Where does the extra kinetic energy of the rocket come from?

Consider a rocket in deep space with no external forces. Using the formula for linear kinetic energy $$\text{KE} = mv^2/2$$ we find that adding $100\ \text{m/s}$ while initially travelling at $1000\ \text{m/s}$ will add a great deal more energy to the ship than adding $100 \ \text{m/s}$ while initially at rest: $$(1100^2 - 1000^2) \frac{m}{2} \gg (100^2) \frac{m}{2}.$$ In both cases, the $\Delta v$ is the same, and is dependent on the mass of fuel used, hence the same mass and number of molecules is used in the combustion process to obtain this $\Delta v$. So I'd wager the same quantity of chemical energy is converted to kinetic energy, yet I'm left with this seemingly unexplained $200,000\ \text{J/kg}$ more energy, and I'm clueless as to where it could have come from.

• Jun 28, 2019 at 12:46
• In a moving rocket the fuel also has the additional kinetic energy acquired by the previously burnt fuel. It gives that extra energy. Nov 22, 2021 at 9:10

You've noted that at high velocities, a tiny change in velocity can cause a huge change in kinetic energy. And that means that the thrust due to burning fuel seems to be able to contribute an arbitrarily high amount of energy, possibly exceeding the chemical energy of the fuel itself.

The resolution is that all of this logic applies to the fuel too! When the fuel is exhausted, it loses much of its speed, so the kinetic energy of the fuel decreases a lot. The extra kinetic energy of the rocket comes from this extra contribution, which can be arbitrarily large.

Of course, the kinetic energy of the fuel didn't come from nowhere. If you don't use gravity wells, that energy came from the fuel you burned previously, which was used to speed up both the rocket and all the fuel inside it. So everything works out -- you don't get anything for free.

For those that want more detail, this is called the Oberth effect, and we can do a quick calculation to confirm it. Suppose the fuel is ejected from the rocket with relative velocity $$u$$, a mass $$m$$ of fuel is ejected, and the rest of the rocket has mass $$M$$. By conservation of momentum, the velocity of the rocket will increase by $$(m/M) u$$.

Now suppose the rocket initially has velocity $$v$$. The change in kinetic energy of the fuel is $$\Delta K_{\text{fuel}} = \frac12 m (v-u)^2 - \frac12 mv^2 = \frac12 mu^2 - muv.$$ The change in kinetic energy of the rocket is $$\Delta K_{\text{rocket}} = \frac12 M \left(v + \frac{m}{M} u \right)^2 - \frac12 M v^2 = \frac12 \frac{m^2}{M} u^2 + muv.$$ The sum of these two must be the total chemical energy released, which shouldn't depend on $$v$$. And indeed, the extra $$muv$$ term in $$\Delta K_{\text{rocket}}$$ is exactly canceled by the $$-muv$$ term in $$\Delta K_{\text{fuel}}$$.

Sometimes this problem is posed with a car instead of a rocket. To understand this case, note that cars only move forward because of friction forces with the ground; all that a car engine does is rotate the wheels to produce this friction force. In other words, while rockets go forward by pushing rocket fuel backwards, cars go forward by pushing the Earth backwards.

In a frame where the Earth is initially stationary, the energy associated with giving the Earth a tiny speed is negligible, because the Earth is heavy and energy is quadratic in speed. Once you switch to a frame where the Earth is moving, slowing the Earth down by the same amount harvests a huge amount of energy, again because energy is quadratic in speed. That's where the extra energy of the car comes from. More precisely, the same calculation as above goes through, but we need to replace the word "fuel" with "Earth".

The takeaway is that kinetic energy differs between frames, changes in kinetic energy differ between frames, and even the direction of energy transfer differs between frames. It all still works out, but you must be careful to include all contributions to the energy.

• "The extra kinetic energy of the rocket comes from this extra contribution, which can be arbitrarily large." - This is wrong. The extra energy doesn't come from the fuel. The same effect is observed if the change in velocity is acquired via some external source, like in the use of a solar sail. Dec 7, 2018 at 17:29
• @richard My explanation is only for firing rockets, of course the extra energy would come from something else if you weren't even using the rocket. In the case of a solar sail, the power boost is because you can harvest more of the photon's kinetic energy when it bounces if you're moving away from it. (Think of the limit of a stationary sail -- in that case you get zero power because the photon bounces off with the same speed it had before.) Dec 7, 2018 at 17:34
• @richard However, as long as you have the solar sail going, energy is continuously going into the rocket fuel. And you can get some of it back if you fire the rocket. Dec 7, 2018 at 17:35
• @knzhou can you take a look here ,maybe you could help :)physics.stackexchange.com/questions/646080/… Jun 17, 2021 at 17:20

Another much clearer way to see the effect of the Oberth effect is when you add the Potential energy to the equation.

