lundyjb said:
So would an object with a gravity of 2g have an acceleration of 2*9.81 (19.62) or is the 9.81 a universal rate of acceleration? (Sorry meant to say rate, not speed of acceleration)
An object doesn't have a fixed gravity, it has a fixed mass. The force felt between two masses is described by the law of universal gravitation:
F=G(m1*m2)/r^2
Where G is the gravitational constant, more or less that "universal rate" called a constant that you were asking about:
G=6.67398 × 10^-11 m^3 kg^-1 s^-2
m1 is the mass of the object (in kilograms)
m2 is the mass of some other object (in kilograms)
r is the distance between the centers of mass of the two objects (in meters)
F is then the mutual force felt by each mass from the other mass (in Newtons)
So m1 feels that force by m2 and m2 feels that force by m1, due to Newton's Third Law (equal and opposite reactions).
Subsequently, the acceleration resulting from a single object (as opposed to the force on a second object), can be found by eliminating m2 from the equation--yet it nevertheless still depends on a distance from the mass, namely r!
a=G*m1/r^2
a is the acceleration (in m/s^2)
This is a result of taking the force equation and using Newton's second law for a hypothetical second object F=m2*a:
F=G(m1*m2)/r^2=m2*a -> a = G*m1/r^2
Edit:
So, using the mass and radius of the Earth we have:
a = G*m1/r^2 = (6.67398 × 10^-11 m^3 kg^-1 s^-2)*(5.972 × 10^24 kg)/(6.373 × 10^6 m)^2 = 9.81 m/s^2
The actual acceleration depends on a person's specific location on the surface of the Earth. It can vary from around 9.78 m/s^2 to 9.83 m/s^2 on Earth's surface, and possibly more at the extremes, depending on latitude and height above/below sea level, among other things.