Date: 05/21/99 at 11:38:31 From: Husni Subject: Gravity Newton's second law of motion states that F = ma, which means that the force in Newtons needed to move an object of m kg with an acceleration of a metres/second^2 equals mass of object * acceleration of object. But why does any free-falling object fall at the rate of 9.8 metres/ second^2 despite its mass? Thanks.
Date: 05/21/99 at 16:05:34 From: Doctor Rick Subject: Re: Gravity Hello, Husni, welcome to Ask Dr. Math. You have noticed the principle of the equality of inertial mass and gravitational mass. Inertial mass is the "m" in Newton's second law, F = ma Inertial mass determines how much acceleration is caused by some force. Gravitational mass is the "m" in Newton's law of gravitation, F = -GMm/r^2 Gravitational mass determines how much force gravity exerts on an object. We can find the acceleration due to gravity by substituting F = ma in the law of gravity. ma = -GMm/r^2 Because the inertial mass of any object is identical to its gravitational mass, the m's cancel and we have a = -GM/r^2 G is a universal constant. With M being the mass of the earth and r its radius, the expression on the right comes out to 9.8 m/s^2. Apart from experiment, there would have been no reason to assume that the two kinds of mass would be the same, making gravitational acceleration constant. People had a right to be surprised at the results of Galileo's experiments on the Tower of Pisa or wherever. In fact, this amazing coincidence, which most people never notice, led to Einstein's theory of general relativity. In this theory, it turns out that gravity is not a force after all, but the effect of "bending of space-time." In other words, gravity is a "fictitious force" akin to "centrifugal force" and the Coriolis "force." Like them, it is very convenient to think of gravity as a force, but your observation gives us reason to believe that it isn't really. If you are acquainted with these fictitious forces, I could say more about how gravity works. But I'd better stop now. This could get very complicated very fast. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum