Consider a small mass $m$ moving in a circular orbit of radius $r$ around a much more massive point mass $M$. Newtonian gravity supplies the inward force required for circular motion.

The gravitational force magnitude is

$$F_g=\frac{GMm}{r^2},$$

while uniform circular motion requires inward net force

$$F_r=m\frac{v^2}{r}.$$

Equating them,

$$\frac{GMm}{r^2}=m\frac{v^2}{r}.$$

The orbiting mass cancels, giving the circular-orbit speed

$$\boxed{v_c=\sqrt{\frac{GM}{r}}}.$$

In this ideal point-mass model, the circular orbital speed depends on the central mass and orbital radius, not on the mass of the orbiting object.

Orbital period

For uniform circular motion,

$$v_c=\frac{2\pi r}{T}.$$

Substituting the gravitational circular speed,

$$\frac{2\pi r}{T}=\sqrt{\frac{GM}{r}}.$$

Therefore

$$\boxed{T=2\pi\sqrt{\frac{r^3}{GM}}},$$

or equivalently,

$$\boxed{T^2=\frac{4\pi^2}{GM}r^3}.$$

This is the circular-orbit form of Kepler's third-law scaling: larger circular orbits have longer periods, with $T^2\propto r^3$ for a fixed central mass.

Circular orbit as continuous free fall

The orbiting object has a nonzero inward acceleration

$$a_r=\frac{v_c^2}{r}=\frac{GM}{r^2}.$$

It is therefore not force-free. Gravity continuously bends the velocity vector toward the central mass. The object keeps missing the center because its tangential speed carries it sideways while it falls inward.

If the speed at a given radius differs from $v_c$, the trajectory is not a circular orbit at that radius. The subsequent motion must be analyzed using the full gravitational dynamics and conserved quantities.

Worked example

An ideal point mass has

$$M=6.0\times10^{24},\mathrm{kg}.$$

Find the circular speed and period at

$$r=7.0\times10^6,\mathrm m.$$

Using

$$G=6.67\times10^{-11},\mathrm{N,m^2/kg^2},$$

$$v_c=\sqrt{\frac{(6.67\times10^{-11})(6.0\times10^{24})}{7.0\times10^6}} \approx7.56\times10^3,\mathrm{m/s}.$$

So

$$v_c\approx7.6,\mathrm{km/s}.$$

The period is

$$T=\frac{2\pi r}{v_c} \approx\frac{2\pi(7.0\times10^6)}{7.56\times10^3} \approx5.82\times10^3,\mathrm s,$$

or about

$$97,\mathrm{min}.$$

Circular orbital motion is therefore a direct application of the same radial dynamics used for any circular trajectory: gravity is the real force providing the required inward acceleration.