← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q8b — Step-by-Step Solution

20 marks · Section B

Central force motion and Kepler's laws · Dynamics & Statics · asked 6× in 13 yrs · Read the full method →

Question

A particle is projected from an apse at a distance $\sqrt c$ from the centre of force with a velocity $\sqrt{\tfrac{2\lambda}{3}c^3}$ and is moving with central acceleration $\lambda(r^5-c^2 r)$. Find the path of motion of this particle. Will that be the curve $x^4+y^4=c^2$?

Technique

Conservation of energy and angular momentum yields the pedal equation $1/p^2=r^2(3c^2-r^4)/(2c^4)$. Verify the candidate curve $x^4+y^4=c^2$ has the same pedal equation by computing it directly via $1/p^2=1/r^2+(dr/d\theta)^2/r^4$.

Solution

Setup. Central force per unit mass with acceleration $a(r)=\lambda(r^5-c^2 r)$ directed toward centre when positive (radially outward when $\lambda(r^5-c^2 r)>0$; i.e., $r>c$ — repulsive then; for $r<c$, $a<0$ means inward acceleration, so attraction). Hmm — the problem may intend “central acceleration” as the magnitude of the acceleration directed inward; the standard convention varies. Let’s take the radial component of force per unit mass as $-\lambda(r^5-c^2 r)$ (so the acceleration vector is $-\lambda(r^5-c^2 r)\hat r$); then $V(r)$ satisfies $-dV/dr=-\lambda(r^5-c^2 r)$ giving $V(r)=\lambda(r^6/6-c^2 r^2/2)$.

Actually, let’s use the problem’s stated direction (toward centre: positive = inward attraction). Then radial force = $-\lambda(r^5-c^2 r)$. Potential: $V(r)$ with $-dV/dr=-\lambda(r^5-c^2 r)\Rightarrow V(r)=\int\lambda(r^5-c^2 r)\,dr=\lambda(r^6/6-c^2 r^2/2)+\text{const}$.

Set the constant to $0$.

Step 1 — Conservation of energy

At apse $r=\sqrt c$: velocity is purely transverse (radial velocity $\dot r=0$). $|v|^2=\frac{2\lambda c^3}{3}$.

$E=\tfrac{1}{2}|v|^2+V(\sqrt c)=\frac{\lambda c^3}{3}+\lambda\Bigl(\frac{(\sqrt c)^6}{6}-\frac{c^2(\sqrt c)^2}{2}\Bigr)=\frac{\lambda c^3}{3}+\lambda\Bigl(\frac{c^3}{6}-\frac{c^3}{2}\Bigr)=\frac{\lambda c^3}{3}+\lambda\Bigl(\frac{c^3-3c^3}{6}\Bigr)=\frac{\lambda c^3}{3}-\frac{\lambda c^3}{3}=0.$

So $E=0$.

Step 2 — Conservation of angular momentum

$h=rv_\theta$. At apse, $v_r=0$ so $v=v_\theta=\sqrt{2\lambda c^3/3}$. Hence

So $h^2=2\lambda c^4/3$.

Step 3 — Energy in $(r,\dot r)$

$E=\tfrac{1}{2}\dot r^2+\tfrac{h^2}{2r^2}+V(r)=0$.

$\dot r^2=-\dfrac{h^2}{r^2}-2V(r)=-\dfrac{2\lambda c^4}{3r^2}-2\lambda\Bigl(\dfrac{r^6}{6}-\dfrac{c^2 r^2}{2}\Bigr)=-\dfrac{2\lambda c^4}{3r^2}-\dfrac{\lambda r^6}{3}+\lambda c^2 r^2$.

Multiply by $3r^2$:

Factor $r^8-3c^2 r^4+2c^4=(r^4-c^2)(r^4-2c^2)$.

So $3r^2\dot r^2=-\lambda(r^4-c^2)(r^4-2c^2)$.

Step 4 — Use the pédal (or $u=1/r$) approach

Alternative: use $1/p^2=1/r^2+(dr/d\theta)^2/r^4$ where $p$ is perpendicular distance from centre to the tangent. And $h^2=p^2 v^2$.

$v^2=2(E-V)=-2V=2\lambda(c^2 r^2/2-r^6/6)=\lambda c^2 r^2-\lambda r^6/3=(\lambda r^2/3)(3c^2-r^4)$.

So $p^2=h^2/v^2=\dfrac{2\lambda c^4/3}{(\lambda r^2/3)(3c^2-r^4)}=\dfrac{2c^4}{r^2(3c^2-r^4)}$.

This is the pedal equation: $p^2 r^2(3c^2-r^4)=2c^4$, or

Step 5 — Verify $x^4+y^4=c^2$

For the curve $x^4+y^4=c^2$, in polar: $r^4(\cos^4\theta+\sin^4\theta)=c^2$, so $r^4=\dfrac{c^2}{\cos^4\theta+\sin^4\theta}$.

Compute $1/p^2=1/r^2+(dr/d\theta)^2/r^4$. Use $g(\theta)=\cos^4\theta+\sin^4\theta=1-\tfrac{1}{2}\sin^2 2\theta$, so $r^4=c^2/g$, $r=c^{1/2}/g^{1/4}$.

$\dfrac{dr}{d\theta}=-\tfrac{c^{1/2}}{4}g^{-5/4}g'(\theta)$. So $\dfrac{1}{r}\dfrac{dr}{d\theta}=-\dfrac{g'}{4g}$.

$\dfrac{(dr/d\theta)^2}{r^4}=\dfrac{1}{r^2}\Bigl(\dfrac{1}{r}\dfrac{dr}{d\theta}\Bigr)^2=\dfrac{(g')^2}{16 g^2 r^2}$.

$\dfrac{1}{p^2}=\dfrac{1}{r^2}+\dfrac{(g')^2}{16 g^2 r^2}=\dfrac{1}{r^2}\Bigl(1+\dfrac{(g')^2}{16 g^2}\Bigr)=\dfrac{16 g^2+(g')^2}{16 r^2 g^2}$.

Compute $g'(\theta)$: $g=1-\tfrac{1}{2}\sin^2 2\theta$, so $g'=-\sin 2\theta\cos 2\theta\cdot 2=-\sin 4\theta$. Hence $(g')^2=\sin^2 4\theta=4\sin^2 2\theta\cos^2 2\theta=4\sin^2 2\theta(1-\sin^2 2\theta)$.

$g^2=(1-\sin^2 2\theta/2)^2=1-\sin^2 2\theta+\tfrac{1}{4}\sin^4 2\theta$.

$16 g^2+(g')^2=16-16\sin^2 2\theta+4\sin^4 2\theta+4\sin^2 2\theta-4\sin^4 2\theta=16-12\sin^2 2\theta=4(4-3\sin^2 2\theta)$.

So $\dfrac{1}{p^2}=\dfrac{4(4-3\sin^2 2\theta)}{16 r^2 g^2}=\dfrac{4-3\sin^2 2\theta}{4 r^2 g^2}$.

Now express in terms of $r$. From $r^4=c^2/g$: $g=c^2/r^4$, $g^2=c^4/r^8$. So $4r^2 g^2=4c^4/r^6$.

Also $\sin^2 2\theta=2(1-g)=2(1-c^2/r^4)=2(r^4-c^2)/r^4$, so $3\sin^2 2\theta=6(r^4-c^2)/r^4$, and $4-3\sin^2 2\theta=(4r^4-6(r^4-c^2))/r^4=(6c^2-2r^4)/r^4=2(3c^2-r^4)/r^4$.

Substitute:

This matches our derived equation exactly! ✓

So the path is $x^4+y^4=c^2$.

Answer