• Images
• Videos
• Documents
• Books
• Articles

History

Main article: History of gravitational theory

In the Western world prior to the 16th century, it was generally assumed that the speed of a falling body would be proportional to its weight—that is, a 10 kg object was expected to fall ten times faster than an otherwise identical 1 kg object through the same medium. The ancient Greek philosopher Aristotle (384–322 BC) discussed falling objects in Physics (Book VII), one of the oldest books on mechanics (see Aristotelian physics). Although, in the 6th century, John Philoponus challenged this argument and said that, by observation, two balls of very different weights will fall at nearly the same speed.¹

In 12th-century Iraq, Abu'l-Barakāt al-Baghdādī gave an explanation for the gravitational acceleration of falling bodies. According to Shlomo Pines, al-Baghdādī's theory of motion was "the oldest negation of Aristotle's fundamental dynamic law [namely, that a constant force produces a uniform motion], [and is thus an] anticipation in a vague fashion of the fundamental law of classical mechanics [namely, that a force applied continuously produces acceleration]."²

Galileo Galilei

See also: Galileo Galilei § Falling bodies

See also: Galileo's Leaning Tower of Pisa experiment

According to a tale that may be apocryphal, in 1589–1592 Galileo dropped two objects of unequal mass from the Leaning Tower of Pisa. Given the speed at which such a fall would occur, it is doubtful that Galileo could have extracted much information from this experiment. Most of his observations of falling bodies were really of bodies rolling down ramps. This slowed things down enough to the point where he was able to measure the time intervals with water clocks and his own pulse (stopwatches having not yet been invented). He repeated this "a full hundred times" until he had achieved "an accuracy such that the deviation between two observations never exceeded one-tenth of a pulse beat." In 1589–1592, Galileo wrote De Motu Antiquiora, an unpublished manuscript on the motion of falling bodies.

Examples

Examples of objects in free fall include:

• A spacecraft (in space) with propulsion off (e.g. in a continuous orbit, or on a suborbital trajectory (ballistics) going up for some minutes, and then down).
• An object dropped at the top of a drop tube.
• An object thrown upward or a person jumping off the ground at low speed (i.e. as long as air resistance is negligible in comparison to weight).

Technically, an object is in free fall even when moving upwards or instantaneously at rest at the top of its motion. If gravity is the only influence acting, then the acceleration³ is always downward and has the same magnitude for all bodies, commonly denoted g {\displaystyle g} .

Since all objects fall at the same rate in the absence of other forces, objects and people will experience weightlessness in these situations.

Examples of objects not in free-fall:

• Flying in an aircraft: there is also an additional force of lift.
• Standing on the ground: the gravitational force is counteracted by the normal force from the ground.
• Descending to the Earth using a parachute, which balances the force of gravity with an aerodynamic drag force (and with some parachutes, an additional lift force).

The example of a falling skydiver who has not yet deployed a parachute is not considered free fall from a physics perspective, since they experience a drag force that equals their weight once they have achieved terminal velocity (see below).

Near the surface of the Earth, an object in free fall in a vacuum will accelerate at approximately 9.8 m/s2, independent of its mass. With air resistance acting on an object that has been dropped, the object will eventually reach a terminal velocity, which is around 53 m/s (190 km/h or 118 mph⁴) for a human skydiver. The terminal velocity depends on many factors including mass, drag coefficient, and relative surface area and will only be achieved if the fall is from sufficient altitude. A typical skydiver in a spread-eagle position will reach terminal velocity after about 12 seconds, during which time they will have fallen around 450 m (1,500 ft).⁵

Free fall was demonstrated on the Moon by astronaut David Scott on August 2, 1971. He simultaneously released a hammer and a feather from the same height above the Moon's surface. The hammer and the feather both fell at the same rate and hit the surface at the same time. This demonstrated Galileo's discovery that, in the absence of air resistance, all objects experience the same acceleration due to gravity. On the Moon, however, the gravitational acceleration is approximately 1.63 m/s2, or only about 1⁄6  that on Earth.

