What is the time it takes for a free fall compared with keplerian time period? Is it half the orbital period or a quarter?
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What time do you consider, the time to the central body when you stop the elliptical motion? At what point do you stop it? – trula Feb 14 '22 at 15:06
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@trula a straight free fall in the r direction – EB97 Feb 14 '22 at 15:30
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That was not my question, if it starts at r and does not collide it will return to the point at the same time a circular moving body would so it is 1/4 of the time, if it collides with a pointl mass as center, a little less if the mass has sam radius. – trula Feb 14 '22 at 15:38
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@trula When it collides with a point mass. In my answers it says half the period but I think like you, a quarter. thanks. – EB97 Feb 14 '22 at 16:20
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Related https://physics.stackexchange.com/q/644602/123208 & https://physics.stackexchange.com/q/14700/123208 – PM 2Ring Feb 14 '22 at 19:41
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You can approximate the linear free fall by a Kepler orbit with an eccentricity very close to 1. However, in this case the focus of the ellipse would practically coincide with the point of return of the orbit as shown in this illustration (taken from this website )

As the orbital period is independent of the eccentricity (if the semi-major axis stays the same), you can see that the linear free fall time to the focus would be half that.
Thomas
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Note that for a fixed value of perigee, the semi-major axis of the ellipse will have to change. In particular, the free-fall time from a distance $a$ will not one-half the orbital period for an object with semi-major axis $a$; it'll be related to an orbit with semi-major axis $a/2$ instead. – Michael Seifert Feb 14 '22 at 19:36
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1@MichaelSeifert In an orbit with semi-major axis $a$ and orbital eccentricity (close to) $1$, the body will fall from a distance $2a$. The time to reach the focus is obviously half of the corresponding orbital period (as both halves of the orbit must take the same time). – Thomas Feb 14 '22 at 19:45