Past exam of the physics B course of the University of Cambridge / 2011 / Paper 2 / Section C / C12 / Solution

Write $\mu =GM$ and $A=\mu m$. For a circular orbit of radius $r$, centripetal balance gives $v^2=\mu /r$. Thus $E=m{v}^2/2-A/r=-A/(2r)$ and $J^2=m^2{r}^2{v}^2=Amr$. Since $e=0$, $a=r_0=r$, these agree with the general expressions.

For an outward coplanar Hohmann transfer, burn prograde at the inner circle to put its radius at the ellipse's pericentre, then burn prograde at apocentre to circularize at the station's radius. The station must be phased to meet the spacecraft there. With $r_p=R$, $r_a=\alpha R$,Energy at the first burn, with $v_0^2=\mu /R$, givesAngular momentum is conserved along the transfer ellipse, so $v_{2-}=v_1/\alpha$, while the outer circular speed is $v_{2+}=v_0/\sqrt{\alpha }$. For $\alpha =3$,The correct second impulse adds $\Delta v=v_{2+}-v_{2-}$. The reversed circularization burn in a Hohmann transfer, using the same impulse magnitude, instead leavesIts magnitude is below the circular speed at $3R$, so that point is the new apocentre. Define $q=r_a{v}_w^2/\mu =(\sqrt{2}-1{)}^2=3-2\sqrt{2}$. For an ellipse at apocentre, $q=1-e_w$, givingThe lunar radius is $R/3$. Since $r_{p,w}<R/3$, the osculating orbit intersects the Moon: the spacecraft crashes before reaching its formal pericentre. The calculation uses impulsive tangential burns and a spherical central gravitational field; it does not assume that a trajectory continues through the solid body.

The original plot uses $R$ as the length unit and $R/3$ as the lunar radius. The initial station position shown is phased for rendezvous: its initial angular lead is $\pi -\pi [(1+\alpha )/(2\alpha ){]}^{3/2}$, modulo $2\pi$.