9. Mechanics — Dynamics

Central orbits

Central orbits describe the motion of a particle under a force always directed towards or away from a fixed point (the center of force), governed by fundamental conservation laws of angular momentum and energy.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Differential Equation of a Central Orbit (Binet's Form and Polar Form)

Let a particle of unit mass move under a central force $P$ per unit mass directed towards the pole $O$. The differential equation of the central orbit in polar coordinates $(r, \theta)$ is given by $$h^2 \left[ \frac{d^2r}{d\theta^2} - \frac{2}{r} \left(\frac{dr}{d\theta}\right)^2 + r \right] = \frac{P}{u^2}$$ or in terms of $u = \frac{1}{r}$ and pedal variable $p$: $$h^2 \left( \frac{d^2u}{d\theta^2} + u \right) = P = \frac{h^2}{p^3} \frac{dp}{dr}$$, where $h = r^2 \dot{\theta}$ is the constant areal velocity (angular momentum per unit mass).

📐 Geometric Interpretation

The trajectory is a plane curve where the radius vector sweeps out equal areas in equal times ($\frac{1}{2} r^2 \dot{\theta} = \frac{h}{2} = \text{constant}$). The pedal equation $p = f(r)$ links the perpendicular from the pole to the tangent with the radius vector.

⚙️ Algebraic & Structural Meaning

Transforms a complex vector equation of motion into a second-order nonlinear ordinary differential equation in $u$ and $\theta$, or a first-order separable relation in $p$ and $r$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Central Force Condition	`Domain of motion $r \in (0, \infty)$`	Strictly Mandatory	If the line of action of the force does not pass through a fixed center, torque is non-zero, and angular momentum $h$ is not conserved.
Smoothness of Orbit Curve	`Differentiable on trajectory interval`	Strictly Mandatory	Discontinuities or sharp corners in $r(\theta)$ lead to undefined second derivatives $\frac{d^2u}{d\theta^2}$, rendering Binet's equation inapplicable.
Non-zero Areal Velocity	$h \neq 0$	Essential for conclusion	If $h = 0$, motion is purely rectilinear (radial), causing the denominator terms involving $h$ to degenerate into radial free-fall.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Inverse-Square Law Force ($P = \frac{\mu}{r^2}$)

$$u = \frac{1}{l} [1 + e \cos(\theta - \alpha)]$$

Rule: For a force inversely proportional to $r^2$ (like gravity), the orbit is always a conic section with origin at a focus, where $l$ is semi-latus rectum and $e$ is eccentricity.

Application: Frequently asked in UP PGT to immediately identify conic orbits from force laws.

Velocity at any point in a Central Orbit

$$v^2 = \hbar^2 \left[ \left(\frac{du}{d\theta}\right)^2 + u^2 \right] = \frac{2}{m} \int P dr + C$$

Rule: Velocity can be found directly from the energy integral without integrating the full differential equation.

Application: Direct numerical questions in KVS/DSSSB PGT calculating speed at periapsis and apoapsis.

Apsidal Distance Condition

$$\frac{dr}{d\theta} = 0 \quad \text{or} \quad \frac{du}{d\theta} = 0$$

Rule: At apses (nearest or furthest points from the center), the tangent is perpendicular to the radius vector, so $p = r$.

Application: Standard question pattern: 'Find the apsidal distance for the given central orbit'.

Pedal Equation to Force Law

$$P = \frac{h^2}{p^3} \frac{dp}{dr}$$

Rule: Given $p = f(r)$, take logarithmic derivative or differentiate directly to find $P$ in seconds.

Law of Force for Equiangular Spiral ($r = a e^{\theta \cot \alpha}$)

$$P = \frac{h^2 \csc^2 \alpha}{r^3}$$

Rule: Memorize standard curves: For $r^n = a^n \cos(n\theta)$, $P \propto r^{2n+3}$ or $u^{2n+3}$. For $r = a e^{\theta \cot \alpha}$, $P \propto \frac{1}{r^3}$.

Application: Direct memory-based question in UP PGT Mathematics.

Apsidal Angle $\psi$

$$\psi = \int_{\theta_1}^{\theta_2} d\theta = \frac{\pi}{\sqrt{2k + 2}} \quad \text{(for } P = \mu u^n \text{)}$$

Rule: Angle between two consecutive apses determines whether the orbit is apsidally stable or closed.

Application: Advanced PGT question testing apsidal angle formula.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Confusing $u$ with $r$ in Binet's Equation

Common Pitfall: Substituting $r$ directly into $h^2(\frac{d^2u}{d\theta^2} + u) = P$ without taking $u = \frac{1}{r}$.

Counterexample: Taking $\frac{d^2r}{d\theta^2}$ instead of $\frac{d^2u}{d\theta^2}$ when $P$ is given in terms of $r$.

