Equations for a falling body
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A set of dynamical 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 very much untrue over larger distances, such as spacecraft trajectories. Please note that in this article any resistance from air (drag) is neglected.
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.
Near the surface of the Earth, use g = 9.81 m/s² (meters per second squared; which might be thought of as "meters per second, per second", or 32 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.
|Distance travelled by an object falling for time :|
|Time taken for an object to fall distance :|
|Instantaneous velocity of a falling object after elapsed time :|
|Instantaneous velocity of a falling object that has travelled distance :|
|Average velocity of an object that has been falling for time (averaged over time):|
|Average velocity of a falling object that has travelled distance (averaged over time):|
|Instantaneous velocity of a falling object that has travelled distance on a planet with mass , with the combined radius of the planet and altitude of the falling object being , this equation is used for larger radii where is smaller than standard at the surface of Earth, but assumes a small distance of fall, so the change in is small and relatively constant:|
|Instantaneous velocity of a falling object that has travelled distance on a planet with mass and radius (used for large fall distances where can change significantly):|
Example: 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%.
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:
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
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.