Orbits under inverse-square central force

At a Glance

• Frequency: 2 sub-parts across 2 of 13 years (2019, 2020)
• Priority tier: T3
• Marks (count): 10 (1), 15 (1)
• Average solve time: ~12 min
• Difficulty mix: medium 2
• Section: B | Dominant type: proof

Why This Chapter Matters

The inverse-square orbit is one of the jewels of classical mechanics, and UPSC tests exactly two aspects of it: the proof that the orbit is a conic (2019, 15 marks) and the least-distance calculation from an impact parameter (2020, 10 marks). The 2019 proof requires the Binet equation, which takes 10–15 minutes if you know it cold, and is essentially unworkable otherwise. Mastering the $u = 1/r$ substitution and the orbit classification by energy gives you a secure 25 marks in the years these questions appear.

Minimum Theory

Central force and constants of motion. Under a central force $\mathbf{F} = f(r)\hat r$, two quantities are conserved: the angular momentum $h = r^2\dot\theta$ (per unit mass) and the total energy $E = \tfrac12(\dot r^2 + r^2\dot\theta^2) + V(r)$. The motion is confined to a plane.

Binet equation. Substitute $u = 1/r$ and use $r^2\dot\theta = h$ to write $\dot r = -h\,du/d\theta$ and $\ddot r = -h^2u^2\,d^2u/d\theta^2$. The radial equation $\ddot r - r\dot\theta^2 = f(r)/m$ becomes the Binet equation:

$\frac{d^2u}{d\theta^2} + u = -\frac{f(1/u)}{h^2 u^2}.$

For an attractive inverse-square force $f(r) = -\mu/r^2$, the right side is $\mu/h^2$ (constant), giving a linear ODE.

The orbit equation. The Binet ODE $u'' + u = \mu/h^2$ has general solution

$u = \frac{\mu}{h^2} + A\cos(\theta - \theta_0) \qquad\Longrightarrow\qquad r = \frac{\ell}{1+e\cos(\theta-\theta_0)},$

where $\ell = h^2/\mu$ (semi-latus rectum) and $e = Ah^2/\mu$ (eccentricity). This is the polar equation of a conic with focus at the centre of force.

Orbit classification by energy. The eccentricity satisfies $e = \sqrt{1 + 2Eh^2/\mu^2}$. Hence:

$e < 1 \iff E < 0 \quad(\text{ellipse, bound orbit})$

$e = 1 \iff E = 0 \quad(\text{parabola, escape threshold})$

$e > 1 \iff E > 0 \quad(\text{hyperbola, unbound orbit})$

Question Archetypes

conic-orbit-proof (1 question(s); 2019)

Recognition Cues

• “Prove the path of a planet moving under acceleration $\mu/r^2$ toward a fixed point is a conic section.”
• “Find conditions for (i) ellipse (ii) parabola (iii) hyperbola.”
• The 15-mark weight means you must prove the Binet equation, solve it, and classify.

Solution Template

1. State that the central force implies planar motion with $r^2\dot\theta = h$ (constant).
2. Substitute $u = 1/r$, write $\dot r = -h\,du/d\theta$ and $\ddot r = -h^2u^2\,u''$.
3. Derive the Binet equation $u'' + u = \mu/h^2$.
4. Write the general solution $u = \mu/h^2 + A\cos(\theta-\theta_0)$.
5. Identify $r = \ell/(1+e\cos\theta)$ as a conic with $\ell = h^2/\mu$, $e = Ah^2/\mu$.
6. State the energy expression $e = \sqrt{1+2Eh^2/\mu^2}$ and classify by sign of $E$.

Worked Example

2019 Paper 1, 2019-P1-Q8b (15 marks)

Prove that the path of a planet moving so that its acceleration is always directed to a fixed point (star) and equals $\mu/(\text{distance})^2$ is a conic section. Find the conditions under which the path becomes (i) ellipse, (ii) parabola and (iii) hyperbola.

Step 1 — Central force: planar motion. The force $\ddot{\mathbf{r}} = -(\mu/r^2)\hat r$ is central, so $\mathbf{r}\times\dot{\mathbf{r}} = \text{const}$. The motion is planar and $r^2\dot\theta = h$ (constant angular momentum per unit mass).

Step 2 — Binet substitution. Set $u = 1/r$. Using $r^2\dot\theta = h$:

$\dot r = -h\frac{du}{d\theta},\qquad \ddot r = -h^2u^2\frac{d^2u}{d\theta^2}.$

The radial equation $\ddot r - r\dot\theta^2 = -\mu/r^2 = -\mu u^2$ becomes

$-h^2u^2u'' - h^2u^3 = -\mu u^2.$

Divide by $-h^2u^2$:

$\frac{d^2u}{d\theta^2} + u = \frac{\mu}{h^2}.$

Step 3 — General solution. This is a linear second-order ODE with constant coefficients and constant forcing. Particular solution: $u = \mu/h^2$. Homogeneous solution: $A\cos(\theta-\theta_0)$. Hence

$u = \frac{\mu}{h^2} + A\cos(\theta-\theta_0).$

Step 4 — Orbit equation. Since $r = 1/u$:

