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Outside a spherical planet, Newtonian mechanics gives the inverse-square central force equationThe 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 areA 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 givesThe 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 $0<\eta \ll 1$, in the same tangential direction. At release $\dot{r}=0$, butThus 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 rootsIts semi-major axis is $a=r_s/(2-b)>r_s$. The speed follows from the same conserved energy: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.