How much more fuel to double the speed?

Question

Suppose it requires a certain amount of fuel to accelerate a car (or a spaceship) from 0 to a speed $v_1$ (in some initial frame of reference). Let's ignore all friction forces, and only consider classical mechanics (at low speeds, we can safely ignore special relativity).

How much more fuel would it require to accelerate from 0 to $v_2$, with $v_2 = 2 \times v1$?

I have two contradictory answers.

On one hand, $E=\frac{1}{2} mv^2$, so doubling the speed multiplies the kinetic energy by 4, so we need to provide 4 times more energy (thus 4 times more fuel).

As an example, let's take $m = 2000Kg$ and $v_1 = 100m/s$.

• To go from 0 to $100m/s$, we need to provide $E=\frac{1}{2} \times 2000 \times 100^2 = 10000000J = 10MJ$.
• To go from 0 to $200m/s$, we need to provide $E=\frac{1}{2} \times 2000 \times 200^2 = 40000000J = 40MJ$, that is 4 times more.

On the other hand, once the car (or spaceship) is at $v_1$ in the initial frame of reference R0, we can consider a new frame of reference R1 where the car/spaceship is at rest. Then we are exactly in the same situation as at the beginning, and we can accelerate one more time from 0 to $v_1$ in R1 (so from 0 to $v_2$ in R0). According to the principle of relativity, there is no reason why accelerating from 0 to $v_1$ in R1 would consume more fuel than when accelerating from 0 to $v_1$ in R0. Thus, we can conclude that the same amount of fuel is consumed to go from 0 to $v_1$ as from $v_1$ to $v_2$ (in R0), therefore twice the speed requires twice more fuel.

To take the same example:

• To go from 0 to $100m/s$ in R0, we need to provide $E=\frac{1}{2} \times 2000 \times 100^2 = 10000000J = 10MJ$.
• To go from 0 to $100m/s$ in R1, we need to provide $E=\frac{1}{2} \times 2000 \times 100^2 = 10000000J = 10MJ$.
• So to go from 0 to $200m/s$ in R0, we need to provide $10 + 10 = 20MJ$, that is twice more.

Which one is wrong?

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This question has been marked as a duplicate of Work done changes between reference frames? However, while related (kinetic energy with a change of reference frame), the question is not the same. For example, the accepted answers says:

The key to unraveling the paradox is to recognize that when the flight attendant pushes the cart forward, he is pushing the rest of the plane backwards by some small velocity.

This does not apply here, and nothing allows to answer the question how much more fuel (2× or 4×) would be required to double the speed.

Answer 1

The issue with your contradictory answers is that you are neglecting the conservation of momentum. Both energy and momentum must be conserved, not just energy.

In order for a car or a spaceship or any other object to accelerate it must exchange momentum with something. In the case of a car it is exchanging momentum with the earth, and in the case of the rocket it is exchanging momentum with the exhaust. Regardless of the specific mechanism used, there is always necessarily some exchange of momentum.

So let's say that your vehicle of mass $m$ is interacting with another object (e.g. the earth or the exhaust) of mass $M$. They start off with velocities $v_i$ and $V_i$ respectively. Some fuel is expended which increases the kinetic energy by an amount $W$. As a result of the interaction the objects change their velocities to $v_f$ and $V_f$ respectively.

Conservation of momentum gives: $$m v_i + M V_i = m v_f + M V_f$$ and conservation of energy gives: $$\frac{1}{2} m v_i^2 + \frac{1}{2} M V_i^2 + W = \frac{1}{2} m v_f^2 + \frac{1}{2} M V_f^2$$ A little bit of algebra gives $$W=\frac{m}{2M}(v_f-v_i)\left[m(v_f-v_i)+M(v_f+v_i-2V_i)\right] $$

