Central force motion and Kepler’s laws

At a Glance

• Frequency: 6 sub-parts across 6 of 13 years (2015, 2016, 2021, 2023, 2024, 2025)
• Priority tier: T2
• Marks (count): 10 (3), 13 (1), 20 (2)
• Average solve time: ~14 min
• Difficulty mix: easy 1, medium 3, hard 2
• Section: B | Dominant type: derivation

Why This Chapter Matters

Central force motion appears in 6 of the last 13 years and consistently earns high marks (the two 20-mark questions in 2023 and 2024 are among the most demanding in the entire corpus). Three problem types are tested: (1) finding the orbit path by solving Binet’s equation, (2) computing the time for a particle to fall from rest to the centre of force, and (3) applying Kepler’s areal-velocity law to find the time over an arc. The hard questions (2023, 2024) require recognising either a pedal-equation relationship or a perfect-square structure in $\dot r^2$. Mastering the energy–angular-momentum setup and Binet’s equation covers all six questions.

Minimum Theory

Setup. A particle of mass $m$ moves in a plane under a central force $\vec F=-f(r)\hat r$ directed toward the origin $O$. The angular momentum $h=r^2\dot\theta$ is conserved. Substituting $u=1/r$, the Binet orbit equation in polar coordinates is: $h^2u^2\!\left(\frac{d^2u}{d\theta^2}+u\right)=\frac{F(r)}{m}=\frac{f(1/u)}{m},$ where $F=f(r)$ is the magnitude of the force per unit mass. This reduces to a second-order ODE for $u(\theta)$.

Pedal equation. For a central orbit, $h^2=p^2v^2$ (where $p$ is the perpendicular from $O$ to the tangent). Combined with energy conservation $\tfrac{1}{2}v^2=E-V(r)$, the pedal equation $p^2$ as a function of $r$ describes the orbit shape independently of time.

Standard Binet solutions.

Apse condition. At an apse (pericentre or apocentre), the radial velocity vanishes: $\dot r=0$, equivalently $du/d\theta=0$. This pins integration constants.

Energy method for radial fall. If a particle falls radially toward $O$ from rest at $r=r_0$: $\tfrac{1}{2}\dot r^2=\Phi(r)-\Phi(r_0),\qquad \Phi(r)=-V(r)=-\int F\,dr.$ The fall time is $T=\int_{r_0}^0 dr/|\dot r|$ — typically evaluated by a trigonometric substitution.

Kepler’s laws. For a gravitational orbit ($F=GM/r^2$):

• First law: orbits are ellipses with the Sun at a focus.
• Second law (areal velocity): the radius vector sweeps equal areas in equal times. Rate: $dA/dt=h/2=\pi ab/T$.
• Third law: $T^2\propto a^3$, where $a$ is the semi-major axis.

Kepler’s areal law for an arc. Time to traverse arc $= (\text{area swept by radius vector})/(\pi ab/T)$.

Question Archetypes

Three patterns cover every central-force question in the corpus.

central-orbit-path (3 question(s); 2015, 2016, 2023)

Recognition Cues

• “A particle is projected from an apse at distance $a$ with velocity $v$; find the orbit.”
• “Force proportional to $r$” / “varies inversely as the cube” / “central acceleration $f(r)$.”
• “Will the path be the curve $x^4+y^4=c^2$?” — verify the pedal equation.

Solution Template

Binet-equation route (most questions):

1. Write the force per unit mass $F(r)$ in terms of $u=1/r$: express $F$ as a function of $u$.
2. Write Binet’s equation: $h^2u^2(u''+u)=F$. Divide to get an ODE for $u(\theta)$.
3. Identify the ODE type: constant-coefficient (if $F\propto u^n$), and solve. For $F\propto u^3$ the ODE is $u''+\lambda u=0$ with $\lambda=1-\mu/h^2$.
4. Apply apse conditions: $u(0)=1/r_0$ (initial distance) and $u'(0)=0$ (apse).
5. Find $h^2$ from the circular-orbit condition: for a circle of radius $a$, $v_c^2/a=F(a)$, giving $h^2=a^2v_c^2$ or directly from the projected velocity.

