 𝛿𝜃

⸫ Rate of description of the sectorial area

= lim𝛿𝑡→0

𝑠𝑒𝑐𝑡𝑜𝑟𝑖𝑎𝑙 𝑎𝑟𝑒𝑎 𝑂𝑃𝑄

𝛿𝑡

= lim𝛿𝑡→0

1

2𝑟

(𝑟+𝛿𝑟) sin 𝛿𝜃

𝛿𝑡

= lim𝛿𝑡→0

{1

2𝑟

(𝑟+𝛿𝑟) sin 𝛿𝜃

𝛿𝜃 .𝛿𝜃

𝛿𝑡}

= 1

2 r2𝑑𝜃

𝑑𝑡 =

1

2 h (const.) → (1) [⸪ r2(

𝑑𝜃

𝑑𝑡)= h]

Thus the rate of description of the sectorial area is

constant and is equal to ℎ

2.

Again for a central orbit, we have

r2(𝑑𝜃

𝑑𝑡)= h

⸫ r2𝑑𝜗

𝑑𝑠 .𝑑𝑠

𝑑𝑡 = h

Or r2 𝑑𝜗/𝑑𝑠. .v = h → (2) [⸪𝑑𝑠

𝑑𝑡 = v i.e the linear

velocity]

But from differential calculus, We have

r𝑑𝜃

𝑑𝑠 = sin∅, where ∅ is the angle between the radius vector

and the tangent.

⸫ r2𝑑𝜃

𝑑𝑠 = r sin∅ = p where p is the length of the

perpendicular drawn from the pole O on the tangent at P.

Putting r2 (𝑑𝜃

𝑑𝑠 ) = p in (2)

We get v p = h

Or v = ℎ

𝑝 → (3)

⸫ v 𝛼1

𝑝

i.e the linear velocity at P varies inversely as the

perpendicular from the fixed point upon the tangent to the

path.

From (3) we have v 2 = (ℎ

𝑝)2

But 1

𝑝2 = 1

𝑟2 + 1

𝑟4 (𝑑𝑟

𝑑𝜃) 2

= u2 + ( 𝑑𝑢

𝑑𝜃)2

⸫ v2 = h2{ u2 + ( 𝑑𝑢

𝑑𝜃)2} → (4)

The eqn. (4) gives the linear velocity at any point of the

path of a central orbit.

Law of force, velocity and the periodic time when the

orbit is an ellipse → A particle moves in an ellipse

under a force which is always directed towards its

focus; to find the law of force and the velocity at any

point of its path.

L

X

We know that the polar equation of an ellipse referred to

its focus S as pole is 𝑙

𝑟 = 1 + e cos𝜃

Or u = 1

𝑙 +

𝑒 𝑐𝑜𝑠𝜃

𝑙 (1)

Where u = 1

𝑟

Differentiating w.r.t ‘𝜃’ we have

𝑑𝑢

𝑑𝜃 = -

𝑒𝑠𝑖𝑛𝜃

𝑙 and

𝑑 2𝑢

𝑑𝜃 2 = -

𝑒 𝑐𝑜𝑠𝜃

𝑙

(1) Law of force → We know that the diff. eqn. of a

central orbit referred to the centre of force as pole is

𝑃

ℎ 2𝑢2 = [ 𝑑2𝑢

𝑑𝜃2 + u]

Where P is the central acceleration assumed to be

attractive.

S

L1

Now here P = h2u2 [ 𝑑2𝑢

𝑑𝜃2 + u]

= h2u2 [1

𝑙 +

𝑒 𝑐𝑜𝑠𝜃

𝑙 -

𝑒 𝑐𝑜𝑠𝜃

𝑙]

= ℎ2𝑢2

𝑙 =

ℎ2

𝑙 𝑟2 = 𝜇

𝑟2 (2)

Where 𝜇 = ℎ

𝑙

2 or h2 = 𝜇𝑙 (3)

⸫ h = √𝜇(semi - latus rectum)

⸫ P 𝛼 1

𝑟2

Hence the acc. varies inversely as the square of the

distance of the particle from the focus. Also the force is

attractive because the value of P is positive.

