# Prove that time for planet to fall into the Sun equals $\left( \frac { 1 }{ 4\sqrt { 2 } } \right)$ of the planet's period [closed]

If a planet was suddenly stopped in its orbit supposed to be circular, show that it would fall onto the sun in a time $\left( \frac { 1 }{ 4\sqrt { 2 } } \right)$ times its time period.

I am getting stuck with this problem. I have applied conservation of energy and I think it may solve it. Please help. Here's my research on the problem:

• Firstly, ${ T }^{ 2 }=\frac { 4{ \pi }^{ 2 }{ r }^{ 3 } }{ GM }$.

• Assuming $v$ to be the velocity of the planet at a distance $s$ from the sun, we get: $\frac { m{ v }^{ 2 } }{ 2 } +\left( \frac { -GMm }{ s } \right) =-\frac { GMm }{ r }$, where $r$ is the radius of the planet's orbit.

• Hi Swapnil and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. – John Rennie Jan 24 '16 at 10:37
• This is a cool question as I think the solution is cool, don't close it :) – innisfree Jan 24 '16 at 11:42
• @innisfree That the question is creative does not mean that it should not be closed. Furthermore, it has no real physical relevance anyhow. – Bernhard Jan 24 '16 at 11:56
• @bernhard It is of physical relevance! it's a handy trick for calculating how long it'll take something to fall into a massive body. – innisfree Jan 24 '16 at 12:01
• I vote reopen - this is a clear question, the solution of which is of broader physical relevance (time for something to fall under gravity) and pedagogical value (learning Kepler's laws) – innisfree Jan 24 '16 at 13:24

• $T$ stands for the orbital period of the planets orbit;
• $r$ stands for the semi-major axis of its orbit;
• $GM$ can be combined into the standard gravitational parameter of the sun/star (you actually also have to include the mass of the planet, but since neither masses change you can view $GM$ as the effective gravitational parameter).
The orbits of planets are usually nearly circular, so it would not matter where along its initial orbit the planet stopped, it would always be approximately $r$. While slowing down its velocity the planet would still be in orbit, but it would become more and more eccentric, with its highest point (aphelion or apoapsis in general) at distance $r$ and its lowest point (perihelion or periapsis in general) closer and closer to the sun. When all velocity (relative to the sun) have been cancelled then the lowest point would be zero. When travelling from the highest point to the lowest point of an orbit it would take half its orbital period (due to symmetry). And the new orbital period can be derived from the changed semi-major axis, which can be calculated by taking the average of the highest and lowest point.