Pedal-equation route (2023 type):

1. Compute $v^2=2(E-V(r))$ using energy conservation.
2. Form the pedal equation $p^2=h^2/v^2$ as a function of $r$.
3. Verify the candidate curve by computing its pedal equation directly (via $1/p^2=1/r^2+(dr/d\theta)^2/r^4$) and matching.

Worked Example(s)

2015 Paper 1, 2015-P1-Q8b (13 marks)

Central force $\propto r$; two apsidal distances $a>b$. Find the orbit.

Force per unit mass $F=kr$ (toward $O$). The equations of motion in Cartesian give $\ddot x=-kx$, $\ddot y=-ky$ — a 2D isotropic harmonic oscillator. The general solution is an ellipse centred at the force centre: $x=A\cos\omega t,\quad y=B\sin\omega t,\quad\omega=\sqrt k.$ The apsidal distances are the max and min of $r=\sqrt{x^2+y^2}$, equal to the semi-axes $A=a$ and $B=b$: $\boxed{\;\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.\;}$

Key distinction from Kepler ($F\propto1/r^2$): for $F\propto r$, the force centre is at the centre of the ellipse (not a focus); apsidal distances are the semi-axes.

2016 Paper 1, 2016-P1-Q5e (10 marks, compulsory)

Central acceleration $F=\mu/r^3$; projected from apse at distance $a$ with velocity $\sqrt2\,v_c$, where $v_c$ is the circular speed at $r=a$. Find the orbit.

Step 1 (Binet). $F=\mu u^3$. Binet’s equation: $h^2u^2(u''+u)=\mu u^3$, so $u''+\left(1-\frac{\mu}{h^2}\right)u=0.$

Step 2 ($h^2$ from initial conditions). Circular speed: $v_c^2/a=\mu/a^3$, so $v_c^2=\mu/a^2$. Projection: $v_0^2=2v_c^2=2\mu/a^2$. At the apse velocity is purely transverse, so $h=av_0$: $h^2=a^2v_0^2=2\mu,\quad\text{hence}\quad 1-\frac{\mu}{h^2}=\frac{1}{2}.$

Step 3 (solve ODE). $u''+\tfrac{1}{2}u=0$: solution is $u=A\cos(\theta/\sqrt2)+B\sin(\theta/\sqrt2)$.

Step 4 (apse conditions). $\theta=0$ is an apse: $u(0)=1/a\Rightarrow A=1/a$; $u'(0)=0\Rightarrow B=0$. $u=\frac{1}{a}\cos\frac{\theta}{\sqrt2}\quad\Longrightarrow\quad\boxed{\;r=a\sec\frac{\theta}{\sqrt2}.\;}$

Common Traps

• $\cos(\theta/\sqrt2)$ becomes zero at $\theta=\pi\sqrt2/2$ — the orbit has an asymptote there (particle escapes). This is a spiral to infinity, not a closed orbit.
• For $F\propto r$: the centre of the ellipse is at $O$; apsidal distances = semi-axes.
• For $F\propto 1/r^2$: the ellipse has $O$ at a focus; apsidal distances are $a(1\pm e)$.
• At an apse: $\dot r=0$, so velocity is wholly transverse, giving $h=r_0 v_0$ directly.

orbital-fall-time (2 question(s); 2021, 2024)

Recognition Cues

• “A planet is suddenly stopped; find the time to fall into the Sun.”
• “Projected at $45°$ with speed $v_c$; find the time to reach the centre.”
• Involves setting up an energy integral and integrating with a trig substitution.

