#### [SOLVED] Velocity of Rocket Exhaust

By Batwayne

i recently learned a bit of rocket propulsion.It wasn't much complex but was explained in simple terms.The only problem i had understanding it was that in calculating the thrust of rocket the velocity of the exhaust was taken relative to the rocket.My problem is : Shouldn't the velocity of the exhaust be taken as the relative to earth.In all the previous examples we had done so , why don't we use the velocity relative to the rocket.Thanks

PS:A simple explanation would be much appreciated #### @John1024 2015-06-30 07:58:33

To see why the exhaust speed is important, let's do a calculation.

1. Let's start with a rocket of mass $m$ going at speed $u$. (We measure all speeds with respect to some inertial reference frame.)

2. Now, suppose it exhausts a tiny amount of propellant of mass $\delta m$ and the propellant is traveling at speed $u_P$. After it exhausts that fuel, the rocket now has mass $m-\delta m$ and is now going at speed $u+\delta u$.

Conservation of momentum says: $$(m - \delta m)(u +\delta u) + \delta m\ u_P = m u$$

Simplifying the above and keeping only first order terms, we obtain:

$$m \delta u = \delta m (u-u_P)$$

In other words, for a given amount of mass, the increase in speed of the rocket, $\delta u$, depends only on the difference between the speed of the rocket, $u$, and the speed of the propellant, $u_P$. The absolute value of either speed is irrelevant.

The difference between the rocket speed and the propellant speed is called the exhaust speed.

For the best chemical rockets, the exhaust speed is around 3,000 meters per second. When electric propulsion is used, exhaust speeds can be up to 20,000 meters per second or more. #### @WetSavannaAnimal 2015-06-30 07:58:22

Imagine the rocket before and after throwing a small ("infinitessimal") amount of fuel out its exhaust. You apply the momentum conservation notion by equating the increase in the rocket's forwards momentum with the momentum of the fuel thrown backwards. The easiest inertial frame to do one's analysis in is that of rocket immediately before the increment $\mathrm{d} m$ of fuel is thrown: in this frame, before the fuel is thrown, the rocket has a momentum of nought; after the fuel is thrown, we have a mass $\mathrm{d} m$ flying backwards at velocity $v_e$ and the rocket is moving forwards a speed $\mathrm{d} v$: the increment in speed wrought by the fuel increment. Momentum conservation is then

$$v_e\,\mathrm{d}m = (m - \mathrm{d} m)\,\mathrm{d} v\quad\Leftrightarrow \frac{v_e}{m} = \frac{\mathrm{d} v}{\mathrm{d}m}$$

which is the Tsiolkovsky equation. It should be very clear that the exhaust velocity $v_e$ to use is the velocity relative to the rocket. Note that there are modifications to this equation to account for the pressures on the rocket when it is steeped in an atmosphere: see my answer here.

The rocket's or exhaust's motion relative to Earth or to anything else is irrelevant: this is simply Galileo's Relativity postulate (oka the "First" postulate of special relativity) - the rocket and its fuel are to be thought of as the below-decks protagonists in the Allegory of Salviati's Ship.

I'm not one generally for Hollywood depictions of science, but I must say I rather liked the way Newton's third law was stated by the Matthew McConaughey character "Joseph Cooper" in the film "Interstellar" when he said "Newton's Third Law: [One] Can't get anywhere without leaving something behind"! This is a great thought to recall when analyzing a problem such as this one