When you perform a rocket burn inside the gravity well of a massive body, the propellant ends up in a lower orbit than if you perform the rocket burn outside of the gravitational well.

The difference in the Potential Energy of the propellant will equal the difference in the Kinetic Energy of your space probe.

• THIS is the explanation I wanted. Thank you! Honestly I think the explanation on wikipedia is misleading bordering on simply incorrect.
– B T
Nov 8, 2018 at 22:55
• No, this is wrong too. You can get the same effect at any speed, irrespective of a planet or gravity, etc. If you add velocity to a given object already in motion, the higher the velocity, the more energy is added for a given amount of velocity increase. Kinetic energy increases exponentially in relation to velocity. Dec 7, 2018 at 17:31
• @richard actually the same explaination works. The result is that the spent fuel has lower energy wrt rest frame, when the rocket is moving really fast. Replace the words Potential Energy with Total Energy and your problem goes away.
– Aron
Dec 7, 2018 at 17:56

Assume the rocket without fuel has weight $M$, the fuel has weight $m$, and the rocket engine works by sending the fuel instantaneously backwards with velocity $v_e$ relative to the initial velocity of the rocket. Thus, by conservation of momentum, the speed gain of the rocket is $$\Delta v_\text{rocket} = \frac{m}{M} v_e.$$

The kinetic energy gain in the system in the COM reference frame is $$\Delta T = \frac{1}{2} M (\Delta v_\text{rocket})^2 + \frac{1}{2} m v_e^2.$$ This is the chemical energy $E_\text{chemical}$ released by burning the fuel (assuming perfect efficiency).

Now what happens when we burn prograde, i.e. acclerate towards the direction of our velocity?

Let's assume that initially the fuel is in the rocket and they are in an orbit with orbital energy $E_0$, which is the sum of the kinetic energy and the potential energy, $$E_0 = T_0 + V_0 = \frac{1}{2} (M+m) v_0^2 - \frac{\gamma(M+m)}{r_0},$$ where $v_0$ is the velocity of the rocket before the burn, $r_0$ is the distance of the rocket to the centre of the central body before the burn, and $\gamma$ is the gravitational parameter of the central body. Now $r_0$ is the parameter which we can choose by choosing when to burn, $E_0$ is a constant determined by our initial orbit, and $v_0$ is then a function of $E_0$ and our choice of $r_0$.

After the burn, the speed of the rocket is $v_0 + \Delta v_\text{rocket}$ and the orbital energy of the rocket is $$E_\text{rocket} = T_\text{rocket} + V_\text{rocket} = \frac{1}{2} M (v_0+\Delta v_\text{rocket})^2 - \frac{\gamma M}{r_0} = \frac{1}{2} M \left( v_0+\frac{m}{M} v_e \right)^2 - \frac{\gamma M}{r_0},$$ and the speed of the fuel is $v_0 - v_e$ and the orbital energy of the fuel is $$E_\text{fuel} = T_\text{fuel} + V_\text{fuel} = \frac{1}{2} m (v_0- v_e)^2 - \frac{\gamma m}{r_0}.$$

As you have seen, the Oberth effect is that the rocket ends with more kinetic energy if the burn is performed at higher $v_0$ and smaller $r_0$ (when keeping the $E_0$ constant).

The total potential energy remains the same, but the total kinetic energy changes, which results in a change in the total energy of the rocket and the fuel, $$(E_\text{rocket} + E_\text{fuel}) - E_0 = (T_\text{rocket} + T_\text{fuel}) - T_0 = \frac{1}{2} \frac{m^2}{M} v_e^2 + \frac{1}{2} m v_e^2 = \frac{1}{2} M (\Delta v_\text{rocket})^2 + \frac{1}{2} m v_e^2.$$ This is the same no matter where the burn is performed! Also it is the same than it is in the initial reference frame of the rocket+fuel system, so it is the chemical energy $E_\text{chemical}$ used in the burn.

Now the question is, how does the energy gain of the rocket depend on the choice of when to burn (i.e. $r_0$, assuming $E_0$ is constant)?

