A 910kg object is released from rest at an altitude of 1200 km above the north pole of the earth. If we ignore atmospheric friction, with what speed does the object strike the surface of the earth?

(the polar radius of the earth is 6357 km)

3 answers

v = sqrt(2 g y) = sqrt(2*9.8*1200)
Unless they want to get picky about the change in g at that height.
You are wrong. The information given implies that this question is specifically testing the student's ability to take the height into account. The correct answer is 4.5 km/s. Could you please demonstrate how to obtain this answer?
My best guess is to determine the acceleration using a=GM/(r+h)^2 and then plug that acceleration into v^2=v0^2+a(y-y0). I had a similar problem in my homework and that got me a better answer than anything else I had tried.