Question on Gravity Keplers Laws???

[++A++]

An object is dropped from a height of 1.27E+7m above the surface of the earth. What is its initial acceleration?

I thought I could use the equation for objects near the surface of the earth in free fall which is :
g (or a) = GM / R^2
But this doesnt give me the right answer. Can anyone point me in the right direction????

[arildno]

Welcome to PF!
You didn't use R=1.27E+7m, did you?
In that case, you'll get the wrong answer..

[arildno]

To give you a hint:
The "R" in the gravitation law is the DISTANCE FROM THE CENTER OF THE EARTH.

[tony873004]

Becareful about what value you use for G.
The gravitational constant G comes in many flavors depending on your units.
You've used meters to represent the radius, and I'm not sure if you're expressing mass in g, kg, or Earth masses. So make sure you pick the right G, or convert your units for use with the G that you're using.

[++A++]

I got it! I added the earths radius (6.37E+6) to the radius above the earths surface and used it in my equation to get the right answer (1.09). Thank you very much for the help!