This $W$ gives the amount of fuel consumed. Notice that it is independent of the reference frame. Suppose that we transform to a frame moving at $u$ relative to the first frame. Then $v'_i = v_i + u$ and similarly for all other velocities. Then, in the formula for $W$ we have: $$W'=\frac{m}{2M}(v'_f-v'_i)\left[m(v'_f-v'_i)+M(v'_f+v'_i-2V'_i)\right] $$$$=\frac{m}{2M}(v_f +u-v_i-u)\left[m(v_f+u-v_i-u)+M(v_f+u+v_i+u-2V_i-2u)\right]$$$$=\frac{m}{2M}(v_f-v_i)\left[m(v_f-v_i)+M(v_f+v_i-2V_i)\right] =W $$

So the amount of fuel consumed is independent of the reference frame. The apparent contradiction comes from neglecting the conservation of momentum. There must be some other object to exchange momentum, whether that is the rocket exhaust, the earth, or (in the case of the "duplicate") an airplane.

As an example, let's take $m = 2000Kg$ and $v_1 = 100m/s$.

• To go from 0 to $100m/s$, we need to provide $E=\frac{1}{2} \times 2000 \times 100^2 = 10000000J = 10MJ$.
• To go from 0 to $200m/s$, we need to provide $E=\frac{1}{2} \times 2000 \times 200^2 = 40000000J = 40MJ$, that is 4 times more.

So, if this is a car then $v_i = V_i = 0 \mathrm{\ m/s}$, $m= 2000 \mathrm{\ kg}$, and $M=5.97 \ 10^{24} \mathrm{\ kg}$ and indeed, we find that for $v_f=100 \mathrm{\ m/s}$ we get $W=10 \mathrm{\ MJ}$ while for $v_f= 200 \mathrm{\ m/s}$ we get $W=40 \mathrm{\ MJ}$. So your first scenario is plausible.

once the car (or spaceship) is at $v_1$ in the initial frame of reference R0, we can consider a new frame of reference R1 where the car/spaceship is at rest. Then we are exactly in the same situation as at the beginning, and we can accelerate one more time from 0 to $v_1$ in R1

So, in the new reference frame we are not exactly in the same situation as before. In this case we have $v_i=0 \mathrm{\ m/s}$ but $V_i=-100 \mathrm{\ m/s}$. Then for $v_f=100 \mathrm{\ m/s}$ we get $W=30 \mathrm{\ MJ}$. So by correctly accounting for the fact that we are exchanging momentum with the Earth which is moving we get that the work is as expected in both frames.

I leave it as an interesting exercise to calculate both scenarios for a rocket with, e.g. $M=1 \mathrm{\ kg}$

Answer 2

The argument is valid that, in the frame of a rocket, when determining how much energy is expended to accelerate to a new velocity different by $\Delta v$, its velocity in another frame should not matter. So, the energy needed for a rocket to go from 0 km/s to 1 km/s or from 1 km/s to 2 km/s should be the same, even though $\frac{m}{2} (v_f^2 - v_i^2)$ triples in this case. But, it is not valid for a car because, in the frame of the car, the road is moving backward faster at higher speeds and it is harder to accelerate by pushing against a faster moving road.

For rockets, the total motion of the Earth, including Large Scale, MW, Solar, and Earth motion, is about 600 km/s. To accelerate a kg on the earth by 1 m/s requires 0.5 Joules and not $((6E5+1)^2 - 6E5^2)/2 = 6E5 Joules. When determining the kinetic energy of a system and the work done to achieve that kinetic energy one needs to specify the frame of reference. In the frame of a rocket in space, the work expended must depend on v to the first power, because it always accelerates from its present rest frame velocity, 0 km/s.

