Circular orbit

A circular orbit maintains constant distance from its centre of attraction. Its velocity changes direction even when its speed stays constant. The inward centripetal acceleration has magnitude $v^2/r$, so the radial force must satisfy $F_r=-m{v}^2/r$. For a Newtonian gravitational field with standard gravitational parameter $\mu$, this gives $v=\sqrt{\mu /r}$.

A circular orbit in a central potential $U(r)=A{r}^s$ requires inward force $U^{\prime }(r)=m{v}^2/r>0$. Thus $2K=r{U}^{\prime }(r)=sU$, giving $K/U=s/2$ for the specified zero of potential. The ratio depends on that choice of zero; adding a constant to $U$ leaves the orbit unchanged but changes $K/U$. This is the circular special case of the virial theorem.

For a fixed spherical gravitating source, place a test particle at $(a,0)$ with initial velocity $(0,\sqrt{\mu /a})$, where $a$ is outside the source. Its exact circular orbit isDifferentiating twice gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}=-\mu \mathbf{r}/a^3$, precisely the Newtonian gravitational field at each point of that circle. There is no outward physical force to cancel gravity in this inertial frame. Gravity supplies the inward centripetal acceleration that continually changes the velocity's direction.

Over a short interval, the displacement from the straight tangent path begins inward with magnitude $\frac{1}{2}(\mu /a^2)t^2$. The curvature of the orbit matters at the same order: the particle moves sideways while continually falling away from its instantaneous tangent. The full circular solution remains at radius $a$ rather than striking the source.

Its speed is $\sqrt{\mu /a}$, its oscillation period is $2\pi \sqrt{a^3/\mu }$, and its specific orbital energy is $-\mu /(2a)$. Removing gravity while keeping the initial velocity would give the straight path $(a,a\omega t)$, rather than a stationary particle. This contrast explains why orbital weightlessness is compatible with a substantial gravitational force: a freely coasting spacecraft and its contents fall together. ^[ref]

In a spherically symmetric gravitational potential $\Phi (r)$, a circular orbit has $r{\Omega }^2={\Phi }^{\prime }(r)$, specific orbital energy $\epsilon =\Phi +r^2{\Omega }^2/2$ and specific angular momentum $\mathbf{h}=r^2\Omega \mathbf{n}$, with $|\mathbf{n}|=1$. Differentiating gives ${\epsilon }^{\prime }=2r{\Omega }^2+r^2\Omega {\Omega }^{\prime }=\Omega (r^2\Omega {)}^{\prime }$. Changes in orbital orientation have $\mathbf{n}\cdot d\mathbf{n}=0$, so $\mathbf{\Omega }\cdot d\mathbf{h}=\Omega d(r^2\Omega )=d\epsilon$. For a particle of variable mass $M$ and total angular momentum $\mathbf{H}=M\mathbf{h}$, this implies $d(M\epsilon )=(\epsilon -\Omega h)dM+\mathbf{\Omega }\cdot d\mathbf{H}$. Moreover $d(\epsilon -\Omega h)/dr=-h{\Omega }^{\prime }$, which is positive when the orbital angular velocity decreases outward.

For unit mass with specific angular momentum $l$ in $V(r)=-e^{-r}$, the effective potential is $-e^{-r}+l^2/(2r^2)$. A positive-radius circular orbit satisfies the displayed equation. Since $r^3{e}^{-r}$ has its unique maximum $27/e^3$ at $r=3$, exactly two radii occur for $0<l^2<27/e^3$. Their radial stability follows from $V_{\mathrm{eff}}^{\prime \prime }=e^{-r}(3/r-1)$: the inner orbit is stable at fixed angular momentum, the outer unstable. The case $l=0$ has no finite-radius circular orbit despite satisfying the upper bound.

For the radial acceleration $-k{r}^n$ with $k>0$, a circular orbit of radius $a$ has squared specific angular momentum $h^2=k{a}^{n+3}$. At fixed $h$, its radial small oscillations have squared frequency $(n+3)k{a}^{n-1}$, so the orbit is stable for $n>-3$. This is a special case of the effective potential stability criterion.

For $V(r)=km{r}^2$, fixed nonzero angular momentum $h$ gives the effective potential $V_{\mathrm{eff}}=h^2/(2m{r}^2)+km{r}^2$. A positive-radius circular orbit exists when $k>0$, at the displayed radius, and $V_{\mathrm{eff}}^{\prime \prime }(r_0)=8km>0$ proves radial stability by the effective potential stability criterion. Its angular speed has magnitude $\sqrt{2k}$ and its small radial oscillation frequency is $\sqrt{8k}$. If $k\le 0$, no such orbit with nonzero angular momentum exists; for $h=0,k>0$, the sole minimum is at the origin, not a positive-radius circular orbit.