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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 $x=\frac{2gR^2}{2gR-u^2}$

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  • $\begingroup$ Since this isn't a physics a forum, you might want to include the fact that $g=\frac{k}{R^2}$. $\endgroup$ Commented Aug 8, 2021 at 11:29
  • $\begingroup$ 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) $\endgroup$ Commented Aug 8, 2021 at 11:41
  • $\begingroup$ 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. $\endgroup$ Commented Aug 8, 2021 at 11:53
  • $\begingroup$ @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. $\endgroup$ Commented Aug 8, 2021 at 12:38
  • $\begingroup$ @AlanAbraham also for part iii) am I supposed to get 11,200m/s after applying the limit? $\endgroup$ Commented Aug 8, 2021 at 12:54

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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!

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