Oberth effect

A rocket works by transferring momentum to its propellant.[3] At a fixed exhaust velocity, this will be a fixed amount of momentum per unit of propellant.[4] For a given mass of rocket (including remaining propellant), this implies a fixed change in velocity per unit of propellant. Because kinetic energy equals mv²/2, this change in velocity imparts a greater increase in kinetic energy at a high velocity than it would at a low velocity. For example, considering a 2 kg rocket:

• at 1 m/s, the rocket starts with 1² = 1 J of kinetic energy. Adding 1 m/s increases the kinetic energy to 2² = 4 J, for a gain of 3 J;
• at 10 m/s, the rocket starts with 10² = 100 J of kinetic energy. Adding 1 m/s increases the kinetic energy to 11² = 121 J, for a gain of 21 J.

This greater change in kinetic energy can then carry the rocket higher in the gravity well than if the propellant were burned at a lower speed.

Rocket engines produce the same force regardless of their velocity. A rocket acting on a fixed object, as in a static firing, does no useful work at all; the rocket's stored energy is entirely expended on accelerating its propellant in the form of exhaust. But when the rocket moves, its thrust acts through the distance it moves. Force multiplied by distance is the definition of mechanical energy or work. So the farther the rocket and payload move during the burn (i.e. the faster they move), the greater the kinetic energy imparted to the rocket and its payload and the less to its exhaust.

This is shown as follows. The mechanical work done on the rocket (${\displaystyle W}$) is defined as the dot product of the force of the engine's thrust (${\displaystyle {\vec {F}}}$) and the displacement it travels during the burn (${\displaystyle {\vec {s}}}$):

${\displaystyle W={\vec {F}}\cdot {\vec {s}}.}$

If the burn is made in the prograde direction, ${\displaystyle {\vec {F}}\cdot {\vec {s}}=\|F\|\cdot \|s\|=F\cdot s}$. The work results in a change in kinetic energy

${\displaystyle \Delta E_{k}=F\cdot s.}$

Differentiating with respect to time, we obtain

${\displaystyle {\frac {\mathrm {d} E_{k}}{\mathrm {d} t}}=F\cdot {\frac {\mathrm {d} s}{\mathrm {d} t}},}$

or

${\displaystyle {\frac {\mathrm {d} E_{k}}{\mathrm {d} t}}=F\cdot v,}$

where ${\displaystyle v}$ is the velocity. Dividing by the instantaneous mass ${\displaystyle m}$ to express this in terms of specific energy (${\displaystyle e_{k}}$), we get

${\displaystyle {\frac {\mathrm {d} e_{k}}{\mathrm {d} t}}={\frac {F}{m}}\cdot v=a\cdot v,}$

where ${\displaystyle a}$ is the acceleration vector.

Thus it can be readily seen that the rate of gain of specific energy of every part of the rocket is proportional to speed and, given this, the equation can be integrated (numerically or otherwise) to calculate the overall increase in specific energy of the rocket.

Integrating the above energy equation is often unnecessary if the burn duration is short. Short burns of chemical rocket engines close to periapsis or elsewhere are usually mathematically modeled as impulsive burns, where the force of the engine dominates any other forces that might change the vehicle's energy over the burn.

For example, as a vehicle falls toward periapsis in any orbit (closed or escape orbits) the velocity relative to the central body increases. Briefly burning the engine (an “impulsive burn”) prograde at periapsis increases the velocity by the same increment as at any other time (${\displaystyle \Delta v}$). However, since the vehicle's kinetic energy is related to the square of its velocity, this increase in velocity has a non-linear effect on the vehicle's kinetic energy, leaving it with higher energy than if the burn were achieved at any other time.[5]

It may seem that the rocket is getting energy for free, which would violate conservation of energy. However, any gain to the rocket's kinetic energy is balanced by a relative decrease in the kinetic energy the exhaust is left with (the kinetic energy of the exhaust may still increase, but it does not increase as much).[2]^: 204 Contrast this to the situation of static firing, where the speed of the engine is fixed at zero. This means that its kinetic energy does not increase at all, and all the chemical energy released by the fuel is converted to the exhaust's kinetic energy (and heat).

At very high speeds the mechanical power imparted to the rocket can exceed the total power liberated in the combustion of the propellant; this may also seem to violate conservation of energy. But the propellants in a fast-moving rocket carry energy not only chemically, but also in their own kinetic energy, which at speeds above a few kilometres per second exceed the chemical component. When these propellants are burned, some of this kinetic energy is transferred to the rocket along with the chemical energy released by burning.[7]

The Oberth effect can therefore partly make up for what is extremely low efficiency early in the rocket's flight when it is moving only slowly. Most of the work done by a rocket early in flight is "invested" in the kinetic energy of the propellant not yet burned, part of which they will release later when they are burned.

• Bi-elliptic transfer
• Gravity assist
• Propulsive efficiency

• Oberth effect
• Explanation of the effect by Geoffrey Landis.

• Rocket propulsion, classical relativity, and the Oberth effect

• Animation (MP4) of the Oberth effect in orbit from the Blanco and Mungan paper cited above.