The initial speed of the rocket+fuel system, $v_0$ is obtained in terms of $r_0$ as $$v_0 = \sqrt{2 \frac{E_0}{M+m} + \frac{2\gamma}{r_0}}.$$

The kinetic energy gain of the rocket (not counting the fuel) when going from $v_0$ to $v + \Delta v_\text{rocket}$ is \begin{align*} \Delta T_\text{rocket} &= \frac{1}{2} M ( v_0 + \Delta v_\text{rocket})^2 - \frac{1}{2} M v_0^2 = M v_0 \Delta v_\text{rocket} + \frac{1}{2} M (\Delta v_\text{rocket})^2 \\ &= M \Delta v_\text{rocket} \sqrt{2 \frac{E_0}{M+m} + \frac{2\gamma}{r_0}} + \frac{1}{2} M (\Delta v_\text{rocket})^2. \end{align*} This formula is a bit complicated but, as you have seen, the gain is biggest when $r_0$ is smallest, that is, when the gravitational potential energy is smallest. Because the increase in the sum of kinetic energies of the rocket and the fuel doesn't depend on $r_0$, the mathematical explanation is that the energy gain comes from the fact that the kinetic energy of the fuel decreases more: $$\Delta T_\text{fuel} = E_\text{chemical} - \Delta T_\text{rocket} = \frac{1}{2} m v_e^2 - M \Delta v_\text{rocket} \sqrt{2 \frac{E_0}{M+m} + \frac{2\gamma}{r_0}}.$$

• "the energy gain comes from the fact that the kinetic energy of the fuel decreases more" - This is the sentence that fixed it in my head. Small changes in velocity at large velocities result in massive changes in kinetic energy. This applies to the fuel too. The fuel burned is giving the remaining fuel increasing kinetic energy which is later transferred to the rocket ON TOP OF the chemical potential energy being used. Aug 3, 2021 at 8:28

The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?

To understand this it is useful to consider a “toy model” rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).

Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning” the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.

So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?

Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.

At low speeds most of the fuel’s energy is “wasted” in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.

I've seen this scenario pop up in discussions more than once (such as here, here, and here), so I decided to put this as a post. If this is too much to read, go to the Some Conclusions section at the end.

The question concerns the compatibility of kinetic energy and Galilean relativity. How is it possible that the same boost $$u\rightarrow u+\Delta v$$ corresponds to different increases in kinetic energy for the same boost $$\Delta v$$ in differing initial velocities $$u$$?

There are a number of scenarios similar to the one OP posted. For example, suppose a person is on a spaceship (in uniform motion) and he throws a ball in the forward direction of the spaceship. If were considering the frame where the spaceship is initially stationary, then the ball would have been moving at, say $$10\text{ m/s}$$. But in the frame where the spaceship is moving forward at $$u = 100\text{ m/s}$$, then the ball is moving at $$110\text{ m/s}$$. The $$\Delta v = 10\text{ m/s}$$ corresponds to different increases in kinetic energy in the two scenarios.

Likewise, we can consider a spaceship giving exhaust in discrete chunks (to simplify the problem) resulting in discrete boosts. The boost $$0\text{ m/s}\rightarrow 100\text{ m/s}$$ corresponds to a different increase in KE than the boost $$1000\text{ m/s}\rightarrow 1100\text{ m/s}$$, even though it is the exact same process.

Both of these scenarios involve some initial potential energy (chemical energy in the person or chemical energy in the fuel of the spaceship) that is converted to kinetic energy resulting in a change in velocity of an object. In the first scenario, a person expends some energy to throw the ball. In the second scenario, a there is an explosion that pushes the rocket forward and expels exhaust.

I'll consider both cases by analyzing the case of a "spring launcher" that launches a box in some direction. The problem is identical to the two scenarios above for all intents and purposes here.

Erroneous Analysis

We can consider the case where we have a "spring launcher" with potential energy $$U$$ that launches a box of mass $$m$$. Below we depict the scenario in the frame where the box is initially at rest. First, I will give the wrong, naive calculation that many people tend to make in this thought experiment.

Assuming the mass of the spring is negligible, a naive calculation using conservation of energy implies $$KE = U$$ and we are lead to say the box is launched with velocity $$v = \sqrt{2U/m}$$.

Now consider the same scenario where the spring launcher and box are initially moving at velocity $$u$$. Again, I will make the wrong, naive calculation to highlight the confusion people may have.

We have an increase in kinetic energy $$\Delta KE = U$$, so $$KE_{\text{final}} = KE_{\text{initial}} + U$$, and so $$v' = \sqrt{ \frac{2U}{m} + u^{2} }.$$ Clearly, we don't have $$v' = v + u$$, which seems to naively show that the spring is not Galilean invariant. A bit differently, we can insist that the box is boosted from $$u$$ to $$u+\Delta v$$ for the same $$\Delta v$$ in all frames, but then the gain in energy would be inconsistent. Either way, we seem to have a problem.

Correct Analysis

The issue in the above analysis is that we fail to take into account Newton's third law. The wall that the spring is attached to has some nonzero and finite mass itself, and by Newton's third law it will be pushed back. A correct analysis would have to take into account both the launched object and the kickback to the launcher.