Free fall in Newtonian mechanics

Main article: Newtonian mechanics

Uniform gravitational field without air resistance

This is the "textbook" case of the vertical motion of an object falling a small distance close to the surface of a planet. It is a good approximation in air as long as the force of gravity on the object is much greater than the force of air resistance, or equivalently the object's velocity is always much less than the terminal velocity (see below).

v ( t ) = v 0 − g t {\displaystyle v(t)=v_{0}-gt\,} and y ( t ) = v 0 t + y 0 − 1 2 g t 2 , {\displaystyle y(t)=v_{0}t+y_{0}-{\frac {1}{2}}gt^{2},}

where

v 0 {\displaystyle v_{0}\,} is the initial vertical component of the velocity (m/s). v ( t ) {\displaystyle v(t)\,} is the vertical component of the velocity at t {\displaystyle t\,} (m/s). y 0 {\displaystyle y_{0}\,} is the initial altitude (m). y ( t ) {\displaystyle y(t)\,} is the altitude at t {\displaystyle t\,} (m). t {\displaystyle t\,} is time elapsed (s). g {\displaystyle g\,} is the acceleration due to gravity (9.81 m/s2 near the surface of the earth).

If the initial velocity is zero, then the distance fallen from the initial position will grow as the square of the elapsed time:

v ( t ) = − g t {\displaystyle v(t)=-gt} and y 0 − y ( t ) = 1 2 g t 2 . {\displaystyle y_{0}-y(t)={\frac {1}{2}}gt^{2}.}

Moreover, because the odd numbers sum to the perfect squares, the distance fallen in successive time intervals grows as the odd numbers. This description of the behavior of falling bodies was given by Galileo.⁶

Uniform gravitational field with air resistance

This case, which applies to 1. skydivers, parachutists or any body of mass, m {\displaystyle m} , and cross-sectional area, A {\displaystyle A} , 2. with Reynolds number Re well above the critical Reynolds number, so that the air resistance is proportional to the square of the fall velocity, v {\displaystyle v} ,

has an equation of vertical motion in Newton's regime

m d v d t = m g − 1 2 ρ C D A v 2 , {\displaystyle m{\frac {\mathrm {d} v}{\mathrm {d} t}}=mg-{\frac {1}{2}}\rho C_{\mathrm {D} }Av^{2}\,,}

where ρ {\displaystyle \rho } is the air density and C D {\displaystyle C_{\mathrm {D} }} is the drag coefficient, assumed to be constant (Re > 1000) although in general it will depend on the Reynolds number.

Assuming an object falling from rest and no change in air density with altitude (ideal gas? 1 / ρ = R T / p {\textstyle 1/\rho =RT/p} ), the solution is:

v ( t ) = v ∞ tanh ⁡ ( g t v ∞ ) , {\displaystyle v(t)=v_{\infty }\tanh \left({\frac {gt}{v_{\infty }}}\right),}

where the terminal speed is given by

v ∞ = 2 m g ρ C D A . {\displaystyle v_{\infty }={\sqrt {\frac {2mg}{\rho C_{D}A}}}\,.}

The object's speed versus time can be integrated over time to find the vertical position as a function of time:

y = y 0 − v ∞ 2 g ln ⁡ cosh ⁡ ( g t v ∞ ) . {\displaystyle y=y_{0}-{\frac {v_{\infty }^{2}}{g}}\ln \cosh \left({\frac {gt}{v_{\infty }}}\right).}

Using the figure of 56 m/s for the terminal velocity of a human, one finds that after 10 seconds he will have fallen 348 metres and attained 94% of terminal velocity, and after 12 seconds he will have fallen 455 metres and will have attained 97% of terminal velocity. Gravity field is (vertical) position-dependent g(y): when y 0 ≪ R {\textstyle y_{0}\ll R} , g / g o = 1 − 2 y 0 / R {\textstyle g/g_{o}=1-2y_{0}/R} . Linear decrease with height, small height compared to Earth's radius R = 6379 km.

