UPSC 2024 Maths Optional Paper 1 Q8c — 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 moves with central acceleration $\mu\!\left(\dfrac{3}{r^3}+\dfrac{d^2}{r^5}\right)$ , projected from distance $d$ at $45°$ to the radius with velocity equal to that in a circular orbit at the same distance. Prove that the time to reach the centre of force is $\dfrac{d^2}{\sqrt{2\mu}}\!\left(2-\dfrac{\pi}{2}\right)$ .

Technique

Energy and angular momentum conservation; effective-potential reduction; recognise a perfect-square structure in $\dot r^2$ ; standard $r^2/(r^2+d^2)$ integral.

Solution

Step 1 — Circular orbit speed.

Central acceleration at radius $d$ : $a(d)=\mu(3/d^3+d^2/d^5)=4\mu/d^3$ .

Circular orbit: $v_c^2/d=4\mu/d^3\;\Rightarrow\;v_c^2=4\mu/d^2$ .

Step 2 — Initial conditions and conserved quantities.

At $r=d$ , projected at $45°$ inward with speed $v_c$ :

\dot r_0=-v_c/\sqrt{2}=-\sqrt{2\mu}/d,\qquad v_\theta=v_c/\sqrt{2}.

Angular momentum: $L=d\cdot v_c/\sqrt{2}=\sqrt{2\mu}$ .

Potential $V(r)=-\tfrac{3\mu}{2r^2}-\tfrac{\mu d^2}{4r^4}$ . Energy:

E=\tfrac{1}{2}v_c^2+V(d)=\tfrac{2\mu}{d^2}-\tfrac{3\mu}{2d^2}-\tfrac{\mu}{4d^2}=\frac{\mu}{4d^2}.

Step 3 — Radial equation.

\dot r^2=2(E-V_{\text{eff}}),\quad V_{\text{eff}}=V(r)+\frac{L^2}{2r^2}=-\frac{\mu}{2r^2}-\frac{\mu d^2}{4r^4}.

\dot r^2=\frac{\mu}{2d^2}+\frac{\mu}{r^2}+\frac{\mu d^2}{2r^4}.

Step 4 — Perfect-square recognition.

Multiply by $2r^4/\mu$ :

\frac{2r^4\dot r^2}{\mu}=\frac{r^4}{d^2}+2r^2+d^2=\left(\frac{r^2+d^2}{d}\right)^2.

Therefore $\dot r=-\dfrac{\sqrt\mu}{d\sqrt{2}}\cdot\dfrac{r^2+d^2}{r^2}$ (negative since $r$ decreases).

Step 5 — Time integral.

T=\int_d^0\frac{dr}{\dot r}=\frac{d\sqrt{2}}{\sqrt\mu}\int_0^d\frac{r^2}{r^2+d^2}\,dr.

\int_0^d\frac{r^2}{r^2+d^2}\,dr=\int_0^d\!\!\left(1-\frac{d^2}{r^2+d^2}\right)dr=\bigl[r-d\arctan(r/d)\bigr]_0^d=d-d\cdot\frac{\pi}{4}=\frac{d(4-\pi)}{4}.

T=\frac{d\sqrt{2}}{\sqrt\mu}\cdot\frac{d(4-\pi)}{4}=\frac{d^2}{\sqrt{2\mu}}\cdot\frac{4-\pi}{2}=\frac{d^2}{\sqrt{2\mu}}\!\left(2-\frac{\pi}{2}\right).

Answer

\boxed{\;T=\frac{d^2}{\sqrt{2\mu}}\!\left(2-\frac{\pi}{2}\right).\;}