A satellite moves in a circular orbit of radius 8R about a planet of radius R that may be treated as a uniform sphere. A thruster fired tangentially…

Let $GM=1$ and $R=1$. The original circular speed at $r=8$ is $v_0=\sqrt{1/8}$, and just after the tangential burn the speed is $v_a=v_0/2$, so $v_a^2=\tfrac1{32}$. Since this is below circular speed, the burn point $r_a=8R$ becomes the apogee. Angular momentum gives $v_pr_p=v_ar_a$ and energy gives $\tfrac12v_p^2-\dfrac1{r_p}=\tfrac12v_a^2-\dfrac1{r_a}=\tfrac1{64}-\tfrac18=-\tfrac7{64}$. With $v_p=8v_a/r_p$ and $v_a^2=\tfrac1{32}$, $\tfrac12v_p^2=\dfrac{1}{r_p^2}$, so $\dfrac1{r_p^2}-\dfrac1{r_p}+\dfrac7{64}=0$, i.e. $7r_p^2-64r_p+64=0$ with roots $r_p=8$ (the apogee itself) and $r_p=\tfrac87$. The perigee $\tfrac87R$ lies outside the planet ($>R$), so the ellipse is actually completed. Hence $v_p/v_a=r_a/r_p=8/(8/7)=7$. In general, reducing the speed to a fraction $k$ of circular speed gives $r_a/r_p=(2-k^2)/k^2$, which is $7$ for $k=\tfrac12$.