Spherical Coordinate Systems(Cartesian, i think it called)

[relion65]

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 I've read from several books and other online resources that's 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)?

 

[caleb5040]

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]

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.

 

[relion65]

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 he's 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]

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?

Acceleration is a vector, not a number.

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 $+g\hat z$.

 

[Ken G]

To underscore that, the questioner seems to imagine that the Earth "comes with" a coordinate system, such that the acceleration of gravity could be positive or negative in some definite way. But in fact coordinate systems are always arbitrary, and many different ones get used on the Earth. The key is that you are clear what direction you are regarding as positive (that's part of the coordinate system), and after that you just need to make sure the sign of gravity corresponds to downward (or toward the center, if you are using a global coordinate system like spherical coordinates). I've seen a hundred problems where that meant a positive gravity, and hundreds more where it meant negative. If I had to choose a more common convention, it would be that gravity has a negative sign, but that's just because it is an attractive force and it is somewhat common to choose a positive sign when distances increase, and a negative sign when they get closer. And as you just heard, the real point is that gravity is a vector, so has strength and direction-- it only acquires a sign when you start talking about the components of gravity, but if you haven't had vector calculus yet, this might be too advanced at this point.