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?

Question

TestHub · Accepted Answer

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

$\frac{{T_{1}}^{2}}{{T_{2}}^{2}} = \frac{r^{3}}{{(\frac{r}{2})}^{3}}$

$T_{2} = \frac{T_{1}}{\sqrt{8}}$

But this is the total time to orbit the sun. To fall down it takes half the total time. So the required time is

$t = \frac{1}{2} \frac{T_{1}}{\sqrt{8}} = T_{1} \frac{\sqrt{2}}{8}$

Physics - Gravitation Question with Solution | TestHub