However, when the air density cannot be assumed to be constant, such as for objects falling from high altitude, the equation of motion becomes much more difficult to solve analytically and a numerical simulation of the motion is usually necessary. The figure shows the forces acting on small meteoroids falling through the Earth's upper atmosphere (an acceleration of 0.1 km/s² is 10 g0). HALO jumps, including Joe Kittinger's and Felix Baumgartner's record jumps, also belong in this category.⁷

Inverse-square law gravitational field

It can be said that two objects in space orbiting each other in the absence of other forces are in free fall around each other, e.g. that the Moon or an artificial satellite "falls around" the Earth, or a planet "falls around" the Sun. Assuming spherical objects means that the equation of motion is governed by Newton's law of universal gravitation, with solutions to the gravitational two-body problem being elliptic orbits obeying Kepler's laws of planetary motion. This connection between falling objects close to the Earth and orbiting objects is best illustrated by the thought experiment, Newton's cannonball.

The motion of two objects moving radially towards each other with no angular momentum can be considered a special case of an elliptical orbit of eccentricity e = 1 (radial elliptic trajectory). This allows one to compute the free-fall time for two point objects on a radial path. The solution of this equation of motion yields time as a function of separation:

t ( y ) = y 0 3 2 μ ( y y 0 ( 1 − y y 0 ) + arccos ⁡ y y 0 ) , {\displaystyle t(y)={\sqrt {\frac {{y_{0}}^{3}}{2\mu }}}\left({\sqrt {{\frac {y}{y_{0}}}\left(1-{\frac {y}{y_{0}}}\right)}}+\arccos {\sqrt {\frac {y}{y_{0}}}}\right),}

where

t {\displaystyle t} is the time after the start of the fall y {\displaystyle y} is the distance between the centers of the bodies y 0 {\displaystyle y_{0}} is the initial value of y {\displaystyle y} μ = G ( m 1 + m 2 ) {\displaystyle \mu =G(m_{1}+m_{2})} is the standard gravitational parameter.

Substituting y = 0 {\displaystyle y=0} we get the free-fall time

t ff = π y 0 3 / ( 8 μ ) {\displaystyle t_{\text{ff}}=\pi {\sqrt {y_{0}^{3}/(8\mu )}}} and t / t ff = 2 / π ( y r ( 1 − y r ) + arccos ⁡ y r ) . {\displaystyle t/t_{\text{ff}}=2/\pi \left({\sqrt {y_{r}\left(1-y_{r}\right)}}+\arccos {\sqrt {y_{r}}}\right).}

The separation can be expressed explicitly as a function of time ⁸

y ( t ) = y 0   Q ( 1 − t t ff ; 3 2 , 1 2 )   , {\displaystyle y(t)=y_{0}~Q\left(1-{\frac {t}{t_{\text{ff}}}};{\frac {3}{2}},{\frac {1}{2}}\right)~,}

where Q ( x ; α , β ) {\displaystyle Q(x;\alpha ,\beta )} is the quantile function of the Beta distribution, also known as the inverse function of the regularized incomplete beta function I x ( α , β ) {\displaystyle I_{x}(\alpha ,\beta )} .

This solution can also be represented exactly by the analytic power series

y ( t ) = ∑ n = 1 ∞ x n n ! ⋅ lim r → 0 ( d n − 1 d r n − 1 [ r n ( 7 2 ( arcsin ⁡ ( r ) − r − r 2 ) ) − 2 3 n ] ) {\displaystyle y(t)=\sum _{n=1}^{\infty }{\frac {x^{n}}{n!}}\cdot \lim _{r\to 0}\left({\frac {\mathrm {d} ^{\,n-1}}{\mathrm {d} r^{\,n-1}}}\left[r^{n}\left({\frac {7}{2}}{\bigl (}\arcsin({\sqrt {r}})-{\sqrt {r-r^{2}}}{\bigr )}\right)^{-{\frac {2}{3}}n}\right]\right)}

= x / lim r → 0 [ ( 7 2 ( arcsin ⁡ ( r ) − r − r 2 ) ) 2 3 ] ′ + x 2 2 ! lim r → 0 ( d 1 d r 1 [ r 2 ( 7 2 ( arcsin ⁡ ( r ) − r − r 2 ) ) − 4 3 ] ) {\displaystyle =x/\lim _{r\to 0}[({\frac {7}{2}}{\bigl (}\arcsin({\sqrt {r}})-{\sqrt {r-r^{2}}}{\bigr )})^{\frac {2}{3}}]'+{\frac {x^{2}}{2!}}\lim _{r\to 0}\left({\frac {\mathrm {d} ^{1}}{\mathrm {d} r^{1}}}\left[r^{2}\left({\frac {7}{2}}{\bigl (}\arcsin({\sqrt {r}})-{\sqrt {r-r^{2}}}{\bigr )}\right)^{-{\frac {4}{3}}}\right]\right)}

