← 2016 Paper 1

UPSC 2016 Maths Optional Paper 1 Q5e — Step-by-Step Solution

10 marks · Section B

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

Question

A particle moves with a central acceleration which varies inversely as the cube of the distance. If it is projected from an apse at a distance $a$ from the origin with a velocity which is $\sqrt2$ times the velocity for a circle of radius $a$, then find the equation to the path.

Technique

Binet’s equation $u''+u=\dfrac{F}{h^2u^2}$ with $F=\mu u^3$; circular-orbit speed $v_c^2=\mu/a^2$; apse condition $u'=0$ to pin the constants.

Solution

Let the central (attractive) acceleration be $F=\dfrac{\mu}{r^3}$ directed toward the centre $O$. Put $u=\dfrac1r$.

Step 1 — Binet’s orbit equation

For a central force the orbit satisfies

where $h=r^2\dot\theta$ is the (constant) angular momentum per unit mass. Dividing by $h^2u^2$,

Step 2 — Velocity condition fixes $h^2$

Circular orbit of radius $a$: the central acceleration supplies the centripetal term, $\dfrac{v_c^2}{a}=\dfrac{\mu}{a^3}$, so $v_c^2=\dfrac{\mu}{a^2}$.

Projection velocity: $v_a^2=2v_c^2=\dfrac{2\mu}{a^2}$.

At an apse the velocity is wholly transverse, so $h=r\,v=a\,v_a$ and

Step 3 — Solve the orbit equation

With (2), equation (1) becomes

whose general solution is $u=A\cos\dfrac{\theta}{\sqrt2}+B\sin\dfrac{\theta}{\sqrt2}$.

Step 4 — Apply the apse conditions

Measure $\theta$ from the apse. There $r=a\Rightarrow u=\tfrac1a$, and the apse condition $\dfrac{dr}{d\theta}=0\Rightarrow\dfrac{du}{d\theta}=0$ at $\theta=0$. From $u'=-\tfrac{A}{\sqrt2}\sin\tfrac{\theta}{\sqrt2}+\tfrac{B}{\sqrt2}\cos\tfrac{\theta}{\sqrt2}$, the condition $u'(0)=0$ gives $B=0$; and $u(0)=\tfrac1a$ gives $A=\tfrac1a$. Hence

Answer