Suppose there is a spring launcher with initial potential energy $$U$$ that is attached to a mass $$m_{1}$$ that launches a mass $$m_{2}$$. Consider a frame where the entire system is initially at rest.

Taking both conservation of momentum and conservation of energy into account, we have \begin{align} m_{1}v_{1} + m_{2}v_{2} &= 0, \\ \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2} &= U. \end{align} Solving for both, and taking into account that mass $$m_{1}$$ is pushed in the $$-x$$-direction and mass $$m_{2}$$ is pushed in the $$+x$$-direction, we obtain \begin{align} v_{1} = -\left[ \frac{2U}{m_{1}\left(1 + \frac{m_{1}}{m_{2}}\right)} \right]^{1/2} \qquad\text{ and }\qquad v_{2} = \left[ \frac{2U}{m_{2}\left(1 + \frac{m_{2}}{m_{1}}\right)} \right]^{1/2}. \end{align}

Now consider the more general case where the entire system is initially moving at velocity $$u$$.

Taking both conservation of momentum and conservation of energy into account, we have \begin{align} m_{1}v_{1}' + m_{2}v_{2}' &= (m_{1} + m_{2})u, \\ \frac{1}{2}m_{1}v_{1}'^{2} + \frac{1}{2}m_{2}v_{2}'^{2} &= U + \frac{1}{2}(m_{1}+m_{2})u^{2}. \end{align} By using the first equation to write $$v_{2}'$$ in terms of $$v_{1}'$$ and plugging this into the second equations, we obtain the quadratic equation $$v_{1}'^{2} \left[ m_{1} + \frac{m_{1}^{2}}{m_{2}} \right] - v_{1}' \left[ 2m_{1}\left(1 + \frac{m_{1}}{m_{2}}\right) \right] + m_{2}\left(1 + \frac{m_{1}}{m_{2}}\right)^{2}u^{2} - (m_{1} + m_{2})u^{2} - 2U = 0.$$ By applying the quadratic formula and doing a lot of simplifying, we obtain $$v_{1}' = u - \left[ \frac{2U}{m_{1}\left(1 + \frac{m_{1}}{m_{2}}\right)} \right]^{1/2}$$ where we chose the negative square root because mass $$1$$ is launched in the $$-x$$-direction. A similar derivation yields $$v_{2}' = u + \left[ \frac{2U}{m_{2}\left(1 + \frac{m_{2}}{m_{1}}\right)} \right]^{1/2}$$ where we chose the positive square root because mass $$2$$ is launched in the $$+x$$-direction.

We see that once we take into account Newton's third law and conservation of momentum, the system exhibits Galilean invariance, energy is conserved, and the velocity boosts are the same no matter what the initial speed of the system was.

Some Conclusions

It's worth pondering how the energy is redistributed across both boxes for different initial velocities $$u$$.

In the case where the initial velocity is $$u=0$$, the release of the spring causes the potential energy $$U$$ to be converted to kinetic energy, and the energy is then shared across both boxes.

The same can be said if the entire box 1 + box 2 system has some small initial velocity $$0 < u < \epsilon$$ (I'll only consider the cases with positive $$u$$; the cases with negative $$u$$ are similar). Both boxes start out with some small kinetic energy, and after the release of the spring, both boxes gain kinetic energy from the stored potential energy of the spring.

However, when the initial velocity exceeds $$u_{0} = \frac{1}{2}\left[\frac{2U}{m_{1}\left(1 + \frac{m_{1}}{m_{2}}\right)}\right]^{1/2},$$ something interesting can be said. If $$u > u_{0}$$, then the release of the spring causes box 1 to lose kinetic energy (because it now loses overall speed from being pushed back). The lost kinetic energy is, of course, transferred to box 2 along with the potential energy released from the spring.

This shows exactly why the nonlinear relationship between kinetic energy and velocity is present. The quadratic relationship means that a rocket ship in the case of a boost $$v\rightarrow v + \Delta v$$ gains more kinetic energy than in the case of a boost $$0\rightarrow \Delta v$$ for the same $$\Delta v$$, and this is expected, because in the former case the exhaust gives off its kinetic energy to the rocket (more energy to the rocket) while in the latter case the exhaust gains kinetic energy by being emitted (less energy to the rocket).

This seemingly unintuitive relationship is now inevitable when we take Newton's third law into account. The potential energy released from the spring (or fuel of the rocket) is needed to separate the two boxes (or in the case of the rocket the fuel is needed to separate the exhaust from the rocket), but how the resulting kinetic energy is distributed depends on the initial velocity of the system.