Solution Template

1. Energy + angular momentum. Compute $E$ and $h$ from the initial conditions.
2. Radial equation. Write $\tfrac{1}{2}\dot r^2=E-V_{\text{eff}}(r)$ where $V_{\text{eff}}=V(r)+h^2/(2r^2)$.
3. Perfect-square check. Try to express $\dot r^2$ as $f(r)^2\cdot g(r)$ for simple $f,g$ — this unlocks an analytic integral.
4. Integrate. $T=\int_{r_0}^0 dr/|\dot r|$. Use $r=r_0\sin^2\phi$ (for Kepler) or $r^2/d^2+\ldots$ substitutions.
5. Ratio. Compare with $T_{\text{orb}}=2\pi\sqrt{a^3/(GM)}$ to give the ratio.

Worked Example(s)

2021 Paper 1, 2021-P1-Q5d (10 marks, compulsory)

A planet in circular orbit of radius $r_0$ is suddenly stopped. Find the time to fall into the Sun and its ratio to the orbital period.

Radial fall. Force: $F=GM/r^2$. Initially at rest at $r_0$, so $h=0$, $E=-GM/r_0$. $\tfrac{1}{2}\dot r^2=\frac{GM}{r}-\frac{GM}{r_0}=GM\cdot\frac{r_0-r}{rr_0}.$ $T=\int_0^{r_0}\sqrt{\frac{rr_0}{2GM(r_0-r)}}\,dr=\sqrt{\frac{r_0}{2GM}}\int_0^{r_0}\sqrt{\frac{r}{r_0-r}}\,dr.$

Trig substitution. Let $r=r_0\sin^2\phi$; the integral becomes $\int_0^{\pi/2}2r_0\sin^2\phi\,d\phi=\pi r_0/2$: $T=\sqrt{\frac{r_0}{2GM}}\cdot\frac{\pi r_0}{2}=\frac{\pi}{2}\sqrt{\frac{r_0^3}{2GM}}.$

Ratio. Kepler’s third law: $T_{\text{orb}}=2\pi\sqrt{r_0^3/(GM)}$. $\frac{T}{T_{\text{orb}}}=\frac{1}{4\sqrt2}=\frac{\sqrt2}{8}.$

$\boxed{\;T=\frac{\pi}{2}\sqrt{\frac{r_0^3}{2GM}}=\frac{T_{\text{orb}}}{4\sqrt2}.\;}$

Alternative (degenerate ellipse). The fall is half a degenerate ellipse with semi-major axis $a=r_0/2$; its period is $T_{\text{degen}}=2\pi\sqrt{(r_0/2)^3/(GM)}=T_{\text{orb}}/(2\sqrt2)$. Fall time $=T_{\text{degen}}/2=T_{\text{orb}}/(4\sqrt2)$ ✓.

2024 Paper 1, 2024-P1-Q8c (20 marks)

Central acceleration $\mu(3/r^3+d^2/r^5)$; projected from distance $d$ at $45°$ with circular speed. Prove fall time $= d^2(2-\pi/2)/\sqrt{2\mu}$.

Step 1 (initial conditions). Circular speed: $v_c^2/d=a(d)=4\mu/d^3$, so $v_c^2=4\mu/d^2$. Projection at $45°$: $h=d\cdot v_c/\sqrt2=\sqrt{2\mu}$, $E=\mu/(4d^2)$.

Step 2 (radial equation). Effective potential $V_{\text{eff}}=-\mu/(2r^2)-\mu d^2/(4r^4)$: $\dot r^2=2(E-V_{\text{eff}})=\frac{\mu}{2d^2}+\frac{\mu}{r^2}+\frac{\mu d^2}{2r^4}.$

Step 3 (perfect square). Multiply by $2r^4/\mu$: $2r^4\dot r^2/\mu=r^4/d^2+2r^2+d^2=(r^2/d+d)^2=(r^2+d^2)^2/d^2$. So $\dot r=-\frac{(r^2+d^2)}{r^2}\cdot\frac{\sqrt\mu}{d\sqrt2}.$

