Free-fall time
Encyclopedia
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 supernova
e -- in which gravity plays a dominant role.
of planetary motion to a degenerate elliptic orbit. Consider a point mass
a distance 
from a point source
of mass
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
. 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 
, or
To see that the semi-major axis is
, 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 
, 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 (
) to the point source of mass 
. In other words, the ellipse becomes a line of length 
. The semi-major axis is half the width of the ellipse along the long axis, which in the degenerate case becomes 
.
If the free-falling body completed a full orbit, it would begin at distance
from the point source mass 
, 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).
This formula also follows from the formula for the falling time as a function of position.
Note that
in the above equation, is the time for the mass to fall in a highly eccentric orbit, make a "hairpin" turn at the central mass at nearly zero radius distance, and then returns to R when it repeats the very sharp turn. This orbit corresponds to nearly linear motion back and from from distance R to distance 0. As noted above, this orbit has only half as long a semimajor axis (R/2) as a circular orbit with radius R (where the semimajor axis is R), and thus the period for the shorter high-eccentricity "orbit" is that for one with an axis of R/2 and a total orbital pathlength of only twice the infall distance. Thus, by Kepler's third law, with half the semimajor axis radius it thus takes only (1/2)3/2 = (1/8)1/2 as long a time period, as the "corresponding" circular orbit that has a constant radius the same as the maximal radius of the eccentric orbit (which goes to essentially zero radius from the primary at its other extreme).
The time to traverse half the distance R, which is the infall time from R along an eccentric orbit, is the Kepler time for a circular orbit of R/2 (not R), which is (1/32)1/2 times the period P of the circular orbit at R. For example, the time for an object in the orbit of the Earth around the Sun, to fall into the Sun if it were suddenly stopped in orbit, would be
, where P is one year. This is about 64.6 days.
is not a point mass, but is distributed in a spherically-symmetric distribution about the center, with an average mass density of 
,
,
where the volume of a sphere is:
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 acceleration of gravity at any given distance
from the center of the sphere depends only upon the total mass contained within 
. 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 of a massless particle at 
can be expressed solely in terms of the total mass 
interior to it. In terms of the average density interior to 
,
where the latter is in SI
units.
This result is the exact same as from the previous section when :
.
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.
≈ 0.42.
Time
Time is a part of the measuring system used to sequence events, to compare the durations of events and the intervals between them, and to quantify rates of change such as the motions of objects....
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
Star formation
Star formation is the process by which dense parts of molecular clouds collapse into a ball of plasma to form a star. As a branch of astronomy star formation includes the study of the interstellar medium and giant molecular clouds as precursors to the star formation process and the study of young...
to helioseismology
Helioseismology
Helioseismology is the study of the propagation of wave oscillations, particularly acoustic pressure waves, in the Sun. Unlike seismic waves on Earth, solar waves have practically no shear component . Solar pressure waves are believed to be generated by the turbulence in the convection zone near...
to supernova
Supernova
A supernova is a stellar explosion that is more energetic than a nova. It is pronounced with the plural supernovae or supernovas. Supernovae are extremely luminous and cause a burst of radiation that often briefly outshines an entire galaxy, before fading from view over several weeks or months...
e -- in which gravity plays a dominant role.
Infall to a point source of gravity
It is relatively simple to derive the free-fall time by applying nothing more than Kepler's Third LawKepler's laws of planetary motion
In astronomy, Kepler's laws give a description of the motion of planets around the Sun.Kepler's laws are:#The orbit of every planet is an ellipse with the Sun at one of the two foci....
of planetary motion to a degenerate elliptic orbit. Consider a point mass
a distance from a point sourcePoint source
A point source is a localised, relatively small source of something.Point source may also refer to:*Point source , a localised source of pollution**Point source water pollution, water pollution with a localized source...
of mass
which falls radially inward to it. Crucially, Kepler's Third LawKepler's laws of planetary motion
In astronomy, Kepler's laws give a description of the motion of planets around the Sun.Kepler's laws are:#The orbit of every planet is an ellipse with the Sun at one of the two foci....
depends only on the semi-major axis
Semi-major axis
The major axis of an ellipse is its longest diameter, a line that runs through the centre and both foci, its ends being at the widest points of the shape...
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
. 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 , orTo see that the semi-major axis is
, 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 , 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 () to the point source of mass . In other words, the ellipse becomes a line of length . The semi-major axis is half the width of the ellipse along the long axis, which in the degenerate case becomes .If the free-falling body completed a full orbit, it would begin at distance
from the point source mass , 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).This formula also follows from the formula for the falling time as a function of position.
Note that
in the above equation, is the time for the mass to fall in a highly eccentric orbit, make a "hairpin" turn at the central mass at nearly zero radius distance, and then returns to R when it repeats the very sharp turn. This orbit corresponds to nearly linear motion back and from from distance R to distance 0. As noted above, this orbit has only half as long a semimajor axis (R/2) as a circular orbit with radius R (where the semimajor axis is R), and thus the period for the shorter high-eccentricity "orbit" is that for one with an axis of R/2 and a total orbital pathlength of only twice the infall distance. Thus, by Kepler's third law, with half the semimajor axis radius it thus takes only (1/2)3/2 = (1/8)1/2 as long a time period, as the "corresponding" circular orbit that has a constant radius the same as the maximal radius of the eccentric orbit (which goes to essentially zero radius from the primary at its other extreme).The time to traverse half the distance R, which is the infall time from R along an eccentric orbit, is the Kepler time for a circular orbit of R/2 (not R), which is (1/32)1/2 times the period P of the circular orbit at R. For example, the time for an object in the orbit of the Earth around the Sun, to fall into the Sun if it were suddenly stopped in orbit, would be
, where P is one year. This is about 64.6 days.Infall of a uniformly dense mass
Now, consider a case where the mass is not a point mass, but is distributed in a spherically-symmetric distribution about the center, with an average mass density of , ,where the volume of a sphere is:
Let us assume that the only force acting is gravity. Then, as first demonstrated by Newton
Isaac Newton
Sir Isaac Newton PRS was an English physicist, mathematician, astronomer, natural philosopher, alchemist, and theologian, who has been "considered by many to be the greatest and most influential scientist who ever lived."...
, and can easily be demonstrated using the divergence theorem
Divergence theorem
In vector calculus, the divergence theorem, also known as Gauss' theorem , Ostrogradsky's theorem , or Gauss–Ostrogradsky theorem is a result that relates the flow of a vector field through a surface to the behavior of the vector field inside the surface.More precisely, the divergence theorem...
, the acceleration of gravity at any given distance
from the center of the sphere depends only upon the total mass contained within . 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 of a massless particle at can be expressed solely in terms of the total mass interior to it. In terms of the average density interior to ,where the latter is in SI
Si
Si, si, or SI may refer to :- Measurement, mathematics and science :* International System of Units , the modern international standard version of the metric system...
units.
This result is the exact same as from the previous section when :
.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 writeHere 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 ≈ 0.42.Reference
- Galactic dynamics Binney, James; Tremaine, Scott. Princeton University Press, 1987.