# Understanding of the $m v^2/2$ formula for kinetic energy

I have some problem to intuitively understand why the kinetic energy grows quadratically with the velocity (at least in non-relativistic case).

Assume the following experiment: we launch an unmanned rocket ship from an asteroid, let it accelerate time $T$ in a direction and then time $2T$ in the backwards direction; so, in time $2T$ it has velocity zero and in $3T$, it should be in the same position as in time $T$, just with opposite velocity. Then we let it fly freely until it hits the asteroid and let's assume that all kinetic energy transforms to heat.

Repeat the same experiment with $T$ replaced by $\lambda T$; the final velocity will be $\lambda$ times bigger and we used $\lambda$-times so much fuel for the accelerations. But this can hardly be converted to $\lambda^2$-times the energy.

Is the solution hidden in the fact that the fuel itself has a non-neglectable weight?

• Yes, the weight is definitely non negligible. The relationship between terminal velocity and amount of fuel will not be linear like you assume. Dec 8 '14 at 10:46
• @Qmechanic Ok, I just wondered where is the main mistake in the virtual experiment I proposed; is it neglecting the weight of the fuel, or also something else? THanks Dec 8 '14 at 13:44
• A confusing example. How about you calculate the energy when you drop it from a known height in a constant gravity field? Dec 8 '14 at 16:13
• @ Peter: It doesn't go linearly, neither quadratically, see for rigorous treatment of the problem the Tsiolkovsky's law about rocket velocity vs. the fuel consumption. en.wikipedia.org/wiki/… Dec 8 '14 at 16:22
• This is an overcomplicated thought experiment. Neglecting the complexities of proper rocket science, if you increase an object's non-relativistic velocity by a factor of $\lambda$, then you increase its kinetic energy by a factor of $\lambda^2$
– Jim
Dec 9 '14 at 12:55

Let's take the things step by step, from linear momentum conservation. Of course, the rigorous treatment is by Tsiolkovsky's law. Here I am saying something less rigorous but intuitive.

To increase the velocity from zero to $\Delta V$ you consume a mass of fuel $\Delta m$. When I say from zero, I mean that I consider a discrete series of quick fuel consumptions. In such a consumption event, in the frame of reference of the rocket its velocity is zero, and we want to acquire a velocity $\Delta V$. Then, the momentum conservation says

$$\Delta {m} .v = (M - \Delta m)\Delta V$$

$$\dfrac{{Δm v^2} + {M (ΔV)^2}}{2} = E_{Δm}$$

where $M$ is the mass of the rocket before consuming $Δm$ of fuel, $v$ is the expulsed gas velocity, and $E_{Δm}$ is the energy that $Δm$ of fuel can release. Let's use absolute values for velocities, for not to carry in all the formulas the minus caused by the opposite direction of the rocket and gas velocities.

Thus, $$v = \dfrac{(M - Δm)ΔV}{Δm},$$ and $$(ΔV)^2 = \dfrac{2E_{Δm}}{ [M^2/Δm - (M - Δm)]} ,$$

from which

$$ΔV = \sqrt{\dfrac{2E_{Δm}}{ [M^2/Δm - (M - Δm)]}} .$$

So, even in my non-rigorous treatment, the velocity and the quantity of burnt fuel are not in linear relationship.

I hope it helps.

You have a couple of problems in your example.

That means that your other problem is that you are assuming that a rocket engine increases the kinetic energy of the rocket ship in a linear relationship with the time it burns. Unfortunately, it does not.

Rockets provide constant thrust, not constant power. The energy provided by a rocket is given by the old standby: $W = F \times d$. Dividing both sides by time, we see that $$\frac{W}{t} = F \times \frac{d}{t}$$ $$P = F \times v$$

The power of a rocket (the rate at which it increases the kinetic energy of a ship) is dependent on the speed. This means the energy of the craft is (roughly) proportional to $t^2$ of the burn, not $t$.