# Free-fall time

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Free-fall time

The free-fall time is the characteristic time that would take a body to collapse under its own gravitational attraction, if no other forces existed to oppose the collapse. As such, it plays a fundamental role in setting the timescale for a wide variety of astrophysical processes -- from star formation to helioseismology to supernovae -- in which gravity plays a dominant role.

Derivation

It is relatively simple to derive the free-fall time using nothing more than Kepler's Third Law of planetary motion. Consider a body a distance $R$ from a point source of mass $M$ which falls radially inward to it. Crucially, Kepler's Third Law depends only on the semi-major axis of the orbit, and does not depend on the eccentricity. A purely radial trajectory is an example of a degenerate ellipse with an eccentricity of 1 and semi-major axis $R/2$. Therefore, the time it would take a body to fall inward, turn around, and return to its original position is the same as the period of a circular orbit of radius $R/2$, or

:$t_\left\{orbit\right\} = frac\left\{2 pi\right\}\left\{sqrt \left\{G M left\left(frac\left\{R\right\}\left\{2\right\} ight\right)^\left\{3/2\right\}=frac\left\{pi R^\left\{3/2\left\{sqrt\left\{2 G M.$

To see that the semi-major axis is $R/2$, we must examine properties of orbits as they become increasingly elliptical. Kepler's First Law states that an orbit is an ellipse with the center of mass as one focus. In the case of a very small mass falling toward a very large mass $M$, the center of mass is within the larger mass. The focus of an ellipse is increasingly off-center with increasing ellipticity. In the limiting case of a degenerate ellipse with an eccentricity of 1, the orbit extends from the initial position of the infalling object ($R$) to the point source of mass $M$. In other words, the ellipse becomes a line of length $R$. The semi-major axis is half the width of the ellipse along the long axis, which in the degenerate case becomes $R/2$.

If the free-falling body completed a full orbit, it would begin at distance $R$ from the point source mass $M$, fall inward until it reached that point source, then turn around and journey back to its original position. In real systems, the point source mass isn't truly a point source and the infalling body eventually collides with some surface. Thus, it only completes half the orbit. But since the infalling part of the orbit is symmetric to the hypothetical outgoing portion of the orbit, we can simply divide the period of the full orbit by two to attain the free-fall time (the time along the infalling portion of the orbit).

:$t_\left\{ff\right\}=t_\left\{orbit\right\}/2=frac\left\{pi R^\left\{3/2\left\{2sqrt\left\{2 G M$

Now, consider a case where the mass $M$ is not a point mass, but is distributed in a spherically-symmetric distribution about the center, with an average mass density of $ho$,

:$ho = frac\left\{3 M\right\}\left\{4 pi R^3\right\}.$ ,

where the volume of a sphere is :$\left\{4/3 pi R^3\right\}.$

Let us assume that the only force acting is gravity. Then, as first demonstrated by Newton, and can easily be demonstrated using the divergence theorem, the force of gravity at any given distance $R$ from the center of the sphere depends only upon the total mass contained within $R$. The consequence of this result is that if one imagined breaking the sphere up into a series of concentric shells, each shell would collapse only subsequent to the shells interior to it, and no shells cross during collapse. As a result, the free-fall time at $R$ can be expressed solely in terms of the total mass $M$ interior to it. In terms of the average density interior to $R$,

:$t_\left\{ff\right\} = frac\left\{1\right\}\left\{4\right\}sqrt \left\{frac\left\{3 pi\right\}\left\{2 G hosimeq 0.54 frac\left\{1\right\}\left\{ sqrt \left\{G hosimeq 66120 frac\left\{1\right\}\left\{ sqrt \left\{ ho$

where the latter is in SI units.

Applications

The free-fall time is a very useful estimate of the relevant timescale for a number of astrophysical processes. To get a sense of its application, we may write

:$t_\left\{ff\right\} simeq frac\left\{35 , mbox\left\{min\left\{sqrt\left\{ ho cdot sqrt\left\{frac\left\{mbox\left\{g\left\{mbox\left\{cm\right\}^3.$

Here we have estimated the numerical value for the free-fall time as roughly 35 minutes for a body of mean density 1 g/cm3.

Comparison

For an object falling from infinity in a capture orbit, the time it takes from a given position to fall to the central point mass is the same as the free-fall time, except for a constant $frac\left\{4\right\}\left\{3pi\right\}$ ≈ 0.42.

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