# Oberth effect

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Not to be confused with Slingshot maneuver.

In astronautics, a powered flyby, or Oberth maneuver, is a maneuver in which a rocket falls into a gravitational well, and then accelerates when its fall reaches maximum speed.[1][2] The resulting maneuver is a more efficient way to gain kinetic energy than applying the same impulse outside of a gravitational well. The gain in efficiency is explained by the Oberth effect, wherein the use of a rocket at higher speeds generates greater mechanical energy than use at lower speeds. In practical terms, this means that the most energy-efficient method for a spacecraft to burn its engine is at the lowest possible orbital periapse, when its orbital velocity (and so, its kinetic energy) is greatest.[1] In some cases, it is even worth spending fuel on slowing the rocket into a gravity well to take advantage of the efficiencies of the Oberth effect.[1] The maneuver and effect are named after Hermann Oberth, the Austro-Hungarian-born German physicist and a founder of modern rocketry, who first described them in 1927.[3]

The Oberth effect is strongest at a point in orbit known as the periapse, where the gravitational potential is lowest, and the speed is highest. This is because firing a rocket engine at high speed causes a greater change in kinetic energy than when fired at lower speed. Because the vehicle remains near periapse only for a short time, for the Oberth maneuver to be most effective, the vehicle must be able to generate as much impulse as possible in the shortest possible time. Thus, the Oberth maneuver is much more useful for high thrust rockets like liquid-propellant rockets, and less useful for low-thrust reaction engines such as ion drives, which take a long time to gain speed. The Oberth effect also can be used to understand the behavior of multi-stage rockets: the upper stage can generate much more usable kinetic energy than the total chemical energy of the propellants it carries.[3]

The Oberth effect occurs because the propellant has more usable energy due to its kinetic energy in addition to its chemical potential energy.[3]:204 The vehicle is able to employ this kinetic energy to generate more mechanical power.

## Description

### Explanation in terms of work

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. 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 can be easily shown. The mechanical work done on the rocket, ${\displaystyle W}$, is defined as the dot product of the force of the engine's thrust, ${\displaystyle {\overrightarrow {F}}}$, and the distance it travels during the burn, ${\displaystyle {\overrightarrow {s}}}$.

${\displaystyle W={\overrightarrow {F}}\cdot {\overrightarrow {s}}}$

If the burn is made in the prograde direction, ${\displaystyle {\overrightarrow {F}}\cdot {\overrightarrow {s}}=\|F\|\cdot \|s\|}$, and we know all the work ends up as a change in kinetic energy, ${\displaystyle \Delta E_{k}}$, leaving us with:

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

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 to calculate the overall increase in specific energy of the rocket.

However, integrating this is often unnecessary if the burn duration is short. For example, as a vehicle falls towards 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.[4]

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).[3]:204 Contrast this to the situation of static firing where the speed of the engine is fixed at zero. This means 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 km/s actually 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. This can 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.

### Impulsive burn

Whereas the integration of the above energy equation can be done, numerically or otherwise, short burns of chemical rocket engines close to periapsis or elsewhere are usually mathematically modelled as impulsive burns, where the force of the engine dominates any other forces that might change the vehicle's energy over the burn.

### Oberth Calculation for Parabolic Orbit

If an impulsive burn of Δv is performed at periapsis in a parabolic orbit then the velocity at periapsis before the burn is equal to the escape velocity (Vesc), and the specific kinetic energy after the burn is:[5]

{\displaystyle {\begin{aligned}e_{k}&={\frac {1}{2}}V^{2}\\[4pt]&={\frac {1}{2}}(V_{\text{esc}}+\Delta v)^{2}\\[4pt]&={\frac {1}{2}}V_{\text{esc}}^{2}+\Delta vV_{\text{esc}}+{\frac {1}{2}}\Delta v^{2}\end{aligned}}}

where ${\displaystyle V=V_{\text{esc}}+\Delta v}$

When the vehicle leaves the gravity field, the loss of specific kinetic energy is:

${\displaystyle {\frac {1}{2}}V_{\text{esc}}^{2}}$

so it retains the energy:

${\displaystyle \Delta vV_{\text{esc}}+{\frac {1}{2}}\Delta v^{2}}$

which is larger than the energy from a burn outside the gravitational field (${\displaystyle {\frac {1}{2}}\Delta v^{2}}$) by:

${\displaystyle \Delta vV_{\text{esc}}}$

When the vehicle has left the gravity well, it is travelling at a speed of

${\displaystyle V=\Delta v{\sqrt {1+{\frac {2V_{\text{esc}}}{\Delta v}}}}}$

For the case where the added impulse Δv is small compared to escape velocity, the 1 can be ignored and the effective delta v of the impulsive burn can be seen to be multiplied by a factor of simply:

${\displaystyle {\sqrt {\frac {2V_{\text{esc}}}{\Delta v}}}}$

Similar effects happen in closed and hyperbolic orbits.

## Parabolic example

If the vehicle travels at velocity v at the start of a burn that changes the velocity by Δv, then the change in specific orbital energy (SOE) is

${\displaystyle v\Delta v+{\frac {1}{2}}(\Delta v)^{2}}$

Once the spacecraft is far from the planet again, the SOE is entirely kinetic, since gravitational potential energy approaches zero. Therefore, the larger the v at the time of the burn, the greater the final kinetic energy, and the higher the final velocity.

The effect becomes more pronounced the closer to the central body, or more generally, the deeper in the gravitational field potential the burn occurs, since the velocity is higher there.

So if a spacecraft is on a parabolic flyby of Jupiter with a periapsis velocity of 50 km/s, and it performs a 5 km/s burn, it turns out that the final velocity change at great distance is 22.9 km/s; giving a multiplication of the burn by 4.6 times.

## References

1. ^ a b c Robert B. Adams, Georgia A. Richardson. "Using the Two-Burn Escape Maneuver for Fast Transfers in the Solar System and Beyond" (PDF). NASA. Retrieved 15 May 2015.
2. ^ Robert Adams (25 Feb 2011). "What Would an Interstellar Mission Look Like?". Discovery News. Retrieved 15 May 2015.
3. ^ a b c d Hermann Oberth (1970). "Ways to spaceflight". Translation of the German language original "Wege zur Raumschiffahrt," (1920). Tunis, Tunisia: Agence Tunisienne de Public-Relations.
4. ^ Atomic Rockets web site: nyrath@projectrho.com Archived July 1, 2007, at the Wayback Machine.
5. ^ following the calculation on rec.arts.sf.science