For a small mass $m$ moving in the gravitational field of a much larger point mass $M$, choose the gravitational potential-energy zero at infinite separation. Then

$$U(r)=-\frac{GMm}{r}.$$

The total mechanical energy is

$$\boxed{E=\frac12mv^2-\frac{GMm}{r}}.$$

This energy determines whether the object is gravitationally bound in the ideal two-body point-mass model.

Bound and unbound motion

At infinite separation,

$$U(\infty)=0.$$

If an object can reach infinity, its kinetic energy there must be nonnegative. Therefore:

• if $E<0$, infinity is energetically inaccessible and the motion is bound;
• if $E=0$, the object is exactly at the escape threshold;
• if $E>0$, the object has enough energy to reach infinity with nonzero residual speed and is unbound.

This classification uses total energy, not just the instantaneous speed. The same speed can represent different energetic situations at different radii because the gravitational potential changes with $r$.

Escape speed

The minimum speed required to escape from radius $r$ is obtained by setting the total energy equal to zero:

$$0=\frac12mv_{\rm esc}^2-\frac{GMm}{r}.$$

Solving,

$$\boxed{v_{\rm esc}=\sqrt{\frac{2GM}{r}}}.$$

The orbiting object's mass cancels. Under the ideal assumptions, escape speed depends only on the central mass and the starting radius.

The formula does not require the initial velocity to point radially outward. Direction affects the shape of the trajectory, but the energy threshold for being able to reach infinity depends on the speed magnitude through $K=\tfrac12mv^2$.

Energy of a circular orbit

For a circular gravitational orbit,

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

Its kinetic energy is therefore

$$K_c=\frac12m\frac{GM}{r}=\frac{GMm}{2r}.$$

Combining this with

$$U=-\frac{GMm}{r}$$

gives

$$\boxed{E_c=-\frac{GMm}{2r}}.$$

A circular orbit is therefore bound. Its potential energy is twice the total energy in magnitude and negative, while its kinetic energy is positive:

$$U=2E_c,\qquad K_c=-E_c.$$

Comparing circular and escape speeds at the same radius,

$$\boxed{v_{\rm esc}=\sqrt2,v_c}.$$

Worked example

For the ideal central mass

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

at radius

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

the escape speed is

$$v_{\rm esc} =\sqrt{\frac{2(6.67\times10^{-11})(6.0\times10^{24})}{7.0\times10^6}} \approx1.07\times10^4,\mathrm{m/s}.$$

Thus

$$v_{\rm esc}\approx10.7,\mathrm{km/s}.$$

The circular speed at the same radius is about $7.6,\mathrm{km/s}$, and

$$\sqrt2(7.6,\mathrm{km/s})\approx10.7,\mathrm{km/s},$$

as expected.

Negative gravitational potential energy encodes binding: energy must be supplied to raise a bound system from its negative total energy to the zero-energy escape threshold.