Question

A planet was suddenly stopped in its orbit supposed to be circular. The time it takes to fall on to the sun is, if time period of planet's revolution is $T$

• A. $\frac{T}{2}$
• B. $\frac{\sqrt{2T}}{4}$
• C. $\frac{\sqrt{2T}}{8}$
• D. $\sqrt{2T}$

Answer 1

Answer: (C)

$\frac{\sqrt{2}T}{8}$

Answer 2

The planet is in a circular orbit of radius $R$ with time period $T$. Its velocity is $v_c = \sqrt{GM/R}$. The time period is given by Kepler's Third Law: $T = 2\pi \sqrt{R^3/GM}$.

When the planet's velocity is suddenly reduced to zero, it begins to fall towards the Sun radially. The motion is governed by the gravitational force $F(r) = GMm/r^2$. Using energy conservation, the velocity $v$ at a distance $r$ from the Sun is given by: $\frac{1}{2}mv^2 - \frac{GMm}{r} = -\frac{GMm}{R}$ $v^2 = 2GM\left(\frac{1}{R} - \frac{1}{r}\right)$ Since $v = -\frac{dr}{dt}$, we have: $dt = -\frac{dr}{\sqrt{2GM\left(\frac{1}{R} - \frac{1}{r}\right)}} = -\sqrt{\frac{Rr}{2GM(r-R)}} dr$ To find the time to fall to the Sun ($r=0$), we integrate from $r=R$ to $r=0$: $t_{fall} = \int_{R}^{0} -\sqrt{\frac{R}{2GM}} \frac{\sqrt{r}}{\sqrt{R-r}} dr = \sqrt{\frac{R}{2GM}} \int_{0}^{R} \frac{\sqrt{r}}{\sqrt{R-r}} dr$ The integral $\int_{0}^{R} \frac{\sqrt{r}}{\sqrt{R-r}} dr$ can be solved using the substitution $r = R\sin^2\theta$, which yields $\frac{\pi R}{2}$. $t_{fall} = \sqrt{\frac{R}{2GM}} \cdot \frac{\pi R}{2} = \frac{\pi R^{3/2}}{2\sqrt{2GM}}$ Now, we relate this to the original time period $T$. From Kepler's Third Law, $T^2 = \frac{4\pi^2 R^3}{GM}$, so $\frac{R^3}{GM} = \frac{T^2}{4\pi^2}$. $t_{fall}^2 = \frac{\pi^2 R^3}{8GM} = \frac{\pi^2}{8} \left(\frac{R^3}{GM}\right) = \frac{\pi^2}{8} \left(\frac{T^2}{4\pi^2}\right) = \frac{T^2}{32}$ $t_{fall} = \frac{T}{\sqrt{32}} = \frac{T}{4\sqrt{2}}$ To match the options, we can rationalize the denominator: $t_{fall} = \frac{T}{4\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}T}{8}$ This matches option (C), assuming that the $\sqrt{T}$ in the options is a typo and should be $T$. The options (B) and (C) are dimensionally incorrect as written, but if interpreted as $\frac{\sqrt{2}T}{4}$ and $\frac{\sqrt{2}T}{8}$, then $\frac{\sqrt{2}T}{8}$ is the correct answer.