Why it fails: Leads to completely erroneous differential equations and mismatch with options.

🛡️ Exam Defense: Always explicitly write $u = 1/r$, $dr = -u^{-2} du$ before substituting into Binet's equation.

⚠️ COMMON ERROR #02

Misinterpreting Apsidal Distance as Mean Radius

Common Pitfall: Assuming apsidal distances occur when $\frac{d^2u}{d\theta^2} = 0$ instead of $\frac{du}{d\theta} = 0$.

Counterexample: Equating curvature instead of radial velocity extrema.

Why it fails: Apses are turning points of radius vector ($dr/d\theta = 0$), not points of inflection.

🛡️ Exam Defense: Always set $\frac{du}{d\theta} = 0$ to find apsidal distances $r$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Binet's Polar Equation	Central force, unit mass, $u = 1/r$	$h^2(\frac{d^2u}{d\theta^2} + u) = P$	Relates inverse radius to central force	Given $r$ as function of $\theta$
Pedal Equation Method	Central force, $p = f(r)$	$P = \frac{h^2}{p^3} \frac{dp}{dr}$	Relates perpendicular on tangent to force	Given relation between $p$ and $r$
Velocity in Central Orbit	Conservative central field	$v^2 = \dot{r}^2 + r^2\dot{\theta}^2$	Total speed squared decomposition	Asked to find velocity at distance $r$
Apsidal Distance	Extremum of radius vector	$u' = 0$ or $\frac{dr}{d\theta} = 0$	Nearest and furthest points from center	Keywords: 'apsidal distance', 'periapsis'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If the law of force towards the pole is $\frac{\mu}{r^3}$, find the differential equation of the orbit and the form of the orbit.

(A) Circle(B) Equiangular spiral(C) Parabola(D) Ellipse

📚 Method A: Formal Textbook Derivation

Step 1: Substitute $P = \mu u^3$ into Binet's equation $h^2(\frac{d^2u}{d\theta^2} + u) = \frac{\mu u^3}{u^2} = \mu u$. Step 2: Simplify to $\frac{d^2u}{d\theta^2} + (1 - \frac{\mu}{h^2})u = 0$. Step 3: Solving this linear differential equation yields exponential/spiral solutions for appropriate constants.

⚡ Method B: 25-Second Exam Shortcut

Recognize the standard result: Force proportional to $r^{-3}$ always corresponds to the equiangular spiral $r = a e^{\theta \cot \alpha}$ in central orbits.

💡 Tip: Memorize force laws mapped to curves: $r^{-2} \to \text{Conic}$, $r^{-3} \to \text{Spiral}$, $r \to \text{Ellipse/Harmonic}$. Target 10-second recall.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Find the law of force towards the pole for the orbit $r^n = a^n \cos(n\theta)$.

(A) $P \propto u^2$(B) $P \propto u^3$(C) $P \propto u^{2n+3}$(D) $P \propto u^{n+1}$

📚 Method A: Formal Textbook Derivation

Step 1: Take log of $r^n = a^n \cos(n\theta) \implies n \log r = n \log a + \log(\cos(n\theta))$. Step 2: Use pedal equation conversion $p = r \sin\psi$. Step 3: Relate $p$ and $r$ to get $p^2 = \frac{r^{n+1}}{a^n}$ or similar, then apply $P = \frac{h^2}{p^3} \frac{dp}{dr}$ to arrive at $P \propto r^{-(2n+3)} = u^{2n+3}$.

⚡ Method B: 25-Second Exam Shortcut

Standard formula for $r^n = a^n \cos(n\theta)$ or $r^n = a^n \sin(n\theta)$ is always $P \propto u^{2n+3}$.

💡 Tip: Direct standard textbook result often asked verbatim in UP PGT.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The differential equation of a central orbit in polar coordinates using $u = \frac{1}{r}$ is given by:

A $h^2(\frac{d^2u}{d\theta^2} + u) = P$

B $h^2(\frac{d^2u}{d\theta^2} - u) = P$

C $h(\frac{d^2u}{d\theta^2} + u) = P/u^2$

D $h^2\frac{d^2u}{d\theta^2} = P$

Derivation & Analysis:

Binet's standard differential equation for central orbits is $h^2(\frac{d^2u}{d\theta^2} + u) = \frac{P}{u^2} = P r^2$, which simplifies to $h^2(\frac{d^2u}{d\theta^2} + u) = P$ when $P$ denotes force per unit mass divided by $u^2$ (or directly expressed as $P$). Option (A) is the standard textbook form.

MCQ 2: If the central orbit is a circle passing through the center of force (pole) with diameter $d$, the law of force is proportional to:

A $r$

B $r^{-2}$

C $r^{-3}$

D $r^{-5}$

Derivation & Analysis:

For a circle passing through the pole, $r = a \cos\theta$. Using Binet's equation or pedal equation $p = \frac{r^2}{a}$, we find $P = \frac{h^2}{p^3}\frac{dp}{dr} \propto r^{-5}$. Thus $P \propto u^5$ or $r^{-5}$.

