← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q8b — Step-by-Step Solution

15 marks · Section B

Orbits under inverse-square central force · Dynamics & Statics · asked 2× in 13 yrs · Read the full method →

Question

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

Technique

Central-force ⇒\Rightarrow planar motion, r2θ˙=hr^2\dot\theta=h; substitute u=1/ru=1/r to obtain the linear Binet equation u′′+u=μ/h2u''+u=\mu/h^2; its solution is the polar conic r=l/(1+ecos⁡θ)r=l/(1+e\cos\theta); classify by e≶1  ⟺  E≶0e\lessgtr1\iff E\lessgtr0.

Solution

Step 1 — Central force ⇒\Rightarrow planar motion with constant areal velocity

Take the star at the origin. The acceleration is directed toward the origin with magnitude μ/r2\mu/r^2:

r⃗¨=−μr2 r^.\ddot{\vec r}=-\frac{\mu}{r^2}\,\hat r.

Since the force is central, the angular momentum per unit mass is conserved:

r⃗×r⃗˙=const ⇒ motion lies in a plane, and r2θ˙=h (const).\vec r\times\dot{\vec r}=\text{const}\ \Rightarrow\ \text{motion lies in a plane, and } r^2\dot\theta=h\ (\text{const}).

Step 2 — Radial equation and the u=1/ru=1/r substitution

In plane polar coordinates the radial acceleration is r¨−rθ˙2\ddot r-r\dot\theta^2. With u=1/ru=1/r and using r2θ˙=hr^2\dot\theta=h, the standard reductions give

r˙=−hdudθ,r¨=−h2u2d2udθ2,rθ˙2=h2u3.\dot r=-h\frac{du}{d\theta},\qquad \ddot r=-h^2u^2\frac{d^2u}{d\theta^2},\qquad r\dot\theta^2=h^2u^3.

The radial equation r¨−rθ˙2=−μ/r2=−μu2\ddot r-r\dot\theta^2=-\mu/r^2=-\mu u^2 becomes

−h2u2d2udθ2−h2u3=−μu2 ⟹  d2udθ2+u=μh2 -h^2u^2\frac{d^2u}{d\theta^2}-h^2u^3=-\mu u^2\ \Longrightarrow\ \boxed{\ \frac{d^2u}{d\theta^2}+u=\frac{\mu}{h^2}\ }

(the Binet equation).

Step 3 — Solve and identify the conic

The general solution is

u=μh2+Acos⁡(θ−θ0),u=\frac{\mu}{h^2}+A\cos(\theta-\theta_0),

with A,θ0A,\theta_0 constants of integration. Hence

1r=μh2(1+Ah2μcos⁡(θ−θ0)) ⟹ r=l1+ecos⁡(θ−θ0),\frac1r=\frac{\mu}{h^2}\Big(1+\frac{Ah^2}{\mu}\cos(\theta-\theta_0)\Big) \ \Longrightarrow\ r=\frac{l}{1+e\cos(\theta-\theta_0)},

where

l=h2μ(semi-latus rectum),e=Ah2μ(eccentricity).l=\frac{h^2}{\mu}\quad(\text{semi-latus rectum}),\qquad e=\frac{Ah^2}{\mu}\quad(\text{eccentricity}).

This is the focus–directrix (polar) equation of a conic with focus at the star. Hence the orbit is a conic section. ■\blacksquare

Step 4 — Eccentricity in terms of energy

The energy per unit mass is E=12(r˙2+r2θ˙2)−μrE=\tfrac12\big(\dot r^2+r^2\dot\theta^2\big)-\dfrac\mu r. Evaluating with the orbit solution (standard result) gives

e= 1+2Eh2μ2 .e=\sqrt{\,1+\frac{2Eh^2}{\mu^2}\,}.

Equivalently, by the vis-viva relation v2=μ(2r−1a)v^2=\mu\big(\tfrac2r-\tfrac1a\big) (with aa the semi-major axis, E=−μ/2aE=-\mu/2a), the orbit type reduces to the sign of EE.

Step 5 — Conditions for the three conics

The type is governed by ee (equivalently by the total energy EE):

Answer

(i) Ellipse:e<1  ⟺  E<0,(ii) Parabola:e=1  ⟺  E=0,(iii) Hyperbola:e>1  ⟺  E>0.\boxed{ \begin{aligned} &\text{(i) Ellipse:} && e<1 \iff E<0,\\ &\text{(ii) Parabola:} && e=1 \iff E=0,\\ &\text{(iii) Hyperbola:} && e>1 \iff E>0. \end{aligned}}
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