Escape velocity and characteristic energy

Draft

Escape is an energy condition, not a speed condition. Stating it that way makes the whole family of results fall out at once, including the ones that matter for interplanetary departure.

The energy condition

From the two-body problem, specific orbital energy is conserved:

\varepsilon = \frac{v^2}{2} - \frac{\mu}{r} = -\frac{\mu}{2a}

The first term is kinetic energy per unit mass, the second is gravitational potential energy per unit mass with the zero taken at infinity. Escape means reaching $r \to \infty$ , where the potential term vanishes. Since $v^2 \ge 0$ always, that is possible if and only if

\varepsilon \ge 0

Everything else on this page is bookkeeping around that inequality.

$\varepsilon$	$a$	$e$	Trajectory
$< 0$	$> 0$	$0 \le e < 1$	Bound: circle or ellipse
$= 0$	$\infty$	$e = 1$	Parabolic, the marginal case
$> 0$	$< 0$	$e > 1$	Hyperbolic, arrives with speed to spare

A negative semi-major axis is not an error. For a hyperbola $a < 0$ , and $\varepsilon = -\mu/2a$ stays positive, as it must.

Escape speed

Set $\varepsilon = 0$ and solve for $v$ :

Escape speed

v_{\text{esc}} = \sqrt{\frac{2\mu}{r}}

Three things about this expression are worth stating explicitly, because each one is a common source of confusion.

It is a speed, not a velocity. Direction does not appear, because energy is a scalar. Any direction that does not intersect the body will do.

It is a function of radius, so “escape velocity” without a stated radius is incomplete. The figure usually quoted is the surface value.

It is exactly $\sqrt{2}$ times the local circular speed. Since $v_{\text{circ}} = \sqrt{\mu/r}$ ,

v_{\text{esc}} = \sqrt{2}\, v_{\text{circ}} \approx 1.414\, v_{\text{circ}}

so escaping from a circular orbit costs a 41.4% speed increase, wherever that orbit is. From a 400 km circular Earth orbit at 7.67 km/s, escape requires 10.85 km/s, an increment of 3.18 km/s.

Escape speeds

Body	Reference radius	$v_{\text{esc}}$ (km/s)
Ceres	surface	0.51
Pluto	surface	1.21
Moon	surface	2.38
Titan	surface	2.64
Mercury	surface	4.25
Mars	surface	5.03
Venus	surface	10.36
Earth	surface	11.18
Earth	400 km altitude	10.85
Neptune	1 bar level	23.5
Jupiter	1 bar level	59.5
Sun	from 1 AU	42.13
Sun	photosphere	617.7

The Sun row from 1 AU is the one that governs interplanetary work. Earth orbits the Sun at 29.78 km/s, so leaving the solar system from Earth’s orbital distance needs only $42.13 - 29.78 = 12.35$ km/s more, and only if the increment is applied in the direction of Earth’s motion. The Earth is already doing most of the work.

Characteristic energy

For an escape trajectory the leftover speed at infinity is the quantity that actually matters, because it sets what the spacecraft can do next. Define the hyperbolic excess velocity $v_\infty$ by evaluating the energy integral at infinity:

\varepsilon = \frac{v_\infty^2}{2} \quad\Longrightarrow\quad v_\infty^2 = v^2 - \frac{2\mu}{r}

The characteristic energy is twice the specific energy:

Characteristic energy

C_3 = v_\infty^2 = -\frac{\mu}{a}

$C_3$ is quoted in km²/s². It is the standard currency for launch vehicle performance to escape trajectories, for one reason: it is additive with the energy the launcher delivers, whereas velocities are not. A launch vehicle performance curve is published as payload mass against $C_3$ , and mission designers read the required $C_3$ off a porkchop plot.

$C_3 = 0$ is exactly the parabolic case: escape with nothing left over. Negative $C_3$ means a bound orbit and is used for high-energy elliptical orbits.

Departure speed for a required C3

Invert the definition at the departure radius:

v_p = \sqrt{C_3 + \frac{2\mu}{r_p}}

Worked example. A Mars departure needing $C_3 = 12$ km²/s², from a 200 km circular parking orbit ( $r_p = 6578$ km, $\mu_\oplus = 398\,600.44$ km³/s²):

v_p = \sqrt{12 + \frac{2(398\,600.44)}{6578}} = \sqrt{12 + 121.19} = 11.54\ \text{km/s}

The circular speed there is $\sqrt{398\,600.44/6578} = 7.78$ km/s, so the trans-Mars injection burn is $11.54 - 7.78 = 3.76$ km/s.

Note what happened to the 12 km²/s². Adding it to 121.19 changed the square root from 11.01 to 11.54 km/s: 12 km²/s² of characteristic energy cost only 0.53 km/s of departure speed, on top of the 3.23 km/s that bare escape would have cost. That is not an accident, and it has a name.

The Oberth effect

Because energy goes as $v^2$ , a fixed velocity increment $\Delta v$ applied at speed $v$ changes specific energy by

\Delta\varepsilon = \frac{(v + \Delta v)^2 - v^2}{2} = v\,\Delta v + \frac{\Delta v^2}{2}

The dominant term is $v\,\Delta v$ . The same propellant buys more energy when burned at higher speed, which means deep in the gravity well, at periapsis.

Consequence 1: Departure burns are performed at perigee of the parking orbit, never at apogee.
Consequence 2: A three-burn escape via a high intermediate apogee can beat a direct escape for large plane changes.
Consequence 3: A Jupiter or solar Oberth manoeuvre buys energy no chemical stage could supply directly.
Limit: The gain is bounded by how deep periapsis can go before atmospheric or thermal limits bind.

The effect is not free energy. The propellant carries kinetic energy of its own before it is burned, and the extra spacecraft energy comes out of the exhaust, which is left in a lower-energy state. Energy is conserved; it is simply partitioned more favourably when the burn happens fast and low.

Escape from a rotating body

A launch site on a rotating planet starts with velocity from the rotation itself. At the equator Earth’s surface moves at

v_{\text{rot}} = \frac{2\pi R_\oplus}{86\,164.09\ \text{s}} = 0.465\ \text{km/s}

using the sidereal day, not the solar day. That contribution is available only for eastward launches, and it scales as $\cos(\text{latitude})$ , which is the reason equatorial launch sites are valuable and why a launch site’s latitude sets the cheapest reachable inclination. See Launch vehicles.

The rotational contribution is small against the roughly 9.4 km/s of ideal velocity a launcher must actually deliver to LEO once gravity, drag and steering losses are counted. See The rocket equation.

References

Bate, R. R., Mueller, D. D., White, J. E. and Saylor, W. W. Fundamentals of Astrodynamics, 2nd ed., Dover, 2020, chapter 1.
Vallado, D. A. Fundamentals of Astrodynamics and Applications, 5th ed., Microcosm Press, 2022.
Oberth, H. Wege zur Raumschiffahrt, R. Oldenbourg, 1929. The original statement of the periapsis-burn argument.
NASA Launch Services Program performance data, published as payload mass against $C_3$ for each vehicle.