Can anyone please provide me the proof of the following?
When a planet is made stop at its orbit supposed to be circular with time period $T$ around the sun then it will hit the sun after a time which is equal to the $\frac{\sqrt{2}}{8} T$.
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Qmechanic
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7Possible duplicate of Prove that time for planet to fall into the Sun equals $\left( \frac { 1 }{ 4\sqrt { 2 } } \right)$ of the planet's period – Das Sep 27 '16 at 11:25
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Possible duplicates: http://physics.stackexchange.com/q/14700/2451 and links therein. – Qmechanic Sep 27 '16 at 12:43
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Suppose that the planet was not completely stopped, but still has some very small velocity. In this case it's orbit is a very-very elongated ellipse. The ellipse is not much different from the straight line connecting the initial position of the planet and the Sun.
Time that takes planet to fall onto the Sun is equal to the half of the period of the planet moving along the ellipse.
The period of rotation along the whole ellipse can be easily calculated using the third Keppler's law:
The square of the orbital period of a planet is directly proportional
to the cube of the semi-major axis of its orbit
You only need to notice, that the semi-major axis of the ellipse is 1/2 of the the semi-major axis of the round orbit.
This solution always looked like a chat for me, but this cheat works here :)
lesnik
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