If a planet were suddenly stopped in its circular orbit, how much time it would take to fall onto the Sun. Assume the planets time-period of revolution as T.

Explanation

For the circular orbit, `T^(2)propr^(3)`. When the planet is falling onto the Sun, we can think of its path as a completely flattened ellipse, whose semi-major axis `=(r)/(2)`.
`rArr(T')^(2)prop((r)/(2))^(3)`
Now, `((T')/(T))^(2)=((r//s)/(r))^(3)`,
`(T')^(2)=(T^2)/(8)`,
`T'=(T)/(2sqrt2)`
The required time is `(T')/(2)`
`rArrt=(T')/(2)=(T)/(4sqrt2)`
(or) `t=(Tsqrt2)/(8)`