= Solution
Outside a spherical planet, <Newtonian mechanics> gives the inverse-square <central force> equation
$$
\ddot{\mathbf r}=-\frac{GM}{r^3}\mathbf r.
$$
The satellite's mass cancels. The <angular momentum> is conserved, so the orbit lies in a plane. In that plane the general equations in <polar coordinates> are
$$
\ddot r-r\dot\theta^2=-\frac{GM}{r^2},\qquad
r\ddot\theta+2\dot r\dot\theta=0.
$$
A <synchronous circular orbit> fixed above a surface point must be equatorial, prograde and have <angular velocity> $\omega=2\pi/T$. Substitution of constant $r$ into the radial equation gives
$$
\boxed{r_s=\left(\frac{GMT^2}{4\pi^2}\right)^{1/3},\qquad
v_s=\frac{2\pi r_s}{T}=\left(\frac{2\pi GM}{T}\right)^{1/3}.}
$$
The radius is measured from the planet's centre; an external orbit requires it to exceed the planet's radius.
Now let the release speed be $v=(1+\eta)v_s$, with $0<\eta\ll1$, in the same tangential direction. At release $\dot r=0$, but
$$
\ddot r(0)=\frac{v^2-v_s^2}{r_s}>0.
$$
Thus the satellite initially moves outward. Its specific total <energy> is $v^2/2-GM/r_s<0$ for this small increase, so the <Kepler orbit> is an <ellipse>, with the release point its <pericentre>. To make this explicit, put $b=v^2/v_s^2\in(1,2)$. The radial turning-point equation from energy and <angular momentum> conservation has roots
$$
r_{{\min}}=r_s,\qquad r_{{\max}}=\frac{b}{2-b}r_s.
$$
Its <semi-major axis> is $a=r_s/(2-b)>r_s$. The speed follows from the same conserved energy:
$$
v(r)^2=GM\left(\frac2r-\frac1a\right).
$$
\b[It is greatest at pericentre and least at apocentre.] In particular, the tangential speed at the two turning points satisfies $r_{{\min}}v_{{\max}}=r_{{\max}}v_{{\min}}$. The orbital period becomes $2\pi\sqrt{a^3/(GM)}>T$, so the orbit is no longer fixed over the same surface point.
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