Approximate Speed of Object Dropped from Height h with Variable Gravity

[pankildesai1]

Suppose an object is dropped from height h, where h < R but gravity is not constant). Show that the speed with which it hits the ground, neglecting friction, is approximately given by:

v = sqrt(2gh)(1 - (h/2r))

where g is the acceleration due to gravity on the surface of the Earth.
HINT: You will need to expand an expression in a MacLaurin series. Be sure to expand an expression with a small value, so higher order terms can be ignored.

 

[malawi_glenn]

And what is your attempt?

 

[pankildesai1]

g(h)= GM/(R+h)^2 which decreases with h increases

g(h) / g = R^2/(R+h)^2

g(h) = g * (1+ (h/R))^-2

now we know that for the fall if we consider energy conservation

1/2 mV^2 = mg(h)h

V^2 = 2*g(h)*h

V^2 = 2*g*h * (1+ (h/R))^-2

V = sqrt(2gh) * (1+(h/R))^-1

now if i expand the function (1+(h/R))^-1

Then I don't get the form which is asked for. Can you please help me to get the correct form?