(2) Velocity: We know that the velocity in a central orbit

is

v2 = h2{ u2 + ( 𝑑𝑢

𝑑𝜃)2}

⸫ v2 = h2 {(1

𝑙 +

𝑒

𝑙𝑐𝑜𝑠𝜃)2+ (-

𝑒𝑠𝑖𝑛𝜃

𝑙)2}

= h2 [ 1

𝑙2 + 2𝑒

𝑙2 𝑐𝑜𝑠𝜃 + 𝑒2

𝑙2 ]

= ℎ2

𝑙[

1+𝑒2

𝑙 +

2𝑒𝑐𝑜𝑠𝜃

𝑙]

= 𝜇[1+𝑒2

𝑙 + 2(u -

1

𝑙)] [from (1)&(3)]

= 𝜇 [2u - 1−𝑒2

𝑙]

If 2a &2b are the lengths of the major and the minor

axes of the ellipse, we have

l = the semi-latus rectum = 𝑏2

𝑎 =

𝑎2(1− 𝑒2)

𝑎 = a(1- e2)

⸫ 1− 𝑒2

𝑙 =

1

𝑎

⸫ v2 = 𝜇[2

𝑟 −

1

𝑎] (4)

Which gives the velocity of the particle at any point of its

path.

Periodic Time: → If T be the total time of describing the

whole area of the ellipse, then we know that the rate of

description of area is 1

2ℎ which is constant.

⸫ 1

2ℎ × 𝑇 = area of the ellipse = 𝜋𝑎𝑏

⸫ T = 2𝜋𝑎𝑏

ℎ =

2 𝜋𝑎𝑏

√𝜇𝑙 =

2𝜋𝑎𝑏

√𝜇.𝑏2

𝑎

= 2𝜋𝑎

32

√𝜇 (5)

Or T = 2𝜋𝑎

32

√𝜇 [⸪ h2 =𝜇𝑙 , l =

𝑏2

𝑎 ]

i.e. the time period for one complete oscillation is

proportional to 𝑎3

2 , a being semi-major axis.

Or T = 2𝜋

√𝜇 [

2

𝑟 -

𝑣2

𝜇 ]-3/2 by (4).

Apse: → An apse is a point on the central orbit at which

the radius vector is drawn from the centre of force to the

point has a maximum or minimum value.

Apsidal distance: → The length of the radius vector at an

apse is called an apsidal distance.

Apsidal angle: → The angle between two consecutive

apsidal distances is called an apsidal angle.

Theorem: At an apse the radius vector is perpendicular

to tangent i.e., at an apse the particle moves at right angles

to radius vector.

Proof: From the definition of an apse, r is maximum or

minimum at apse i.e., u=1/r is minimum or maximum at

an apse.

⸫ at an apse, 𝑑𝑢

𝑑𝜃 = 0

But we know that 1

𝑝2 = {u2 + ( 𝑑𝑢

𝑑𝜃)2}

⸫ at an apse, 1

𝑝2 = u2 = 1

𝑟2

Or p = r or r sin∅ = r or Sin ∅ = 1 or ∅ = 90°. This

proves that at an apse the radius vector is perpendicular to

the tangent or in other words at an apse the particle moves

at right angles to the radius vector

O

C P1

V

V

P A

B

OA= OC

Property of the apse line: If the central acceleration P is

a single valued function of the distance, every apse-line

divides the orbit into and equal symmetrical portions, and

thus there can only be two apsidal distances.

Proof: We know that for a central orbit

𝑃

𝑢2 = h2[ 𝑑2𝑢

𝑑𝜃2 + u]

Multiplying both sides by 2(𝑑𝑢

𝑑𝜃)and integrating w.r.t 𝜃

We have v2 = h2[(𝑑𝑢

𝑑𝜃)2+ u2]

= 2∫𝑃

𝑢2du + C (1)

v2 = C - 2∫ 𝑃𝑑𝑟 [ ⸫1

𝑢 = r or -

1

𝑢2du = dr]

The eqn. (1) shows that if P is a single valued function of

the distance r, then the velocity of the particle is the same

at the same distance r and is independent of the direction

of motion.