MCQ 3: The areal velocity $h$ of a particle moving in a central orbit is given by:

A $r^2 \dot{\theta}$

B $r \dot{\theta}^2$

C $r \ddot{\theta}$

D $\frac{1}{2} r^2 \theta$

Derivation & Analysis:

Areal velocity is defined as $\frac{dA}{dt} = \frac{1}{2} r^2 \dot{\theta} = \text{constant} = \frac{h}{2}$, so $h = r^2 \dot{\theta}$.

MCQ 4: At an apse of a central orbit, the value of $\frac{dr}{d\theta}$ is:

A 1

B 0

C Infinity

D Undefined

Derivation & Analysis:

An apse is a point where the radius vector is a maximum or minimum, which mathematically corresponds to $\frac{dr}{d\theta} = 0$ (or $\frac{du}{d\theta} = 0$).

MCQ 5: If the central force is $\mu u^2$ ($P = \frac{\mu}{r^2}$), the orbit is a:

A Circle

B Parabola

C Conic section (Ellipse/Parabola/Hyperbola)

D Equiangular spiral

Derivation & Analysis:

An inverse-square law of force ($P = \mu / r^2$) always yields a conic section with focus at the center of force.

MCQ 6: The pedal equation of a central orbit is given by $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} - \frac{r}{a^2b^2}$... wait, more simply, if $p^2 = ar$, then the law of force $P$ is proportional to:

A $r^{-3}$

B $r^{-4}$

C $r^{-2}$

D $r$

Derivation & Analysis:

Using $P = \frac{h^2}{p^3} \frac{dp}{dr}$. If $p^2 = ar$, then $2p \frac{dp}{dr} = a \implies \frac{dp}{dr} = \frac{a}{2p}$. Thus $P = \frac{h^2}{p^3} \left(\frac{a}{2p}\right) \propto \frac{1}{p^4} \propto r^{-2}$? Wait, let's check: if $p^2 = ar$, then $p^3 = a^{3/2} r^{3/2}$ or similar. More precisely, $p^4 = a^2 r^2 \implies P \propto r^{-4}$ when evaluated carefully.

MCQ 7: If the orbit is a circle $r = 2a \cos\theta$ with center of force at the pole, the law of force is:

A $\frac{\mu}{r^2}$

B $\frac{\mu}{r^3}$

C $\frac{\mu}{r^5}$

D $\mu r$

Derivation & Analysis:

For $r = 2a \cos\theta$, $u = \frac{1}{2a}\sec\theta$. Applying Binet's equation yields $P \propto u^5 = \frac{1}{r^5}$.

MCQ 8: Assertion (A): In any central orbit, angular momentum about the center of force is conserved. Reason (R): The central force passes through the center, producing zero torque.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Since the line of action of the central force passes through the pole, the torque $\vec{\tau} = \vec{r} \times \vec{F} = \vec{0}$, which implies conservation of angular momentum ($\vec{h} = \text{constant}$). Thus both are true and (R) explains (A).

MCQ 9: The angle $\psi$ between the radius vector and the tangent to the central orbit is given by:

A $\sin\psi = \frac{p}{r}$

B $\cos\psi = \frac{p}{r}$

C $\tan\psi = r \frac{d\theta}{dr}$

D $\psi = \theta$

Derivation & Analysis:

By definition of the pedal equation, $p = r \sin\psi$, which gives $\sin\psi = \frac{p}{r}$.

Q10

MCQ 10: If the velocity of a particle in a central orbit at distance $r$ from the center is $v$, then the transverse component of velocity is:

A $r \dot{\theta}$

B $\dot{r}$

C $\sqrt{\dot{r}^2 + r^2\dot{\theta}^2}$

D $r^2 \dot{\theta}$

Derivation & Analysis:

In polar coordinates, velocity has two orthogonal components: radial component $\dot{r}$ along the radius vector, and transverse component $r \dot{\theta}$ perpendicular to the radius vector.

Common questions

Binet's form converts reciprocal radius $u = 1/r$ into a linear second-order differential equation with constant coefficients for standard force laws, making integration and solution identification much faster than dealing with $r$ and its derivatives directly.

Confusing apsidal distance conditions ($\frac{du}{d\theta} = 0$) with general orbital inflection points, or miscalculating the exponent of $r$ when converting pedal equations to force laws.

Memorize the standard force law mappings: $P \propto r^{-2}$ (Conics/Kepler), $P \propto r^{-3}$ (Equiangular Spiral), $P \propto r$ (Harmonic oscillator/Ellipse centered at origin), and $P \propto r^{-5}$ (Circle through pole).

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