# Spherical Coordinate Systems(Cartesian, i think it called)

Me and my friend have been arguing about the coordinate system used for the earth... specifically gravity. he's trying to tell me the value of gravity is -9.8ms/2, when ive read from several books and other online resources thats it 9.8ms/2... a positive number. Hes keeps going on and on and on about how it has to be negative to go down, when, i know for a fact from being a java programmer, that coordinate systems work in the way that positive numbers make u move downward... now the earth and 2d coordinate system used for bouncing balls might be a different thing, but same principal. basically, all i want to know, is gravity -9.8ms/2(a negative number), or 9.8ms/2(a positive number)?

Do you mean the gravitational force or the acceleration due to gravity?

I think the coordinate system is arbitrary, so you could define gravitational acceleration to be positive or negative depending on how you set up your coordinate system.

If you mean the constant g, though, then that would be positive number (http://en.wikipedia.org/wiki/Standard_gravity). I think that's the convention.

D H
Staff Emeritus
g0 is a positive number, but whether gravitational acceleration is positive or negative depends on whether you have defined up or down to be positive. There's nothing saying you must have down positive. In fact, it is quite common to have up positive, in which case g will be negative.

well, i mean for the earth... he says that on the earth, it has to be negative to go downward. i was using the whole coordinate system thing to explain it better to him because hes also a java programmer. And yes, acceleration of gravity as defined below, which i also showed him. I mean, do u see a negative sign there? because i dont... but, i figured this would be the place to get it resolved. So the final question: Here on earth, where gravity pulls u towards the center of the earth and hols u firmly down to the surface, is accerlation of gravity a negative or positive number?
g = Gm(earth)/r2(earth) = 9.8ms/2

D H
Staff Emeritus
If you look at a tiny portion of the surface of the Earth, this vector is nearly constant in magnitude and direction. You can treat this as constant without much loss of accuracy over this small area. Then you can define a local coordinate system in which one of the axes is directed along or against this local acceleration vector. If you choose this vertical axis (call it the z axis) to be positive upward then the acceleration due to gravity is $-g\hat z$. If you choose the vertical axis to be positive downward then acceleration due to gravity is $+g\hat z$.