By Nomoturtle


2016-10-18 01:00:09 8 Comments

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.

4 comments

@Dale 2020-03-29 23:02:26

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.

@knzhou 2016-10-18 01:10:50

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.

@richard 2018-12-07 17:29:56

"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.

@knzhou 2018-12-07 17:34:46

@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.)

@knzhou 2018-12-07 17:35:12

@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.

@JiK 2016-10-18 08:13:52

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: $$ \Delta T_\text{fuel} = E_\text{chemical} - \Delta T_\text{rocket} = \frac{1}{2} m v_e^2 - M \Delta v_\text{rocket} \sqrt{2 \frac{E_0}{M+m} + \frac{2\gamma}{r_0}}. $$

@Aron 2016-10-18 04:16:26

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.

@B T 2018-11-08 22:55:30

THIS is the explanation I wanted. Thank you! Honestly I think the explanation on wikipedia is misleading bordering on simply incorrect.

@richard 2018-12-07 17:31:55

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.

@Aron 2018-12-07 17:56:04

@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.

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