Step 4 (integrate). $T=\frac{d\sqrt2}{\sqrt\mu}\int_0^d\frac{r^2}{r^2+d^2}\,dr=\frac{d\sqrt2}{\sqrt\mu}\left[r-d\arctan\frac{r}{d}\right]_0^d=\frac{d\sqrt2}{\sqrt\mu}\cdot d\!\left(1-\frac{\pi}{4}\right)=\frac{d^2\sqrt2}{\sqrt\mu}\cdot\frac{4-\pi}{4}.$ $\boxed{\;T=\frac{d^2}{\sqrt{2\mu}}\!\left(2-\frac{\pi}{2}\right).\;}$

Common Traps

• The trig substitution for Kepler ($r=r_0\sin^2\phi$) regularises the singularity at $r=r_0$ where $\dot r=0$.
• The perfect-square step in 2024 is the key: $r^4/d^2+2r^2+d^2=(r^2+d^2)^2/d^2$. Without it the integral has no closed form.
• “Projected at $45°$” means the velocity makes $45°$ with the radial direction, so the radial and transverse components are both $v_c/\sqrt2$.

kepler-areal (1 question(s); 2025)

Recognition Cues

• “Earth travels over half its orbit [the far half / near half]; how much longer than $T/2$?”
• Kepler’s second law: equal areas in equal times.

Solution Template

1. Identify the two parts of the orbit (split by the minor axis or another chord).
2. Compute the area swept by the radius vector from the focus for each part. For the far half: $A_{\text{far}}=\tfrac{1}{2}\pi ab+\text{triangle}$.
3. Time $\propto$ area: $t=A\cdot(T/\pi ab)$.
4. Substitute $e$ and $T$ numerically.

Worked Example(s)

2025 Paper 1, 2025-P1-Q5c (10 marks, compulsory)

Prove that the Earth takes approximately 2 days longer over the half of its orbit remote from the Sun (separated by the minor axis), given $e=1/60$.

Ellipse: semi-axes $a,b$, Sun $S$ at focus $(−ae,0)$ (taking centre at origin). The minor axis ($x=0$) cuts the orbit into near and far halves. Endpoints of minor axis: $L=(0,b)$, $L'=(0,-b)$.

Area of far half. The far arc ($x<0$) plus radii $SL$, $SL'$ sweeps: $A_{\text{far}}=\underbrace{\tfrac{\pi ab}{2}}_{\text{half-ellipse}(x<0)}+\underbrace{\triangle SLL'}_{\text{added since }S\text{ is on near side}}.$ Triangle $SLL'$ has base $2b$ and height $ae$: $\triangle SLL'=abe$.

Time for far half. $t_{\text{far}}=\frac{A_{\text{far}}}{\pi ab/T}=T\left(\frac{1}{2}+\frac{e}{\pi}\right)=\frac{T}{2}+\frac{Te}{\pi}.$

Extra time. $\Delta t=\frac{Te}{\pi}=\frac{365.25\times(1/60)}{\pi}\approx1.94\approx\boxed{2\text{ days.}}$

Common Traps

• The triangle $SLL'$ is added for the far half (focus is on the near side). For the near half, the triangle would be subtracted.
• Use $t=A/(h/2)=A\cdot T/(\pi ab)$; don’t confuse $h=\pi ab/T$ (per year) with the angular momentum $h$ in orbit equations.
• The answer $\Delta t=Te/\pi$ holds exactly; the “2 days” is the numerical approximation for Earth’s parameters.

Marks-Aware Writing

10-mark and 13-mark questions (2015-Q8b, 2016-Q5e, 2021-Q5d, 2025-Q5c): Structure the solution in clearly labelled steps — setup, key conservation law, ODE/integral, boundary conditions, boxed result. For Binet: write the equation, simplify, solve; 2–3 displayed equations.

20-mark questions (2023-Q8b, 2024-Q8c): These require methodical work — energy, angular momentum, radial equation, and a non-trivial integration step. The 2023 pedal equation verification and the 2024 perfect-square identification are the crux moves — describe them explicitly. Examiners allocate 5–6 marks just for setting up the energy–angular-momentum conservation correctly.

Practice Set