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UPSC 2021 Maths Optional Paper 1 Q5d — 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

Planet in circular orbit around Sun is suddenly stopped. Find time to fall into Sun. Also find ratio of falling time to revolution period.

Technique

Energy conservation for radial fall; substitute $r=r_0\sin^2\phi$ to handle the singular integrand at $r=r_0$ (top of fall).

Solution

Setup. Planet at distance $r_0$ from Sun (radius of original circular orbit). When stopped, it falls radially inward under gravity $F=-GMm/r^2$.

Step 1 — Energy method for fall

At distance $r$: $\dfrac{1}{2}m\dot r^2-\dfrac{GMm}{r}=E$.

Initial (at $r=r_0$, $\dot r=0$): $E=-\dfrac{GMm}{r_0}$.

So $\dfrac{1}{2}\dot r^2=\dfrac{GM}{r}-\dfrac{GM}{r_0}=GM\dfrac{r_0-r}{r r_0}$.

$\dot r=-\sqrt{\dfrac{2GM(r_0-r)}{r r_0}}$ (negative, falling).

Step 2 — Separate and integrate

$T_{\text{fall}}=\int dt=\int_{r_0}^0\dfrac{dr}{\dot r}=\int_0^{r_0}\sqrt{\dfrac{r r_0}{2GM(r_0-r)}}\,dr=\sqrt{\dfrac{r_0}{2GM}}\int_0^{r_0}\sqrt{\dfrac{r}{r_0-r}}\,dr$.

Step 3 — Evaluate the integral

Substitute $r=r_0\sin^2\phi$, $dr=2r_0\sin\phi\cos\phi\,d\phi$. When $r=0,\phi=0$; $r=r_0,\phi=\pi/2$.

$r_0-r=r_0\cos^2\phi$. $\sqrt{r/(r_0-r)}=\sin\phi/\cos\phi=\tan\phi$.

$\int_0^{r_0}\sqrt{r/(r_0-r)}\,dr=\int_0^{\pi/2}\tan\phi\cdot 2r_0\sin\phi\cos\phi\,d\phi=2r_0\int_0^{\pi/2}\sin^2\phi\,d\phi=2r_0\cdot\pi/4=\pi r_0/2$.

So $T_{\text{fall}}=\sqrt{\dfrac{r_0}{2GM}}\cdot\dfrac{\pi r_0}{2}=\dfrac{\pi}{2}\sqrt{\dfrac{r_0^3}{2GM}}$.

Step 4 — Compare with orbital period

Kepler’s third law: $T_{\text{rev}}=2\pi\sqrt{\dfrac{r_0^3}{GM}}$.

Ratio: $\dfrac{T_{\text{fall}}}{T_{\text{rev}}}=\dfrac{(\pi/2)\sqrt{r_0^3/(2GM)}}{2\pi\sqrt{r_0^3/(GM)}}=\dfrac{1}{4}\cdot\dfrac{1}{\sqrt 2}=\dfrac{1}{4\sqrt 2}=\dfrac{\sqrt 2}{8}$.

Answer