Que 1: A particle moves under a force

m 𝜇{3au4 – 2(a2 – b2)u5}, a>b and is projeted from an

apse at a distance (a + b) with velocity √𝜇

𝑎 + 𝑏 . Show that

the eqn. of its path is r = a+ b cos𝜃

Sol: We have P= h2u2 [u + 𝑑2𝑢

𝑑𝜃2] (1)

Now P = 𝜇{3au4 – 2(a2 – b2)u5}

⸫ from (1) h2 [u + 𝑑2𝑢

𝑑𝜃2] = 𝑃

𝑢2 = 𝜇{3𝑎𝑢4 – 2(𝑎2 – 𝑏2)𝑢5}

𝑢2

Or h2 [u + 𝑑2𝑢

𝑑𝜃2] = 𝜇{3𝑎𝑢2 − 2(𝑎2 − 𝑏2)𝑢3}

Multiplying both sides by 2𝑑𝑢

𝑑𝜃 and integrating we have

v2 = h2 [u2 + ( 𝑑𝑢

𝑑𝜃)2] = 2 𝜇 {au3- 2(a2- b2)

𝑢4

4 }+ A

v2 = h2 [u2 + ( 𝑑𝑢

𝑑𝜃)2] = 𝜇 {2au3- (a2- b2) u4} + A (2)

where A is a constant.

But initially at an apse

r = a + b i.e u = 1

𝑎+𝑏, v =

√𝜇

𝑎 + 𝑏 ,

𝑑𝑢

𝑑𝜃 = 0

⸫ from (2) we get

𝜇

(𝑎+𝑏)2 = h2 [1

(𝑎+ 𝑏 )2] = 𝜇[

2𝑎

(𝑎+𝑏)3 -

(𝑎2−𝑏2)

(𝑎+𝑏)4 ]

from first & 2nd term

⸫ h2 = 𝜇

from 1st & last term

𝜇

(𝑎+𝑏)2 = 𝜇[ 2𝑎2+2𝑎𝑏− 𝑎2+𝑏2

(𝑎+𝑏)4 ] + A

Or 𝜇

(𝑎+𝑏)2 = 𝜇[ (𝑎+𝑏)2

(𝑎+𝑏)4 ] + A

⸫ A= 0

Putting the value of h2 and A in (2) we get

𝜇[u2 + ( 𝑑𝑢

𝑑𝜃)2] = 𝜇 {2au3- (a2- b2) u4}

Or [u2 + ( 𝑑𝑢

𝑑𝜃)2] = 2au3- (a2- b2) u4

Or (𝑑𝑢

𝑑𝜃)2 = - u2 + 2au3- (a2- b2) u4 (3)

But u = 1

𝑟 so that

𝑑𝑢

𝑑𝜃 = -

1

𝑟2

𝑑𝑟

𝑑𝜃

From (3) (- 1

𝑟2

𝑑𝑟

𝑑𝜃) 2 = -

1

𝑟2 + 2𝑎

𝑟3 – (𝑎2−𝑏2

𝑟4 )

Or 1

𝑟4 (𝑑𝑟

𝑑𝜃)2 =

1

𝑟4 [-r2 + 2ar - (a2- b2)]

Or (𝑑𝑟

𝑑𝜃)2 = [-r2 + 2ar - a2 + b2] = b2- (r2-2ar+a2)

= b2- (r-a)2

⸫ 𝑑𝑟

𝑑𝜃 = ±√𝑏2 − (𝑟 − 𝑎)2

d𝜃 = - 𝑑𝑟

√{𝑏2−(𝑟−𝑎)2}

Integrating, we get

𝜃 + A = cos-1[𝑟−𝑎

𝑏] (4)

Initially r =a + b, 𝜃 = 0

A =0

then from (4)

⸫ cos-1[𝑟−𝑎

𝑏] = 𝜃

Or [𝑟−𝑎

𝑏] = cos𝜃

Or r – a = b cos𝜃

⸫ r = a + b cos𝜃 ans.

Que 2: A particle moves with central acc. 𝜇(𝑟 + 𝑎4

𝑟3)

being projected from an apse at a distance ‘a’ with a

velocity 2√𝜇 a. Prove that it describes the curve

r2 (2 + cos√3 𝜃) = 3a2

Sol: Here P = 𝜇(𝑟 + 𝑎4

𝑟3) = 𝜇[1

𝑢 + 𝑎4𝑢3] ( as u=

1

𝑟)

The differential equation of the path is

P = h2u2 {𝑑2𝑢

𝑑𝜃 2 + u} → (1)