$r = \frac{h^2/\mu}{1 + (Ah^2/\mu)\cos(\theta-\theta_0)} = \frac{\ell}{1+e\cos(\theta-\theta_0)},$

where $\ell = h^2/\mu$ and $e = Ah^2/\mu$. This is the focus-directrix polar equation of a conic with the star at the focus. Hence the orbit is a conic. $\blacksquare$

Step 5 — Classification. The energy per unit mass is $E = \tfrac12(\dot r^2 + r^2\dot\theta^2) - \mu/r$. Using the orbit solution, the standard result is

$e = \sqrt{1 + \frac{2Eh^2}{\mu^2}}.$

$\boxed{(i)\ E<0 \implies e<1 \text{ (ellipse).}\qquad (ii)\ E=0 \implies e=1 \text{ (parabola).}\qquad (iii)\ E>0 \implies e>1 \text{ (hyperbola).}}$

Common Traps

• The Binet substitution requires $r^2\dot\theta = h$ throughout — state this conservation law explicitly before dividing.
• The reductions $\dot r = -h\,u_\theta$ and $\ddot r = -h^2u^2 u_{\theta\theta}$ must be derived, not just quoted. They take two lines using the chain rule and the $r^2\dot\theta = h$ relation.
• The constant $e = Ah^2/\mu$ — state this identification clearly so the examiner can see $r = \ell/(1+e\cos\theta)$.
• Give conditions in both $e$ form and energy $E$ form — both appear in the question, and both earn marks.

orbit-least-distance (1 question(s); 2020)

Recognition Cues

• “Particle starts at great distance with velocity $V$”; “perpendicular from centre on initial path is $p$.”
• “Find the least distance from the centre” or “show $V^2\lambda = \sqrt{\mu^2+p^2V^4}-\mu$.”
• Two conserved quantities: angular momentum $h = pV$ and energy $E = \tfrac12 V^2$.

Solution Template

1. Angular momentum from initial data: $h = pV$ (impact parameter times speed at infinity).
2. Energy at infinity: $E = \tfrac12 V^2$ (potential energy $\to 0$ as $r\to\infty$).
3. At closest approach $r = \lambda$, $\dot r = 0$ so all KE is transverse: $v = h/\lambda$.
4. Energy conservation at $r=\lambda$: $\tfrac12(pV/\lambda)^2 - \mu/\lambda = \tfrac12 V^2$.
5. Rearrange to a quadratic in $\lambda$, solve for the positive root.

Worked Example

2020 Paper 1, 2020-P1-Q7c-ii (10 marks)

A particle starts at a great distance with velocity $V$. Let $p$ be the length of the perpendicular from the centre of a star on the tangent to the initial path of the particle. Show that the least distance of the particle from the centre of the star is $\lambda$, where $V^2\lambda = \sqrt{\mu^2+p^2V^4}-\mu$.

Step 1 — Angular momentum. At great distance, speed $V$ and impact parameter $p$, so

$h = pV.$

Step 2 — Energy at infinity. As $r\to\infty$, potential $-\mu/r\to 0$:

$E = \tfrac12 V^2.$

Step 3 — At closest approach. At $r = \lambda$, $\dot r = 0$, so speed $= h/\lambda = pV/\lambda$. Energy conservation:

$\frac{1}{2}\!\left(\frac{pV}{\lambda}\right)^2 - \frac{\mu}{\lambda} = \frac{1}{2}V^2.$

Step 4 — Solve. Multiply by $2\lambda^2$: $p^2V^2 - 2\mu\lambda = V^2\lambda^2$, i.e.

$V^2\lambda^2 + 2\mu\lambda - p^2V^2 = 0.$

Taking the positive root:

$\lambda = \frac{-\mu + \sqrt{\mu^2 + p^2V^4}}{V^2},$

$\boxed{V^2\lambda = \sqrt{\mu^2+p^2V^4} - \mu.}$

Common Traps

• Impact parameter $p$ is the perpendicular distance from the centre to the initial line of motion (not to any point on the orbit) — hence $h = pV$ exactly.
• At closest approach $\dot r = 0$; all velocity is transverse: $v_\perp = h/\lambda$. A common error is writing $v_\perp = V$ instead.
• There are two roots of the quadratic; choose the positive one (a distance must be positive). The negative root is unphysical.
• Check: as $\mu\to 0$, $V^2\lambda \to pV^2$, so $\lambda\to p$ — the particle goes straight through with no deflection, as expected.

Marks-Aware Writing

A 15-mark answer (conic proof) must: derive $r^2\dot\theta = h$; carry out the Binet substitution step by step; solve the ODE and identify the conic; provide the energy classification in both $e$ and $E$ terms. The derivation of the Binet equation alone is worth about 6 marks — do not skip steps.

A 10-mark answer (least distance) must: identify $h = pV$ and $E = \tfrac12 V^2$; write the energy equation at closest approach; solve the quadratic and choose the correct root. Showing the $\mu\to 0$ sanity check adds a sentence and earns partial-marks insurance.

Practice Set

• 2021-P1-Q5d (10 m) — Orbit problem; apply conservation laws. ()
• 2025-P1-Q5c (10 m) — Central orbit calculation. ()