Motion question on escape velocity

Assume that the magnitude and direction due to gravity at a point outside Earth at distance $$x$$ from the Earth's centre is equal to $$-\frac{k}{x^2}$$, where $$k$$ is a constant.

i) Neglecting atmospheric resistance, prove that is an object is projected from the Earth's surface with speed $$u$$, it's speed $$v$$ in any position is given by $$v^2=u^2-2gR^2(\frac1R-\frac1x)$$ where $$R$$ is the Earth's radius and $$g$$ is the magnitude of acceleration due to Earth's gravity at the Earth's surface.

ii) Show that the greatest height $$H$$ above the Earth's surface is given by $$H=\frac{u^2R}{2gR-u^2}$$

iii) Hence if the radius of the Earth is $$6400$$km and the acceleration due to gravity is $$9.8ms^{-1}$$ find the speed required by the object to escape Earth's gravitational influence.

I've done part i) and iii), I'm struggling with part ii)
For Part ii) I've let $$v=0$$ but I get $$x=\frac{2gR^2}{2gR-u^2}$$

• Since this isn't a physics a forum, you might want to include the fact that $g=\frac{k}{R^2}$. Commented Aug 8, 2021 at 11:29
• I think the maths department at our school expects us to be able to work that out organically because I had to work that out for myself in part i) Commented Aug 8, 2021 at 11:41
• For part 3 you would use the condition in part ii and set $\lim_{H\to\infty}$. I am also stuck on part ii. I don't think it is possible to solve the differential equation from part i. Could we use the fact that the maximum height will be attained when the velocity is $0$? I'm not sure if this statement is necessarily true in all differential equations, however. Commented Aug 8, 2021 at 11:53
• @AlanAbraham in another question on this site math.stackexchange.com/questions/1645277/… someone has asked to show the same thing but the answer stops at where I cannot continue so I assume that is the correct condition. Commented Aug 8, 2021 at 12:38
• @AlanAbraham also for part iii) am I supposed to get 11,200m/s after applying the limit? Commented Aug 8, 2021 at 12:54

I'm pretty sure you made this mistake: Substituting $$H=x$$ instead of $$R+H=x$$. With $$x=R+H$$, and with $$v=0$$, we have $$u^2=2gR^2\left(\dfrac{1}{R}-\dfrac{1}{R+H}\right)=\dfrac{2gRH}{R+H} \implies H=\dfrac{u^2R}{2gR-u^2}$$. Hope it helps!