Or h2 [ 𝑑2𝑢

𝑑𝜃 2 + u] =

𝑃

𝑢2 = 𝜇

𝑢2[1

𝑢 + 𝑎4𝑢3]

= 𝜇 [1

𝑢 3 + a4u]

Multiplying both sides by 2𝑑𝑢

𝑑𝜃 and integrating we get

v2 = h2 [u2 + ( 𝑑𝑢

𝑑𝜃)2] = 2𝜇 [−

1

2𝑢 2 +

𝑎4 𝑢2

2] + C → (2)

Now using initial conditions when u =1

𝑎,

𝑑𝑢

𝑑𝜃 = 0

and v = 2𝑎√𝜇

We get 4a2𝜇 = ℎ2

𝑎2 = 2 𝜇[- 𝑎2

2 +

𝑎4

2𝑎2] +C

From 1st & 2nd term

⸫ h2 = 4a4 𝜇

From1st & last term

⸫ C = 4a2 𝜇

Hence by putting values of h2 and C in equation (2).It

becomes.

4a4 𝜇 [u2 + ( 𝑑𝑢

𝑑𝜃)2] = 𝜇 [−

1

𝑢 2 + 𝑎4𝑢2]+4a2 𝜇

𝜇 cancel throughout

4a4 [u2 + ( 𝑑𝑢

𝑑𝜃)2] = [−

1

𝑢 2 + 𝑎4𝑢2]+4a2

4a4 ( 𝑑𝑢

𝑑𝜃)2 = −

1

𝑢 2 + 𝑎4𝑢2+4a2 – 4a4u2

= −1+𝑎4𝑢4+4𝑎2𝑢2− 4𝑎4𝑢4

𝑢 2

= −1−3𝑎4𝑢4+4𝑎2𝑢2

𝑢2

Adding and subtracting 4

3

= −1−3𝑎4𝑢4+4𝑎2𝑢2

𝑢2 + 4

3 -

4

3

= −1−(3𝑎4𝑢4−4𝑎2𝑢2 )

𝑢2 + 4

3 -

4

3

= -1 + 4

3 -

[1−(3𝑎4𝑢4−4𝑎2𝑢2 )

𝑢2 - 4

3

= [1

3 - (√3 a2u2 -

2

√3)2] /u2

Or 2a2 (𝑑𝑢

𝑑𝜃) = ±

√[1

3−(√3 𝑎2𝑢2 −

2

√3)2]

𝑢

Now put √3 a2u2 - 2

√3 = t

Or 2a2du.u = -d𝜃 √[1

3− (√3 𝑎2𝑢2 −

2

√3)2

2√3𝑎2𝑢𝑑𝑢 = dt

⸫ - 2𝑎2 𝑢𝑑𝑢

√[ 1

3−(√3 𝑎2𝑢2 −

2

√3)2

= 𝑑𝜃

or - 𝑑𝑡

√1

3 −𝑡2

= √3𝑑𝜃

Integrating, we get

Cos-1√3t = √3𝜃 + A

Putting the value of ‘t’.

Or Cos-1√3(√3 a2u2 - 2

√3) = √3𝜃 + A → (3)

By initial condition when u = 1

𝑎, let 𝜃= 0

Or Cos-11 = √3(0) + A

⸫ A = 0

The equation (3) becomes

Or Cos-1√3(√3 a2u2 – 2

√3) = √3𝜃

𝑜𝑟 √3(√3 a2u2 – 2

√3) = cos√3𝜃

Or 3a2u2 – 2 = cos√3𝜃

Or 3a2u2 = 2 + cos√3𝜃

Or 3a2 = 2+𝑐𝑜𝑠√3𝜃

𝑢2 [⸪ u = 1

𝑟 ]

⸫ 3a2 = r2 (2 + cos√3𝜃)

Proved.

Newtonian Law of Attraction: This law may be

expressed as follows; between every two particles, of

masses m1and m2 placed at a distance r apart, the mutual

attraction is

𝛾𝑚1𝑚2

𝑟2

Units of force, where 𝛾 is a constant, depending, on the

units of mass and length employed, and is known as the

constant of gravitation.

Article: A particle moves in a path so that its acc. is

always directed to a fixed point and is equal to

𝜇

(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2; Show that the path is a conic section and

distinguish between three cases that arise.

