Gravitational force Equations

November 20, 2020

Gravitational Forces Equation

Near the surface of the Earth, use = 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 by the appropriate scaling factor. A coherent set of units for , , and is essential. Assuming SI units, is measured in meters per second squared, so must be measured in meters, in seconds and 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 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.

The equations[edit]

Measured fall time of a small steel sphere falling from various heights. The data is in good agreement with the predicted fall time of , where h is the height and g is the acceleration of gravity.


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, in the above equations may be replaced by G(M+m)/r2 where 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.

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

where 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[edit]

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.

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