# Oberth effect

In astronautics, the Oberth effect is where the use of a rocket engine when travelling at high speed generates more useful energy than one at low speed. The Oberth effect occurs because the propellant has more usable energy due to its kinetic energy on top of its chemical potential energy. The vehicle is able to employ this kinetic energy to generate more mechanical power. It is named after Hermann Oberth, the Austro-Hungarian-born, German physicist and a founder of modern rocketry, who first described the effect.[1]

In interplanetary spaceflight the Oberth effect is used in a powered flyby or Oberth maneuver where the application of an impulse, typically from a rocket engine, close to a gravitational body (where the gravity potential is low, and the speed is high) can result in a higher change in kinetic energy and final speed (i.e. higher specific energy) than the same impulse applied farther from the body for the same initial orbit. For the Oberth effect to be most effective, the vehicle must be able to generate as much impulse as possible at the lowest possible altitude; thus the Oberth effect is often far less useful for low-thrust reaction engines such as ion drives, which are limited in their ability to generate a large amount of impulse in a small amount of time.

The Oberth effect also can be used to understand the behaviour of multi-stage rockets; the upper stage can generate much more usable kinetic energy than might be expected from simply considering the chemical energy of the propellants it carries.

Historically, a lack of understanding of this effect led early investigators to conclude that interplanetary travel would require completely impractical amounts of propellant, as without it, enormous amounts of energy would be needed.[1]

## Description

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 to hypersonic speed. But when the rocket moves, its thrust acts through the distance it moves. Force acting through a 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 can be defined as

$\Delta E_k = F \cdot s$

where $E_k$ is the kinetic energy, $F$ is the force (the thrust of the rocket which is considered constant), and $s$ is the distance. Differentiating with respect to time, we obtain

$\frac{\operatorname{d}E_k}{\operatorname{d}t} = F \cdot \frac{\operatorname{d}s}{\operatorname{d}t}$

or

$\frac{\operatorname{d}E_k}{\operatorname{d}t} = F \cdot v$

where $v$ is the velocity. Dividing by the instantaneous mass $m$ to express this in terms of specific energy ($e_k$), we get

$\frac{\operatorname{d}e_k}{\operatorname{d}t} = \frac F m \cdot v = a \cdot v$

where $a$ 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 ($\Delta v$). However, since the vehicle's kinetic energy is related to the square of its velocity, this increase in velocity has a disproportionate effect on the vehicle's kinetic energy; leaving it with higher energy than if the burn were achieved at any other time.[2]

It may seem that the rocket is getting energy for free, which would violate conservation of energy. However, any gain to the rocket's energy is balanced by an equal decrease in the energy the exhaust is left with. When expended lower in the gravitational field, even if the exhaust is left with more kinetic energy, it is left with less total energy. The effect would be even stronger if the exhaust speed could be made equal to the speed of the rocket, then the exhaust would be left without kinetic energy, so the total energy of the exhaust would be as low as its potential energy. Contrast this to the situation of static firing: as the speed of the engine is zero its specific energy does not increase at all, with all chemical energy of the fuel being converted to the exhaust's kinetic energy.

At very high speed the mechanical power imparted to the rocket can even exceed the total power liberated in the combustion of the propellants, and 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 make up for what seems like an 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.

## 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

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

Once the space craft is far from the planet again, the SOE is entirely kinetic, since gravitational potential energy tends to 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.

## Detailed proof

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:

\begin{align} e_k &= \frac{1}{2} V^2 \\ &= \frac{1}{2} (V_\text{esc} + \Delta v )^2 \\ &= \frac{1}{2} V_\text{esc} ^ 2 + \Delta v V_\text{esc} + \frac{1}{2} \Delta v^2 \end{align}

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

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

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

so it retains the energy:

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

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

$\Delta v V_\text{esc}$

It can then be easily shown that the impulse is multiplied by a factor of:

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

Substituting 50 km/s escape velocity and 5 km/s burn we get a multiplier of 4.6.

Similar effects happen in closed and hyperbolic orbits.