+ x 3 3 ! lim r → 0 ( d 2 d r 2 [ r 3 ( 7 2 ( arcsin ⁡ ( r ) − r − r 2 ) ) − 2 ] ) + ⋯ {\displaystyle +{\frac {x^{3}}{3!}}\lim _{r\to 0}\left({\frac {\mathrm {d} ^{2}}{\mathrm {d} r^{2}}}\left[r^{3}\left({\frac {7}{2}}{\bigl (}\arcsin({\sqrt {r}})-{\sqrt {r-r^{2}}}{\bigr )}\right)^{-2}\right]\right)+\cdots } ⁹

Evaluating this yields:¹⁰¹¹

y ( t ) / y 0 = x − 1 5 x 2 − 3 175 x 3 − 23 7875 x 4 − 1894 3 , 031875 x 5 − 3293 21 , 896875 x 6 − 2 , 418092 62 , 077 , 640625 x 7 − ⋯ {\displaystyle y(t)/y_{0}=x-{\frac {1}{5}}x^{2}-{\frac {3}{175}}x^{3}-{\frac {23}{7875}}x^{4}-{\frac {1894}{3,031875}}x^{5}-{\frac {3293}{21,896875}}x^{6}-{\frac {2,418092}{62,077,640625}}x^{7}-\cdots }

= x − 1 5 x [ x + ( 3 7 x 2 + 23 315 x 3 + 1894 121 , 275 x 4 + 3293 875 , 875 x 5 + 2 , 418092 2 , 483 , 105625 x 6 + ⋯ ) / 5 ] {\displaystyle =x-{\frac {1}{5}}x[x+({\frac {3}{7}}x^{2}+{\frac {23}{315}}x^{3}+{\frac {1894}{121,275}}x^{4}+{\frac {3293}{875,875}}x^{5}+{\frac {2,418092}{2,483,105625}}x^{6}+\cdots )/5]}

= x − 1 5 x [ x + ( 3 7 + ( 23 63 + 1894 24 , 255 x + 3293 175 , 175 x 2 + 2 , 418092 480 , 621125 x 3 ) x / 5 + ⋯ ) x 2 / 5 ] , {\displaystyle =x-{\frac {1}{5}}x[x+({\frac {3}{7}}+({\frac {23}{63}}+{\frac {1894}{24,255}}x+{\frac {3293}{175,175}}x^{2}+{\frac {2,418092}{480,621125}}x^{3})x/5+\cdots )x^{2}/5],}

where

x = [ 3 2 ( π 2 − t 2 μ y 0 3 ) ] 2 / 3 = [ 3 π / 4 ⋅ ( 1 − t / t ff ) ] 2 / 3 . {\displaystyle x=\left[{\frac {3}{2}}\left({\frac {\pi }{2}}-t{\sqrt {\frac {2\mu }{{y_{0}}^{3}}}}\right)\right]^{2/3}=[3\pi /4\cdot (1-t/t_{\text{ff}})]^{2/3}.} ¹²

In general relativity

Further information: General relativity

In general relativity, an object in free fall is subject to no force and is an inertial body moving along a geodesic. Far away from any sources of space-time curvature, where spacetime is flat, the Newtonian theory of free fall agrees with general relativity. Otherwise the two disagree; e.g., only general relativity can account for the precession of orbits, the orbital decay or inspiral of compact binaries due to gravitational waves, and the relativity of direction (geodetic precession and frame dragging).

The experimental observation that all objects in free fall accelerate at the same rate, as noted by Galileo (1590?) and then embodied in Newton's theory as the equality of gravitational and inertial masses (1687) was later confirmed to high accuracy by modern forms of the Eötvös experiment (orig. 1885). It is the basis of the equivalence principle, from which basis Einstein's theory of general relativity initially took off.

See also

• Equations for a falling body
• G-force
• High-altitude military parachuting
• Reduced-gravity aircraft
• Terminal velocity
• Weightlessness

External links

• Freefall formula calculator www.fxsolver.com
• The Way Things Fall www.phy6.org/stargaze: an educational website

Wikimedia Commons has media related to Free fall.

