Gravity - Mathematical equations for a falling body - Encyclopedia II

The equations below describe the motion of a falling body, assuming that the acceleration due to gravity is a constant, g (in which case Newton's law of gravitation simplifies to F = mg where m is the mass of the earth). 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).

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. He measured elapsed time with a water clock, using an "extremely accurate balance" to measure the amount of water. ²

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. For example, a person jumping headfirst from an airplane will never exceed a speed of about 200 mph due to air resistance. 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 = 9.8 m/s² (metres per second per second), approximately. For other planets, multiply g by the appropriate scaling factor. It is essential to use consistent units for g, d, t and v. Assuming SI units, g is measured in metres per second per second, so d must be measured in metres, t in seconds and v in metres per second. To convert metres per second to kilometres per hour (km/h) multiply by 3.6. In all cases the body is assumed to start from rest.

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 metres; and so on.

For any mass distribution there is a scalar field, the gravitational potential (a scalar potential), which is the gravitational potential energy per unit mass of a point mass, as function of position. It is

where the integral is taken over all mass. Minus its gradient is the gravity field itself, and minus its Laplacian is the divergence of the gravity field, which is everywhere equal to -4πG times the local density.

Thus when outside masses the potential satisfies Laplace's equation (i.e., the potential is a harmonic function), and when inside masses the potential satisfies Poisson's equation with, as right-hand side, 4πG times the local density.

The acceleration measured on the rotating surface of the Earth is not quite the same as the acceleration that is measured for a free-falling body because of the centrifugal force. In other words, 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.

Adapted from the Wikipedia article "Mathematical equations for a falling body", under the G.N U Free Docmentation License. Please also see http://en.wikipedia.org/wiki/Main_Page