Sol: Here P = 𝜇

𝑟2

Hence the differential eqn. of the path is

ℎ2

𝑝3

𝑑𝑝

𝑑𝑟 = P =

𝜇

𝑟2 → (1)

Integrating we have

- ℎ2

2𝑝2 = - 𝜇

𝑟 + B

ℎ2

𝑝2 = 2𝜇

𝑟 + C

Also we know that h = p v

v2 = ℎ2

𝑝2 = 2𝜇

𝑟 + C → (2)

Now the (p,r) equation of an ellipse and hyperbola,

referred to a focus are respectively.

𝑏2

𝑝2 = 2𝑎

𝑟 – 1, and

𝑏2

𝑝2 = 2𝑎

𝑟 + 1 → (3)

Where 2a & 2b are the transverse and conjugate axes.

Hence, when C is negative, (2) is an ellipse; when C is

positive , it is a hyperbola.

Also when C = 0, (2) becomes 𝑝2

𝑟 = constant, and this is

the (p, r) equation of a parabola referred to its focus.

Hence (2) always represents a conic section, whose focus

is at the centre of force, which is an

Ellipse negative

Parabola according as C is zero

Or hyperbola or positive

i.e. according as v2 <=> 2𝜇

𝑟, according as the square of

the velocity at any point P is <=> 2𝜇

𝑆𝑃,where S is the

focus.

Again, comparing eqns. (2) & (3) we have, in the case of

the ellipse,

ℎ2

𝑏2 = 𝜇

𝑎 =

𝐶

−1

⸫ h = √𝜇𝑏2

𝑎 = √𝜇 × 𝑠𝑒𝑚𝑖 − 𝑙𝑎𝑡𝑢𝑠 𝑟𝑒𝑐𝑡𝑢𝑚, and C= -

𝜇

𝑎

Hence in case of the ellipse v2 = 𝜇[ 2

𝑟 -

1

𝑎] → (4)

So, for the hyperbola, v2 = 𝜇[ 2

𝑟 +

1

𝑎],

And, for the parabola, v2 = 2𝜇

𝑟

It will be noted that in each case the velocity at any point

does not depend on the direction of the velocity.

Cor1: If a particle is projected at a distance R with

velocity V in any direction the path is an ellipse, parabola

or hyperbola, according as V2 <=> 2𝜇

𝑟.

Now the square of the velocity that would be acquired in

falling from infinity to the distance R.

=2∫ (−𝜇

𝑟2

𝑅

∞) dr = [

2𝜇

𝑟]∞

𝑅 = 2𝜇

𝑅

Hence the path is an ellipse, parabola or hyperbola,

according as the velocity at any point is <=>

that acquired is falling from infinity to that point

Cor2: The velocity V1 for the description of a circle of

radius R is given by

V1

2

𝑅 = normal acceleration =

𝜇

𝑅2, so that V12 =

𝜇

𝑅

And ⸫ V1 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑓𝑟𝑜𝑚 𝑖𝑛𝑓𝑖𝑛𝑖𝑡𝑦

√2

Kepler’s laws

The three important laws discovered by the famous

astronomer Kepler which govern the motion of various

planets about the sun are stated as follows:-

(1) Each planet describes an ellipse having the sun as

one of its foci

(2) The areas described by the radii drawn from the

planet to the sun are in the same orbit proportional

to the times of describing them

(3) The squares of the periodic times of planets are

proportional to the cubes of the major axis of their

orbits.

Deductions

(1) From first Kepler's law we conclude that the

acceleration of each planet towards the sun varies

inversely as the square of its distance from the sun.

(2) From the second how we observe that the rate of

description of sectorial area is constant or

r2𝜃=constant. Hence the force on planet is entirely

directed to the sun.

(3) From third law, T= 2𝜋𝑎

32

√𝜇 that 𝜇 is constant i.e.

acceleration at unit distance from the sun is the

same for all planets

Properties:

Y P Z

SSsS

(1) The sum of the focal distances of a point on an

ellipse is equal to 2a, where 2a is the length of its

major axis. Thus SP + HP = 2a

θ ∅

(2) The product of perpendiculars drawn from the foci

on any tangent to an ellipse is constant and equal to

square of the semi- minor axis of the ellipse i.e.

SY.HZ = b2

(3) The length of latus rectum is 2𝑏2

𝑎, where

b2 = a2 (1 – e2),e being the eccentricity of the ellipse.

