UPSC 2015 Maths Optional Paper 1 Q8b — Step-by-Step Solution

13 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 in a plane under a force, towards a fixed centre, proportional to the distance. If the path of the particle has two apsidal distances $a,b$ ( $a>b$ ), then find the equation of the path.

Technique

Central force $\propto r$ ⇒ isotropic harmonic oscillator ⇒ orbits are ellipses centred at the force centre; apsidal distances = lengths of semi-axes.

Solution

Setup. Central force $\vec F=-k r\,\hat r$ (toward origin, magnitude proportional to $r$ ). This is a 2D isotropic harmonic oscillator. Equations of motion:

\ddot{\vec r}=-k\vec r.

In Cartesian:

\ddot x=-kx,\qquad\ddot y=-ky.

General solution:

x(t)=A\cos(\omega t+\alpha),\qquad y(t)=B\cos(\omega t+\beta),\qquad\omega=\sqrt k.

This describes an ellipse centred at the origin (parametric form).

Step 1 — Orient axes along principal directions

By rotating coordinates, choose the ellipse’s principal axes as the $x,y$ axes. Then phases differ by $\pi/2$ :

x=A\cos\omega t,\qquad y=B\sin\omega t.

Equation: $\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$ — ellipse with semi-axes $A,B$ centred at the origin.

Step 2 — Apsidal distances

An apse is a point where $\dot r=0$ (radial velocity vanishes). For a central-force orbit, apses are points where $r$ achieves extrema.

For an ellipse centred at the origin, the maximum and minimum of $r=\sqrt{x^2+y^2}$ :

Maximum: at the end of the semi-major axis. If $A>B$ , this is at $(\pm A,0)$ , with $r=A$ .
Minimum: at the end of the semi-minor axis. At $(0,\pm B)$ , with $r=B$ .

Given two apsidal distances $a>b$ : $a=A,\;b=B$ .

Step 3 — Equation of path

Answer

\boxed{\;\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.\;}