Work to reach $v_f$ in the rocket frame of reference

Rather than give philosophical explanations, we can directly find what the rocket equation implies for the energy requirements to attain final velocity $v_f$. We will assume the rocket expels gas at constant velocity $v_e$. The rocket equation is often written as: $$ m_f = m_i e^{-v_f/v_e}, $$ where $m_i$ is the rocket's initial mass and $m_f$ is the final mass. The mass of the gas expelled is $$ m_{gas} = m_i - m_f = m_i(1 - e^{-v_f/v_e}).$$ The energy, or work, expended, in the frame of the rocket, is the amount of energy to accelerate the propellant mass used to velocity $v_e$: $$ W = \frac{m_{gas }v_e^2}{2} = \frac{m_i}{2}(1-e^{-v_f/v_e})v_e^2$$

We can examine the case where not too much of the initial mass has been used up as propellant, so that means $v_f \ll v_e$, and the work simplifies to approximately: $$ W \approx \frac{m_i v_f v_e}{2}$$

The energy expended goes as the first power of $v_f$.

Instantaneous Power in the rest frame of reference

But wait you say, both frames of reference agree on how much propellant is used and how much energy is held in a liter of propellant. So, they must agree on the energy being spent per second. What is the power consumption as seen in the rest frame? Well, the energy carried away in rocket propellant exhaust per second is, $$ P_p = - \frac{\dot{m}}{2}(v_e-v)^2,$$ where the mass of the rocket is decreasing so $\dot{m}$ is negative. And the power going into accelerating the rocket is, $$ P_r = \frac{\partial{(mv^2/2)}}{\partial{t}} = mva + \frac{\dot{m}v^2}{2}$$ The conservation of momentum requires $ ma = -\dot{m} v_e$ which we can use in the first term of $P_r$, $$P_r = -\dot{m}\left(v_ev - \frac{v^2}{2}\right)$$ We can rewrite $P_p$ as, $$P_p = -\dot{m}\left(\frac{v_e^2}{2} - v_ev + \frac{v^2}{2}\right).$$ Finally, adding the power going into the rocket and the fuel together we get the total power supplied by burning the propellant: $$ P_T = P_r + P_p = \frac{-\dot{m}v_e^2}{2}$$ So even in the rest frame, the power spent to accelerate at a slowly varying rate $a = \frac{-\dot{m}v_e}{m}$ does not depend on the velocity.

As an aside, the unfortunate thing about rocket kinematic is that while $v < v_e$ one is expending more energy than $\frac{1}{2}m_i v^2$ and when $v > v_e$ most of the mass of the rocket is used as fuel. But at least, going from 1 m/s to 2 m/s is not 3 times harder than going from 0 m/s to 1 m/s.

Answer 3

On the other hand, once the car (or spaceship) is at 1 in the initial frame of reference R0, we can consider a new frame of reference R1 where the car/spaceship is at rest. Then we are exactly in the same situation as at the beginning

For a rocket, yes you can do this. Since you've spent the energy to bring the reaction mass up to speed, this is valid.

For a car, you can't do this. The car pushes against the earth. In the frame where the car is at rest, the earth is moving backward at high speed. Because of that difference, it takes more energy for the same acceleration.

Answer 4

Suppose it requires a certain amount of fuel to accelerate a car (or a spaceship) from 0 to a speed v1 (in some initial frame of reference). Let's ignore all friction forces, and only consider classical mechanics (at low speeds, we can safely ignore special relativity).

How much more fuel would it require to accelerate from 0 to v2 , with v2=2×v1 ?

If the mass of the fuel is small compared to the mass of the car and the fuel burns at a constant rate, then the amount of fuel burned is linearly proportional to the time.

The acceleration required to achieve a given velocity is:

$$a = \frac{v-v_o}{t-t_o}$$

If the acceleration remains constant, then:

$$ a = \frac {v_1}{ t_1} = \frac {v_2}{ t_2}$$

If $v_2 = 2v_1$ then $t_2 = 2t_1$

So if it takes $t$ to burn $m_{fuel}$ then twice the time requires twice the fuel.

This holds true for cars at typical speeds up to 200 mph.

For rockets at high speeds, the fuel mass is no longer negligible and the fuel burned is no longer linearly proportional to time.

The Tsiolkovsky equation is used for rockets and is a function of fuel mass and exhaust velocities:

$$\Delta v = v_{exhaust}ln(\frac{m_o}{m_f})$$

https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation