Equations for a falling body

A set of equations describe the resultant trajectories when objects move owing to a constant gravitational force under normal Earth-bound conditions. For example, Newton's law of universal gravitation simplifies to F = mg, where m is the mass of the body. This assumption is reasonable for objects falling to earth over the relatively short vertical distances of our everyday experience, but is untrue over larger distances, such as spacecraft trajectories.

History

Galileo was the first to demonstrate and then formulate these equations. He used a ramp to study rolling balls, the ramp slowing the acceleration enough to measure the time taken for the ball to roll a known distance.^[1]^[2] He measured elapsed time with a water clock, using an "extremely accurate balance" to measure the amount of water.^{[note 1]}

The equations ignore air resistance, which has a dramatic effect on objects falling an appreciable distance in air, causing them to quickly approach a terminal velocity. The effect of air resistance varies enormously depending on the size and geometry of the falling object — for example, the equations are hopelessly wrong for a feather, which has a low mass but offers a large resistance to the air. (In the absence of an atmosphere all objects fall at the same rate, as astronaut David Scott demonstrated by dropping a hammer and a feather on the surface of the Moon.)

The equations also ignore the rotation of the Earth, failing to describe the Coriolis effect for example. Nevertheless, they are usually accurate enough for dense and compact objects falling over heights not exceeding the tallest man-made structures.

Overview

An initially stationary object which is allowed to fall freely under gravity falls a distance proportional to the square of the elapsed time. This image, spanning half a second, was captured with a stroboscopic flash at 20 flashes per second. During the first 0.05s the ball drops one unit of distance (about 12 mm), by 0.10s it has dropped at total of 4 units, by 0.15s 9 units, and so on.

Near the surface of the Earth, the acceleration due to gravity g = 9.81 m/s² (meters per second squared; which might be thought of as "meters per second, per second", or 32.2 ft/s² as "feet per second per second") approximately. For other planets, multiply g by the appropriate scaling factor. A coherent set of units for g, d, t and v is essential. Assuming SI units, g is measured in meters per second squared, so d must be measured in meters, t in seconds and v in meters per second.

In all cases, the body is assumed to start from rest, and air resistance is neglected. Generally, in Earth's atmosphere, all results below will therefore be quite inaccurate after only 5 seconds of fall (at which time an object's velocity will be a little less than the vacuum value of 49 m/s (9.8 m/s² × 5 s) due to air resistance). Air resistance induces a drag force on any body that falls through any atmosphere other than a perfect vacuum, and this drag force increases with velocity until it equals the gravitational force, leaving the object to fall at a constant terminal velocity.

Atmospheric drag, the coefficient of drag for the object, the (instantaneous) velocity of the object, and the area presented to the airflow determine terminal velocity.

Apart from the last formula, these formulas also assume that g negligibly varies with height during the fall (that is, they assume constant acceleration). The last equation is more accurate where significant changes in fractional distance from the center of the planet during the fall cause significant changes in g. This equation occurs in many applications of basic physics.

Equations

Distance $\ d\$ travelled by an object falling for time $\ t\$ :	$\ d={\frac {1}{2}}gt^{2}$
Time $\ t\$ taken for an object to fall distance $\ d\$ :	$\ t=\ {\sqrt {{\frac {2d}{g}}}}$
Instantaneous velocity $\ v_{i}\$ of a falling object after elapsed time $\ t\$ :	$\ v_{i}=gt$
Instantaneous velocity $\ v_{i}\$ of a falling object that has travelled distance $\ d\$ :	$\ v_{i}={\sqrt {2gd}}\$
Average velocity $\ v_{a}\$ of an object that has been falling for time $\ t\$ (averaged over time):	$\ v_{a}={\frac {1}{2}}gt$
Average velocity $\ v_{a}\$ of a falling object that has travelled distance $\ d\$ (averaged over time):	$\ v_{a}={\frac {{\sqrt {2gd}}}{2}}\$
Instantaneous velocity $\ v_{i}\$ of a falling object that has travelled distance $\ d\$ on a planet with mass $\ M\$ , with the combined radius of the planet and altitude of the falling object being $\ r\$ , this equation is used for larger radii where $\ g\$ is smaller than standard $\ g\$ at the surface of Earth, but assumes a small distance of fall, so the change in $\ g\$ is small and relatively constant:	$\ v_{i}={\sqrt {{\frac {2GMd}{r^{2}}}}}\$
Instantaneous velocity $\ v_{i}\$ of a falling object that has travelled distance $\ d\$ on a planet with mass $\ M\$ and radius $\ r\$ (used for large fall distances where $\ g\$ can change significantly):	$\ v_{i}={\sqrt {2GM{\Big (}{\frac {1}{r}}-{\frac {1}{r+d}}{\Big )}}}\$

Measured fall time of a small steel sphere falling from various heights. The data is in good agreement with the predicted fall time of

{\sqrt {2h/g}}

, where h is the height and g is the acceleration of gravity.

