Orbital Energy & Escape

Start from the one force. Gravity supplies the centripetal force, GMm/r² = mv²/r, and multiplying both sides by r/2 gives ½mv² = GMm/2r without ever finding v. The potential energy, with its zero at infinite separation, is U = −GMm/r: twice as large, and negative. Adding them gives E = −GMm/2r. So every circular orbit keeps its books in the ratio K : U : E = 1 : −2 : −1, whatever the planet, the radius or the satellite's mass. Take one kilogram at r = 2R from Earth's centre, where GM/R = 62.6 MJ kg⁻¹: K = 15.6 MJ, U = −31.3 MJ and E = −15.6 MJ, at a speed of 5.59 km s⁻¹. On a graph of energy against r, the curve of circular-orbit energy, −GMm/2r, runs exactly halfway between the U curve and zero, and all three curves close on zero as r grows. The negative total is the part that matters: it is the energy the satellite is short of being free.

For the satellite and planet together, the net work done by every force other than gravity equals the change in E. Between two circular orbits that is W = E₂ − E₁ = (GMm/2)(1/r₁ − 1/r₂), and the ledger splits it: U changes by 2W and K by −W. Take a 2000 kg satellite from a 300 km parking orbit, r₁ = 6.671 × 10⁶ m, to the geostationary radius, r₂ = 4.216 × 10⁷ m. Its total energy climbs from −59.8 GJ to −9.45 GJ, so the engines must supply a net 50.3 GJ. The potential energy rises by 100.6 GJ, twice that, while the kinetic energy falls by 50.3 GJ and the speed drops from 7.73 to 3.07 km s⁻¹. Half of the potential energy gained is bought by the engines; the other half is paid out of the satellite's own motion. These totals depend only on the two end orbits. How the burns are arranged changes the fuel used, never the ledger.

Calculus turns the drag paradox into numbers. A drag force F_D opposes the velocity, so it removes energy at the rate dE/dt = −F_D v. While the orbit stays nearly circular, E = −GMm/2r, so dE/dt = (GMm/2r²) dr/dt, and with GM = v²r the radius falls at dr/dt = −2F_D r/(mv). The speed follows from v = √(GM/r): dv/dt = −(v/2r) dr/dt = +F_D/m. The satellite gains speed at exactly the rate drag alone would take it away. Over one period T = 2πr/v the radius drops by 4πr F_D/(m g(r)), where g(r) = GM/r² is the local field, so the fraction of the radius lost per orbit is 4π times the ratio of drag to local weight. A 1200 kg satellite 350 km up, with r = 6.721 × 10⁶ m and 2.0 mN of drag, loses 16 m of radius and gains 9.1 mm s⁻¹ on each 91-minute orbit: about 250 m a day, from a force about one five-millionth of its weight.

Gravity has unlimited range, yet the work it does on a body moving from r out to infinity is finite: ∫ (−GMm/r′²) dr′ from r to ∞ = −GMm/r. Kinetic energy can therefore cover the whole debt, and a freely coasting body never turns back if ½mv² − GMm/r ≥ 0. The least such speed is v_esc = √(2GM/r): 11.19 km s⁻¹ from Earth's surface and 2.38 km s⁻¹ from the Moon's, ignoring air and rotation. Three features matter. The mass cancels, so a pebble and a probe need the same speed. The direction never enters, because energy is a scalar: straight up, sideways or at 40° all escape, provided the path misses the planet and its atmosphere. And from a circular orbit v_esc = √2 × √(GM/r), so a satellite needs 41% more speed, which doubles K and lifts E from −K to zero. From a 300 km parking orbit that means a burn from 7.73 to 10.93 km s⁻¹.

In unpowered motion E is fixed and K = E − U(r) cannot be negative, so the body is confined to radii where U(r) ≤ E. That one inequality classifies every path. If E < 0 the state is bound, and −GMm/r ≤ E caps the distance at rₘₐₓ = GMm/|E|. A body moving straight out reaches that ceiling and falls back; one moving sideways with the same energy turns round sooner, because part of its kinetic energy stays tied up in sideways motion. A circular orbit at r has enough energy to coast straight out to 2r, yet stays at r. Fire a probe straight up from an airless Earth at 10.0 km s⁻¹ and E = −12.6 MJ kg⁻¹, so it stops at 4.98R and returns. If E = 0 the body just escapes, its speed shrinking towards zero. If E > 0 it leaves with speed to spare, v∞ = √(v² − v_esc²): launched at 12.0 km s⁻¹, only 0.81 km s⁻¹ above escape speed, it still coasts away at 4.34 km s⁻¹.

The accounts assume one isolated spherical body, no drag and burns short enough to count as instantaneous. Each assumption has a limit. Escaping Earth is not escaping the Sun: at Earth's distance the Sun's escape speed is 42.12 km s⁻¹, and Earth's orbital motion supplies 29.78 km s⁻¹ of it, so a probe leaving along that motion still needs a further 12.34 km s⁻¹ once clear of Earth, which from the ground takes √(11.19² + 12.34²) = 16.7 km s⁻¹. Air steals energy from anything fast and low, so no real launch coasts up from the surface at 11 km s⁻¹; rockets climb under thrust, and escape speed matters only where the engine stops. Earth's rotation lends up to 0.465 km s⁻¹ to an eastward launch. And the model is Newtonian: set √(2GM/r) equal to the speed of light and r = 2GM/c², 2.95 km for the Sun's mass, a regime where general relativity replaces this reasoning.