Question

If a planet was suddenly stopped in its orbit, suppose to be circular, find how much time will it take in falling onto the sun?

• A. $\left(\frac{\sqrt{2}}{8}\right)$ times the period of the planet's revolution
• B. $4\sqrt{2}$ times the period of the planet's revolution
• C. $3\sqrt{2}$ times the period of the planet's revolution
• D. 9 times the period of the planet's revolution

Answer 1

Answer: (A)

Assuming the new path of the planet as an elongated ellipse of semi major axis $\frac{r}{2}$ and using kepler's law we get
$\frac{T_1^2}{T_2^2}=\frac{r^3}{\left(\frac{r}{2}\right)^3}$
$T_{2}=\frac{T_{1}}{\sqrt{8}}$
But this is the total time to orbit the sun. To fall down it takes half the total time. So the required time is
$t=\frac{1}{2}\frac{T_{1}}{\sqrt{8}}=T_{1}\frac{\sqrt{2}}{8}$