Project Gravita

This is really a beautiful problem. This uses a bit of knowledge of mathematics too. Solve this problem if you are comfortable of properties of ellipses. Let us calculate the total energy of the particle at the time of launch. $TE = KE+PE$ $TE= \dfrac{m}{2}\dfrac{GM}{R}-\dfrac{GMm}{R}$ $TE=\dfrac{-GMm}{2R}$ We know that the particle will move around in an orbit of ellipse. Let eccentricity and semi major axis and semi minor axis of the ellipse are $e$ and $a$ and $b$ respectively. Now it is well known that $TE=\dfrac{-GMm}{2a}$ Hence we get $a=R$ We know that Center of Earth will lie on one of the focii of the ellipse. Also if we draw the line joining end of minor axis and the focus the length of the segment is $a$ . So we can say that if we join the point of launch and point of landing, that would be semi minor axis. From geometry and pythagoras theorem we get $2b=\sqrt{2}R$ Hence $b=\dfrac{R}{\sqrt{2}}$ From geometry of ellipse $e^{2} = 1 - \dfrac{b^{2}}{a^{2}}$ $e=\dfrac{1}{\sqrt{2}}$ Again from geometry of ellipse, length of focus from far off vertex is $d = a+ae$ Hence $d=R(1+\dfrac{1}{\sqrt{2}})$

Let the length of semimajor axis be a and eccentricity of the ellipse be e $\frac{-GMm}{2R}+\frac{m}{2}(\sqrt{\frac{GM}{R}})^{2}=-\frac{GMm}{2a}$

$a=R$ As the separation between the ends of semi-major minor axis and the focus is equal to the length of the semi major axis we can clearly say that launching and landing sites are at the end of the semi minor axis. $Rcos\frac{\pi}{4}=ae$

$a+ae=R\frac{\sqrt{2}+1}{\sqrt{2}}$

Maybe you can switch this image in place of the one posted now, which is hard to discern.

Edit Okay, it's been switched. All good.