Gravitational attraction problem

[uha1]

Homework Statement

A particle is projected vertically upward from the Earth's surface with initial speed Vo. Prove that the maximun height H reached above the Earth's surface is H= Vo^2R/(2gR-Vo^2)

The Attempt at a Solution

R is the Earth's radius. If the gravitational attraction at the Earth's surface is mg, then the attraction at some height r above the surface will be mgR^2/(R+r)^2. The attraction can be integrated from r = 0 to r = H to get the total work involved, which must equal the kinetic energy mVo^2/2. HOWever, when we integral from
r=0 to r=H of mgR^2/(R+r)^2 and set it equal to mVo^2/2,
H= does not give the value Vo^2R/(2gR-Vo^2) ??

 

[Mentz114]

It's easier to use
final potential energy = initial kinetic energy.

 

[uha1]

It's easier to use
final potential energy = initial kinetic energy.

If we do it that way, it gives us...


mv^2(1/2) = mgh

h= v^2*(1/2)/g

and it still doesn't satisfy the answer

 

[Mentz114]

You're assuming the potential is the same at R and R+h.
But they stand in the ratio R/(R+h)