References

1. Cohen, Morris R.; Drabkin, I. E., eds. (1958). A Source Book in Greek Science. Cambridge, MA: Harvard University Press. p. 220. ↩

2. Pines, Shlomo (1970). "Abu'l-Barakāt al-Baghdādī , Hibat Allah". Dictionary of Scientific Biography. Vol. 1. New York: Charles Scribner's Sons. pp. 26–28. ISBN 0-684-10114-9. (cf. Abel B. Franco (October 2003). "Avempace, Projectile Motion, and Impetus Theory", Journal of the History of Ideas 64 (4), pp. 521–546 [528].) 0-684-10114-9 ↩

3. Feynman, R.Ph.; Leighton, R.B.; Sands, M. (1964). "The Feynman Lectures on Physics Vol. I Ch. 8: Motion". www.feynmanlectures.caltech.edu. https://www.feynmanlectures.caltech.edu/I_08.html#8-5_acceleration ↩

4. "Free fall graph" (PDF). Green Harbor Publications. 2010. Retrieved 14 March 2016. http://www.greenharbor.com/fffolder/speedtime.pdf ↩

5. "Free fall graph" (PDF). Green Harbor Publications. 2010. Retrieved 14 March 2016. http://www.greenharbor.com/fffolder/speedtime.pdf ↩

6. Olenick, R.P.; Apostol, T.M.; Goodstein, D.L. (2008). The Mechanical Universe: Introduction to Mechanics and Heat. Cambridge University Press. p. 18. ISBN 978-0-521-71592-8. 978-0-521-71592-8 ↩

7. An analysis of such jumps is given in Mohazzabi, P.; Shea, J. (1996). "High altitude free fall" (PDF). American Journal of Physics. 64 (10): 1242. Bibcode:1996AmJPh..64.1242M. doi:10.1119/1.18386. http://www.jasoncantarella.com/downloads/AJP001242.pdf ↩

8. Obreschkow, Danail (7 June 2024). "From Cavitation to Astrophysics: Explicit Solution of the Spherical Collapse Equation". Phys. Rev. E. 109 (6): 065102. arXiv:2401.05445. Bibcode:2024PhRvE.109f5102O. doi:10.1103/PhysRevE.109.065102. PMID 39021019. https://link.aps.org/doi/10.1103/PhysRevE.109.065102 ↩

9. x 1 / 1 ! ⋅ lim r → 0 [ N u m ( r ) / D e n ( r ) ] = x ⋅ [ 0 / ( 7 / 2 ⋅ ( 0 − 0 ) ) 2 / 3 ] {\displaystyle x^{1}/1!\cdot \lim _{r\to 0}[Num(r)/Den(r)]=x\cdot [0/(7/2\cdot (0-0))^{2/3}]} = x lim r → 0 [ d N u m / d r : d D e n / d r ] = x lim r → 0 [ d r 1 / d r : D e n ′ ] {\displaystyle x\lim _{r\to 0}[\operatorname {d} \!{Num}/\operatorname {d} \!r:\operatorname {d} \!{Den}/\operatorname {d} \!r]=x\lim _{r\to 0}[\operatorname {d} \!{r^{1}}/\operatorname {d} \!r:{Den}']} ↩

10. Foong, S K (2008). "From Moon-fall to motions under inverse square laws". European Journal of Physics. 29 (5): 987–1003. Bibcode:2008EJPh...29..987F. doi:10.1088/0143-0807/29/5/012. S2CID 122494969. https://doi.org/10.1088%2F0143-0807%2F29%2F5%2F012 ↩

11. Mungan, Carl E. (2009). "Radial Motion of Two Mutually Attracting Particles" (PDF). The Physics Teacher. 47 (8): 502–507. Bibcode:2009PhTea..47..502M. doi:10.1119/1.3246467. https://apps.dtic.mil/sti/pdfs/ADA534896.pdf ↩

12. At t=0 x = ( 3 / 4 ⋅ π ) 2 / 3 {\textstyle x=(3/4\cdot \pi )^{2/3}} and y=y0, at t = t f f {\textstyle t=t_{ff}} x=0 and y=0. ↩