Dealing with Impact on Non-Flat Surfaces?

Question

A 1kg object is released 500 km above earth. What is its impact speed as it hits the ground, ignoring air resistance?

What would the impact speed be if the earth was flat?

Answer 1

The potential energy a height h above the earth is GmM/r where r=Rearth+h and the change in potential energy =1/2mv^2
GmM/Rearth-GmM/(Rearth+h)=1/2mv^2
If the earth were flat means that h<

Answer 2

The force of gravity F=-G*M*m/x^2 and directed to the Earth, x is a distance between Earth and object centers any minute before collision, M is Earth’s mass, given m=1kg, G is gravity constant; here comes Boom! - the work of gravity
W=-G*M*m*intergal{for x=R+h until R} of x^(-2) *dx =
G*M*m/x = G*M*m* (1/R – 1/(R+h)) = G*M*m*h/(R^2+h*R) = (G*M*m/R^2) * h/(1+h/R), where given h=5*10^5, R=5.370*10^6m is Earth’s radius, g= 9.81m/s^2 = G*M*m/R^2;
thus W=mgh/(1+h/R) = kinetic energy = 0.5*m*v^2, hence speed of impact v=sqrt(2gh/(1+h/R)) = 2996 m/s; thus for m=1kg or not = 1kg the speed will be the same, yet impact W depends on mass m.
now let’s take a spade and flatten the spot of impact, and see what would happen. Boom! Do you see any difference Owl_Finch?