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UPSC 2020 Maths Optional Paper 1 Q7c-ii — Step-by-Step Solution

10 marks · Section B

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

Question

A particle starts at a great distance with velocity $V$. Let $p$ be the length of the perpendicular from the centre of a star on the tangent to the initial path of the particle. Show that the least distance of the particle from the centre of the star is $\lambda$, where $V^2\lambda=\sqrt{\mu^2+p^2V^4}-\mu$. Here $\mu$ is a constant.

Technique

Central orbit conservation laws: angular momentum $h=pV$ from initial data; energy $E=\tfrac12V^2$ at infinity; set $\dot r=0$ at $r=\lambda$ and solve the resulting quadratic.

Solution

The particle moves under the star’s attraction. Per unit mass, the attractive central force is the inverse-square law $F=\dfrac{\mu}{r^2}$ (directed toward the centre), with potential energy $U(r)=-\dfrac{\mu}{r}$. Two conserved quantities govern the motion: angular momentum and energy (both per unit mass).

Step 1 — Angular momentum (from the initial data)

At a great distance the speed is $V$ and the perpendicular from the centre on the line of motion (the impact parameter) is $p$. Angular momentum per unit mass:

This is constant throughout the motion.

Step 2 — Energy (evaluate at infinity)

At “a great distance” $r\to\infty$, $U\to0$, speed $=V$, so total energy per unit mass:

Step 3 — Condition at least distance

At the closest approach $r=\lambda$, the radial velocity vanishes ($\dot r=0$), so the whole speed is transverse:

Energy conservation at $r=\lambda$:

Step 4 — Solve for $\lambda$

Multiply through by $2\lambda^2$:

Quadratic in $\lambda$; take the positive root (distance $>0$):

Hence

Answer