Examples

The first equation shows that, after one second, an object will have fallen a distance of 1/2 × 9.8 × 1² = 4.9 meters. After two seconds it will have fallen 1/2 × 9.8 × 2² = 19.6 meters; and so on. The second to last equation becomes grossly inaccurate at great distances. If an object fell 10,000 meters to Earth, then the results of both equations differ by only 0.08%; however, if it fell from geosynchronous orbit, which is 42,164 km, then the difference changes to almost 64%.

Based on wind resistance, for example, the terminal velocity of a skydiver in a belly-to-earth (i.e., face down) free-fall position is about 195 km/h (122 mph or 54 m/s). This velocity is the asymptotic limiting value of the acceleration process, because the effective forces on the body balance each other more and more closely as the terminal velocity is approached. In this example, a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on.

Higher speeds can be attained if the skydiver pulls in his or her limbs (see also freeflying). In this case, the terminal velocity increases to about 320 km/h (200 mph or 90 m/s), which is almost the terminal velocity of the peregrine falcon diving down on its prey. The same terminal velocity is reached for a typical .30-06 bullet dropping downwards—when it is returning to earth having been fired upwards, or dropped from a tower—according to a 1920 U.S. Army Ordnance study.

Competition speed skydivers fly in the head down position and reach even higher speeds. The current world record is 1,357.6 km/h (843.6 mph/Mach 1.25) by Felix Baumgartner who skydived from 38,969.4 m (127,852.4 ft) above earth on 14 October 2012. The record was set due to the high altitude where the lesser density of the atmosphere decreased drag.

For astronomical bodies other than Earth, and for short distances of fall at other than "ground" level, g in the above equations may be replaced by G(M+m)/r² where G is the gravitational constant, M is the mass of the astronomical body, m is the mass of the falling body, and r is the radius from the falling object to the center of the body.

Removing the simplifying assumption of uniform gravitational acceleration provides more accurate results. We find from the formula for radial elliptic trajectories:

The time t taken for an object to fall from a height r to a height x, measured from the centers of the two bodies, is given by:

t={\frac {\arccos {\Big (}{\sqrt {{\frac {x}{r}}}}{\Big )}+{\sqrt {{\frac {x}{r}}\ (1-{\frac {x}{r}})}}}{{\sqrt {2\mu }}}}\,r^{{3/2}}

where $\mu =G(m_{1}+m_{2})$ is the sum of the standard gravitational parameters of the two bodies. This equation should be used whenever there is a significant difference in the gravitational acceleration during the fall.

Acceleration relative to the rotating Earth

Centripetal force causes the acceleration measured on the rotating surface of the Earth to differ from the acceleration that is measured for a free-falling body: the apparent acceleration in the rotating frame of reference is the total gravity vector minus a small vector toward the north-south axis of the Earth, corresponding to staying stationary in that frame of reference.

Notes

↑ See the works of Stillman Drake, for a comprehensive study of Galileo and his times, the Scientific Revolution.

References

↑ Jespersen, James; Fitz-Randolph, Jane. From Sundials to Clocks: Understanding Time and Frequency (PDF). National Institute of Standards and Technology Monograph 155 (Report) (1999 ed.). U.S. Department of Commerce Technology Administration and National Institute of Standards and Technology. pp. 188–190.
↑ MacDougal, D.W. (2012). "Chapter 2 - Galileo's Great Discovery: How Things Fall". Newton’s Gravity: An Introductory Guide to the Mechanics of the Universe, Undergraduate Lecture Notes in Physics (PDF). New York: Springer Science+Business Media. doi:10.1007/978-1-4614-5444-1_2.

External links

Falling body equations calculator

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[3] See the works of Stillman Drake, for a comprehensive study of Galileo and his times, the Scientific Revolution.

[1] Jespersen, James; Fitz-Randolph, Jane. From Sundials to Clocks: Understanding Time and Frequency (PDF). National Institute of Standards and Technology Monograph 155 (Report) (1999 ed.). U.S. Department of Commerce Technology Administration and National Institute of Standards and Technology. pp. 188–190.

[2] MacDougal, D.W. (2012). "Chapter 2 - Galileo's Great Discovery: How Things Fall". Newton’s Gravity: An Introductory Guide to the Mechanics of the Universe, Undergraduate Lecture Notes in Physics (PDF). New York: Springer Science+Business Media. doi:10.1007/978-1-4614-5444-1_2.