(4) CH = CS = ae

(5) The tangent and normal at any point are each

equally inclined to focal radii of that point. i.e.

< 𝑌𝑃𝑆 = <ZPH.

Que 1: A particle describes an ellipse under a force 𝜇

(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2 towards the focus. If it was projected with

velocity V from appoint distant r from the centre of force.

Show that its periodic time is

2𝜋

√𝜇 [

2

𝑟 −

𝑉2

𝜇]−3/2

Sol: Let a be the length of the semi-major axis of the

ellipse described by the particle then the velocity V at a

point distant r from the centre of force is given by

V2 = 𝜇[ 2

𝑟 -

1

𝑎]

⸫ 𝑉

𝜇

2 = [

2

𝑟 -

1

𝑎]

Or 1

𝑎 = [

2

𝑟 -

𝑉

𝜇

2]

Or a = [ 2

𝑟 -

𝑉

𝜇

2]-1

⸫ The periodic time = 2𝜋𝑎

32

√𝜇 =

2𝜋

√𝜇 [

2

𝑟 -

𝑉

𝜇

2]-3/2

Proved.

Que 2: If v1 and v2 are the linear velocities of a planet

when it is respectively nearest and farthest from the sun.

Prove that (1 – e) v1 = (1 + e) v2.

Sol:

Taking S the focus as sun.

1 A A1 C

H

S

Let v1 and v2 be the velocities of the planet at the points A

and A1 which are nearest and farthest from the sun at s,

then at A.SO

r = SA = CA – CS = a – ae = a (1- e), v = v1

and at A1 r = SA1 = CS + CA1 = ae + a = a (1 + e), v = v2

Also v2 = 𝜇[ 2

𝑟 -

1

𝑎]

⸫ v12 = 𝜇[

2

𝑎(1−𝑒) -

1

𝑎]

= 𝜇[2−1+𝑒

𝑎(1−𝑒) ] = 𝜇[

1+𝑒

𝑎(1−𝑒) ]

And v22 = 𝜇[

2

𝑎(1+𝑒) -

1

𝑎]

= 𝜇[2−1−𝑒

𝑎(1+𝑒) ]

= 𝜇[1−𝑒

𝑎(1+𝑒) ]

Dividing we get

𝑣1

2

𝑣22 =

(1+𝑒)(1+𝑒)

(1−𝑒)(1−𝑒) =

(1+𝑒)2

(1−𝑒)2

Or v1

v2 =

(1+𝑒)

(1−𝑒)

⸫ (1-e) v1= (1+e) v2

Proved.

Que 3 : A particle moves with central acceleration

𝜇

(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒)2 It is projected with velocity V at a distance R.

Show that its path is a rectangular hyperbola if the

angle of projection is sin-1 𝜇

𝑉𝑅(𝑉2− 2𝜇

𝑅)

12

Sol:

We know that v = ℎ

𝑝 → (1)

From the fig. let 𝛼 be the required angle of projection

p = R sin𝛼 and v = V (given)

Also h = √𝜇 𝑙 = √𝜇.𝑏2

𝑎 =√(

𝜇 𝑎2

𝑎) = √(𝜇𝑎)

Because in a rectangular hyperbola, b = a

From (1)

⸫ V = √𝜇 𝑎

𝑅 𝑠𝑖𝑛𝛼

P

R

𝛼

𝛼

S

v ∅

Or sin𝛼 = √𝜇 𝑎

𝑅 𝑉 → (2)

In the case of a hyperbolic orbit

v2 = 𝜇 [ 2

𝑟 +

1

𝑎]

When r =R, v =V (given)

⸫ V2 = 𝜇 [ 2

𝑅 +

1

𝑎]

Or V2 - 2𝜇

𝑅 =

𝜇

𝑎

Or a = 𝜇

(𝑉2− 2𝜇

𝑅 )

Or √𝑎 = √𝜇

(𝑉2− 2𝜇

𝑅 )

From (2)

Sin𝛼 = √𝜇

𝑉𝑅 √

𝜇

(𝑉2− 2𝜇

𝑅 )

= 𝜇

𝑉𝑅(𝑉2− 2𝜇

𝑅)

12

𝛼 = sin-1[𝜇

𝑉𝑅(𝑉2− 2𝜇

𝑅)

12

Proved.