Acceleration given distance from earth (r)

In the equation you posted, you're missing the gravitational constant "G." However, I suspect you used it in your calculation because I got the same answer when I included G. You seem to be assuming that your answer of .31 m/s² is wrong, but it seems fine to me.

I wouldn't use small g to denote the acceleration in orbit though. Small g usually refers to the acceleration due to gravity on earth's surface. Instead, you could call it "a" for acceleration.

 

Sure, you can use Newton's law of gravitation no problem. That's what the equation you used is derived from. Here's how:

[itex]\displaystyle F=G\frac{mM_E}{r^2}[/itex]

[itex]\displaystyle a=F/m[/itex]

Substitute F:

[itex]\displaystyle a=G\frac{mM_E}{r^2}/m[/itex]

and m cancels to give:

[itex]\displaystyle a=G\frac{M_E}{r^2}[/itex]

Then, plugging in the radius given by the problem, and putting in the numerical values for G and M_E (mass of earth):

[itex]\displaystyle a=(6.673×10^{-11})\frac{5.972×10^{24} kg}{(3.59×10^7 m)^2} = 0.31\frac{m}{s^2}[/itex]

Are you convinced that you got the right answer yet? :)