Gravitational Energy: Turning Radius and Escape Speed

Use a hypothetical planet with $\mu=1.6\times10^{14}\,\mathrm{m^3/s^2}$ and $R=6.0\times10^6\,\mathrm m$. An object starts at $r_0=1.0\times10^7\,\mathrm m$, moving radially outward at $v_0=4000\,\mathrm{m/s}$. Find its greatest radius and altitude before it returns.

First calculate energy per kilogram, keeping the potential term negative:

$$\varepsilon=\frac{4000^2}{2}-\frac{1.6\times10^{14}}{1.0\times10^7}=-8.0\times10^6\,\mathrm{J/kg}.$$

Because the motion is radial, the entire speed is zero at the outer turning point. Conservation then gives $\varepsilon=-\mu/r_{\max}$, so

$$r_{\max}=-\frac{\mu}{\varepsilon}=2.0\times10^7\,\mathrm m.$$

The maximum altitude is $h_{\max}=r_{\max}-R=1.4\times10^7\,\mathrm m$. The object rises by $r_{\max}-r_0=1.0\times10^7\,\mathrm m$, a different distance. Check: its potential energy per kilogram increases from −16 million to −8 million J/kg, exactly matching the 8 million J/kg of kinetic energy lost.

Zero speed at the top does not mean equilibrium. Gravity still points inward, so the object begins returning. Our calculation stops at the planet's surface on its return; the exterior model is not a description of motion through the planet.

Do not transfer the zero-speed step to an arbitrary orbit. At the farthest point of a nonradial bound orbit, radial speed is zero but tangential speed generally is not. The same 4000 m/s directed tangentially at this starting radius gives a circular orbit instead. Energy alone does not determine the trajectory; direction and angular momentum matter.

At the same starting radius, find the ideal minimum escape speed. Then find the limiting speed far away if the object launches radially outward at 6000 m/s.

At the threshold, the object can approach infinite separation with speed approaching zero. Because $U\to0$, its total energy is zero. Setting the initial specific energy to zero gives

$$v_{\mathrm{esc}}=\sqrt{\frac{2\mu}{r_0}}\approx5657\,\mathrm{m/s}.$$

This threshold is independent of the test mass, under our assumptions. For 6000 m/s, the initial specific energy is $\varepsilon=2.0\times10^6\,\mathrm{J/kg}$. At infinity, potential energy approaches zero, leaving

$$v_\infty=\sqrt{2\varepsilon}=2000\,\mathrm{m/s}.$$

Check directly: $v_\infty^2=v_0^2-2\mu/r_0$. Do not subtract the two speeds; subtract their squared contributions in the energy equation. No finite outer turning radius exists for these outward radial launches with $\varepsilon\ge0$. At the exact threshold the object keeps moving outward at every finite radius and slows toward zero only asymptotically; it does not reach infinity in a finite time.

Negative energy means the separation has an upper bound, not that a safe satellite orbit is guaranteed. A radial launch can return into the planet. For nonnegative energy, an outward collision-free trajectory is needed to escape; an inward launch may hit the surface first. Escape also does not mean gravity switches off at some altitude.

Compare a 250 kg satellite in two possible circular orbits around the same planet: $r_1=1.0\times10^7\,\mathrm m$ and $r_2=2.0\times10^7\,\mathrm m$. For circular motion only, the inward-force equation gives $v^2=\mu/r$. Substitute into kinetic and total energy:

$$K_{\mathrm{circ}}=\frac{m\mu}{2r},\qquad E_{\mathrm{circ}}=-\frac{m\mu}{2r}.$$

In the lower orbit, $K_1=2.0\times10^9\,\mathrm J$, $U_1=-4.0\times10^9\,\mathrm J$ and $E_1=-2.0\times10^9\,\mathrm J$. In the higher orbit these values are respectively $1.0\times10^9\,\mathrm J$, $-2.0\times10^9\,\mathrm J$ and $-1.0\times10^9\,\mathrm J$.

The higher state has 1 billion joules more total energy, despite having 1 billion joules less kinetic energy. Its potential energy increased by 2 billion joules. Its circular speed is about 2828 m/s rather than 4000 m/s. Comparing only speeds misses the larger potential-energy change.

This is a comparison of two states, not a maneuver plan or fuel calculation. An engine burn initially changes velocity at the satellite's current position. The circular-speed formula does not describe every point of the transfer trajectory, and the net mechanical-energy difference alone does not specify thrust, fuel, or timing.