Answer
See the detailed answer below.
Work Step by Step
a) We know that the final velocity is half the initial velocity, so to find the final one we need to find the initial one.
The net force exerted on the spaceship initially is given by
$$\sum F_G=\dfrac{GM \color{red}{\bf\not}m_{ship}}{r_0^2}= \color{red}{\bf\not}m_{ship}a_r$$
where $a_r=v^2/r_0$
$$ \dfrac{GM }{r_0^{ \color{red}{\bf\not}2}}= \dfrac{v_0^2}{ \color{red}{\bf\not}r_0} $$
Hence,
$$v_0=\sqrt{\dfrac{GM}{r_0}}\tag 1$$
Thus, the final velocity is given by
$$v_{new}=\dfrac{1}{2}v_0=\dfrac{1}{2}\sqrt{\dfrac{GM}{r_0}}$$
$$\boxed{v_{new}= \sqrt{\dfrac{GM}{4r_0}}}\tag 2$$
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b) The maximum radius is the initial radius $r_0$, as we see in the figure below.
We can see that the energy of the system is conserved since we assume that is an isolated system.
Thus,
$$E_i=E_f$$
$$K_i+U_{ig}=K_f+U_{gf}$$
$$\frac{1}{2} \color{red}{\bf\not}m_{ship}v_i^2+\dfrac{-GM \color{red}{\bf\not}m_{ship}}{r_0 }=\frac{1}{2} \color{red}{\bf\not}m_{ship}v_f^2+\dfrac{-GM \color{red}{\bf\not}m_{ship}}{r_{min} }$$
$$\frac{1}{2} v_i^2+\dfrac{-GM }{r_0 }=\frac{1}{2} v_f^2+\dfrac{-GM }{r_{min} }$$
$\times 2$;
$$ v_i^2+\dfrac{-2GM }{r_0 }= v_f^2+\dfrac{-2GM }{r_{min} }$$
$$ v_i^2- v_f^2=\dfrac{-2GM }{r_{min} }+\dfrac{ 2GM }{r_0 }\tag 3$$
And since the system is isolated, the linear momentum around the planet is conserved.
$$L_i=L_f$$
$$ \color{red}{\bf\not}m_{ship}v_i\;r_0= \color{red}{\bf\not}m_{ship}v_fr_{min}$$
$$ v_ir_0= v_f\;r_{min}$$
Thus,
$$v_f=\dfrac{v_ir_0}{r_{min}}$$
Plugging into (3);
$$v_i^2- \left[\dfrac{v_ir_0}{r_{min}}\right]^2 =\dfrac{-2GM }{r_{min} }+\dfrac{ 2GM }{r_0 } $$
$$v_i^2- \left(\dfrac{r_0^2}{r_{min}^2}\right)v_i^2 =\dfrac{-2GM }{r_{min} }+\dfrac{ 2GM }{r_0 } $$
$$v_i^2\left[ 1- \dfrac{r_0^2}{r_{min}^2}\right] =\dfrac{-2GM }{r_{min} }+\dfrac{ 2GM }{r_0 } $$
Noting that $v_i$ here is the new velocity we found above.
Plugins from (2)
$$\dfrac{ \color{red}{\bf\not}G \color{red}{\bf\not}M}{4r_0}\left[ 1- \dfrac{r_0^2}{r_{min}^2}\right] =\dfrac{-2 \color{red}{\bf\not}G \color{red}{\bf\not}M }{r_{min} }+\dfrac{ 2 \color{red}{\bf\not}G \color{red}{\bf\not}M }{r_0 } $$
$\times 4r_0$;
$$ 1- \dfrac{r_0^2}{r_{min}^2} =\dfrac{-8r_0}{r_{min} }+\dfrac{ 8 \color{red}{\bf\not}r_0 }{ \color{red}{\bf\not}r_0 } $$
$$ \dfrac{r_{min}^2-r_0^2}{r_{min}^{ \color{red}{\bf\not}2}} =\dfrac{-8r_0+8r_{min} }{ \color{red}{\bf\not}r_{min} }$$
$$r_{min}^2-r_0^2=-8r_0r_{min}+8r_{min}^2$$
$$7r_{min}^2 -8r_0r_{min} +r_0^2=0$$
Hence, using the quadratic formula;
$$r_{min}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$
where here $a=7$, $b=-8r_0$, and $c=r_0^2$
$$r_{min}=\dfrac{ -(-8r_0) \pm \sqrt{ (-8r_0)^2-4 (r_0^2)(7)}}{2(7)}$$
Therefore,
$$\boxed{r_{min}=\dfrac{r_0}{7}}\;\;\;\;\;\;{\rm Or}\;\;\;\;\;\; r_{min}=r_0$$
This means that the maximum radius is $r_0$ and the minimum radius is $r_0/7$
